Chapter 5 Continuity And Differentiability

Given:

The function $f(x) = 2x + 3$.

The point at which continuity is to be checked is $x = 1$.

To Check:

The continuity of the function $f(x)$ at $x=1$.

Solution:

A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:

1. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x)$ exists.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this case, $c = 1$. We need to check the continuity of $f(x) = 2x + 3$ at $x=1$.

Step 1: Check if $f(1)$ is defined.

Substitute $x = 1$ into the function $f(x)$:

$f(1) = 2(1) + 3$

$f(1) = 2 + 3$

$f(1) = 5$

Since $f(1)$ is a real number (5), it is defined.

Step 2: Check if $\lim\limits_{x \to 1} f(x)$ exists.

For the limit to exist at $x=1$, the left-hand limit (LHL) and the right-hand limit (RHL) must exist and be equal.

Calculate the LHL:

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (2x + 3)$

As $x$ approaches 1 from the left side, $2x+3$ approaches $2(1)+3 = 5$.

LHL $= 5$

Calculate the RHL:

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (2x + 3)$

As $x$ approaches 1 from the right side, $2x+3$ approaches $2(1)+3 = 5$.

RHL $= 5$

Since LHL = RHL = 5, the limit $\lim\limits_{x \to 1} f(x)$ exists and is equal to 5.

Step 3: Check if $\lim\limits_{x \to 1} f(x) = f(1)$.

From Step 1, we have $f(1) = 5$.

From Step 2, we have $\lim\limits_{x \to 1} f(x) = 5$.

Comparing these values, we see that $\lim\limits_{x \to 1} f(x) = f(1)$.

Since all three conditions for continuity are met, the function $f(x) = 2x + 3$ is continuous at $x=1$.

Answer:

The function $f(x) = 2x + 3$ is continuous at $x=1$.

Given:

The function $f(x) = x^2$.

The point at which continuity is to be examined is $x = 0$.

To Examine:

Whether the function $f(x)$ is continuous at $x=0$.

Solution:

A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:

1. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x)$ exists.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this case, $c = 0$. We need to examine the continuity of $f(x) = x^2$ at $x=0$.

Step 1: Check if $f(0)$ is defined.

Substitute $x = 0$ into the function $f(x)$:

$f(0) = 0^2$

$f(0) = 0$

Since $f(0)$ is a real number (0), it is defined.

Step 2: Check if $\lim\limits_{x \to 0} f(x)$ exists.

For the limit to exist at $x=0$, the left-hand limit (LHL) and the right-hand limit (RHL) must exist and be equal.

Calculate the LHL:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} x^2$

As $x$ approaches 0 from the left side, $x^2$ approaches $(0)^2 = 0$.

LHL $= 0$

Calculate the RHL:

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} x^2$

As $x$ approaches 0 from the right side, $x^2$ approaches $(0)^2 = 0$.

RHL $= 0$

Since LHL = RHL = 0, the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 0.

Step 3: Check if $\lim\limits_{x \to 0} f(x) = f(0)$.

From Step 1, we have $f(0) = 0$.

From Step 2, we have $\lim\limits_{x \to 0} f(x) = 0$.

Comparing these values, we see that $\lim\limits_{x \to 0} f(x) = f(0)$ (since $0 = 0$).

Since all three conditions for continuity are met, the function $f(x) = x^2$ is continuous at $x=0$.

Answer:

The function $f(x) = x^2$ is continuous at $x=0$.

Given:

The function $f(x) = |x|$.

The point at which continuity is to be discussed is $x = 0$.

The function $f(x) = |x|$ can be defined piecewise as:

$f(x) = \begin{cases} x & , & \text{if } x \geq 0 \\ -x & , & \text{if } x < 0 \end{cases}$

To Discuss:

The continuity of the function $f(x)$ at $x=0$.

Solution:

A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:

1. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x)$ exists (i.e., $\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x)$).

3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this case, $c = 0$. We need to discuss the continuity of $f(x) = |x|$ at $x=0$.

Step 1: Check if $f(0)$ is defined.

For $x=0$, we use the definition $f(x) = x$ since $0 \geq 0$.

$f(0) = 0$

Since $f(0)$ is a real number (0), it is defined.

Step 2: Check if $\lim\limits_{x \to 0} f(x)$ exists.

We need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $x=0$.

Calculate the LHL at $x=0$:

For $x < 0$, $f(x) = -x$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (-x)$

As $x$ approaches 0 from the left side (i.e., $x$ is a small negative number), $-x$ approaches $-(0) = 0$.

LHL $= 0$

Calculate the RHL at $x=0$:

For $x > 0$, $f(x) = x$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (x)$

As $x$ approaches 0 from the right side (i.e., $x$ is a small positive number), $x$ approaches 0.

RHL $= 0$

Since LHL = RHL = 0, the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 0.

Step 3: Check if $\lim\limits_{x \to 0} f(x) = f(0)$.

From Step 1, we have $f(0) = 0$.

From Step 2, we have $\lim\limits_{x \to 0} f(x) = 0$.

Comparing these values, we see that $\lim\limits_{x \to 0} f(x) = f(0)$ (since $0 = 0$).

Since all three conditions for continuity are met, the function $f(x) = |x|$ is continuous at $x=0$.

Answer:

The function $f(x) = |x|$ is continuous at $x=0$.

Given:

The function $f(x)$ defined as:

$f(x) = \begin{cases} x^3+3, & , & \text{if } x \neq 0 \\ 1, & , & \text{if } x = 0 \end{cases}$

The point at which continuity is to be checked is $x = 0$.

To Show:

That the function $f(x)$ is not continuous at $x = 0$.

Solution:

For a function $f(x)$ to be continuous at a point $x=c$, the following three conditions must be satisfied:

1. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x)$ exists.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this problem, we need to check the continuity of $f(x)$ at $x=0$. Here, $c=0$.

Step 1: Evaluate $f(0)$.

According to the definition of the function, when $x = 0$, $f(x) = 1$.

$f(0) = 1$

Since $f(0)$ is a finite real number, it is defined.

Step 2: Evaluate the limit of $f(x)$ as $x$ approaches 0, i.e., $\lim\limits_{x \to 0} f(x)$.

For $x$ approaching 0, but $x \neq 0$, the function is defined as $f(x) = x^3 + 3$.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (x^3 + 3)$

As $x$ approaches 0, $x^3$ approaches $0^3 = 0$.

$\lim\limits_{x \to 0} (x^3 + 3) = 0 + 3 = 3$

So, $\lim\limits_{x \to 0} f(x) = 3$. The limit exists.

Step 3: Compare $f(0)$ and $\lim\limits_{x \to 0} f(x)$.

From Step 1, $f(0) = 1$.

From Step 2, $\lim\limits_{x \to 0} f(x) = 3$.

We observe that $f(0) \neq \lim\limits_{x \to 0} f(x)$, since $1 \neq 3$.

Since the third condition for continuity is not met, the function $f(x)$ is not continuous at $x=0$.

Conclusion:

The function $f(x) = \begin{cases} x^3+3, & , & \text{if } x \neq 0 \\ 1, & , & \text{if } x = 0 \end{cases}$ is not continuous at $x=0$.

Given:

The function $f(x) = k$, where $k$ is a constant.

To Check:

The points where the function $f(x)$ is continuous.

Solution:

A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:

1. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x)$ exists.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Let's consider an arbitrary real number $c$. We will check the continuity of $f(x) = k$ at $x=c$.

Step 1: Check if $f(c)$ is defined.

Substitute $x = c$ into the function $f(x)$:

$f(c) = k$

Since $k$ is a constant (a real number), $f(c)$ is defined.

Step 2: Check if $\lim\limits_{x \to c} f(x)$ exists.

For a constant function $f(x) = k$, as $x$ approaches any value $c$, the value of the function remains $k$.

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} k = k$

The limit exists and is equal to $k$.

Step 3: Check if $\lim\limits_{x \to c} f(x) = f(c)$.

From Step 1, we have $f(c) = k$.

From Step 2, we have $\lim\limits_{x \to c} f(x) = k$.

Comparing these values, we see that $\lim\limits_{x \to c} f(x) = f(c)$ (since $k = k$).

Since all three conditions for continuity are met for any arbitrary real number $c$, the constant function $f(x) = k$ is continuous at every real number.

Answer:

The constant function $f(x) = k$ is continuous at all real numbers, i.e., it is continuous everywhere on $\mathbb{R}$.

Given:

The identity function $f(x) = x$ on real numbers.

To Prove:

That the function $f(x) = x$ is continuous at every real number.

Proof:

To prove that the function $f(x) = x$ is continuous at every real number, we consider an arbitrary real number, let's call it $c$. We need to check the three conditions for continuity at $x=c$:

1. $f(c)$ must be defined.

2. $\lim\limits_{x \to c} f(x)$ must exist.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Step 1: Evaluate $f(c)$.

For the function $f(x) = x$, when $x = c$, the function value is:

$f(c) = c$

Since $c$ is a real number, $f(c)$ is a well-defined real number. Thus, the first condition is met.

Step 2: Evaluate the limit of $f(x)$ as $x$ approaches $c$, i.e., $\lim\limits_{x \to c} f(x)$.

For the function $f(x) = x$, the limit as $x$ approaches $c$ is simply:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} x$

As $x$ gets arbitrarily close to $c$, the value of $x$ gets arbitrarily close to $c$.

$\lim\limits_{x \to c} x = c$

The limit exists and is equal to $c$. Thus, the second condition is met.

Step 3: Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$.

From Step 1, we have $f(c) = c$.

From Step 2, we have $\lim\limits_{x \to c} f(x) = c$.

Comparing these two values, we see that $\lim\limits_{x \to c} f(x) = f(c)$, since both are equal to $c$. Thus, the third condition is met.

Since all three conditions for continuity at $x=c$ are satisfied for any arbitrary real number $c$, the function $f(x) = x$ is continuous at every real number.

Hence, proved.

Given:

The function $f(x) = |x|$.

The function can be defined piecewise as:

$f(x) = \begin{cases} x & , & \text{if } x \geq 0 \\ -x & , & \text{if } x < 0 \end{cases}$

To Check:

Whether the function $f(x) = |x|$ is a continuous function over its domain (all real numbers).

Solution:

A function is continuous if it is continuous at every point in its domain. The domain of $f(x) = |x|$ is all real numbers, $\mathbb{R}$.

We need to check the continuity at every point $x=c \in \mathbb{R}$. We consider two cases: when $c = 0$ and when $c \neq 0$.

Case 1: Continuity at $x = 0$.

A function $f(x)$ is continuous at $x=0$ if $f(0)$ is defined, $\lim\limits_{x \to 0} f(x)$ exists, and $\lim\limits_{x \to 0} f(x) = f(0)$.

1. Evaluate $f(0)$: For $x=0$, $f(x) = x$. So, $f(0) = 0$. $f(0)$ is defined.

2. Evaluate $\lim\limits_{x \to 0} f(x)$: We need to check the left-hand limit (LHL) and the right-hand limit (RHL).

LHL: $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (-x) = -0 = 0$ (since for $x < 0$, $f(x) = -x$).

RHL: $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (x) = 0$ (since for $x > 0$, $f(x) = x$).

Since LHL = RHL = 0, the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 0.

3. Compare $f(0)$ and $\lim\limits_{x \to 0} f(x)$: We have $f(0) = 0$ and $\lim\limits_{x \to 0} f(x) = 0$. Since $f(0) = \lim\limits_{x \to 0} f(x)$, the function is continuous at $x=0$.

Case 2: Continuity at $x = c$, where $c \neq 0$.

We need to consider $c > 0$ and $c < 0$.

If $c > 0$, then for all $x$ sufficiently close to $c$, $x > 0$. Thus, $f(x) = x$.

1. $f(c) = c$. Defined.

2. $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} x = c$. Limit exists.

3. $f(c) = \lim\limits_{x \to c} f(x)$ (since $c=c$). So, $f(x)$ is continuous for all $c > 0$.

If $c < 0$, then for all $x$ sufficiently close to $c$, $x < 0$. Thus, $f(x) = -x$.

1. $f(c) = -c$. Defined.

2. $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (-x) = -c$. Limit exists.

3. $f(c) = \lim\limits_{x \to c} f(x)$ (since $-c=-c$). So, $f(x)$ is continuous for all $c < 0$.

From both cases, we see that the function $f(x) = |x|$ is continuous at $x=0$ and also continuous at all points $x \neq 0$. Therefore, it is continuous at every real number.

Answer:

Yes, the function $f(x) = |x|$ is a continuous function. It is continuous at every real number.

Given:

The function $f(x) = x^3 + x^2 - 1$.

To Discuss:

The continuity of the function $f(x)$.

Solution:

The given function is $f(x) = x^3 + x^2 - 1$.

This function is a polynomial function. A polynomial function is a function of the form $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_i$ are real coefficients and $n$ is a non-negative integer.

The function $f(x) = x^3 + x^2 - 1$ fits this definition with $n=3$, $a_3=1$, $a_2=1$, $a_1=0$, and $a_0=-1$.

The domain of a polynomial function is the set of all real numbers, $\mathbb{R}$.

A fundamental property in calculus is that all polynomial functions are continuous at every point in their domain.

Since the domain of $f(x) = x^3 + x^2 - 1$ is $\mathbb{R}$, and it is a polynomial function, it is continuous at every real number.

Alternatively, we can demonstrate continuity at an arbitrary real number $c$. A function $f(x)$ is continuous at $x=c$ if:

1. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x)$ exists.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Let $c$ be any real number.

1. Evaluate $f(c)$: $f(c) = c^3 + c^2 - 1$. Since $c$ is a real number, $c^3 + c^2 - 1$ is a well-defined real number. So, $f(c)$ is defined.

2. Evaluate $\lim\limits_{x \to c} f(x)$: Using the properties of limits (the limit of a sum is the sum of the limits, the limit of a power is the power of the limit, and the limit of a constant is the constant):

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (x^3 + x^2 - 1)$

$= \lim\limits_{x \to c} x^3 + \lim\limits_{x \to c} x^2 - \lim\limits_{x \to c} 1$

$= (\lim\limits_{x \to c} x)^3 + (\lim\limits_{x \to c} x)^2 - 1$

Since $\lim\limits_{x \to c} x = c$ for any real number $c$:

$= c^3 + c^2 - 1$

The limit exists and is equal to $c^3 + c^2 - 1$.

3. Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$: We have $f(c) = c^3 + c^2 - 1$ and $\lim\limits_{x \to c} f(x) = c^3 + c^2 - 1$. Thus, $\lim\limits_{x \to c} f(x) = f(c)$.

Since all three conditions for continuity are satisfied for an arbitrary real number $c$, the function $f(x) = x^3 + x^2 - 1$ is continuous at every real number.

Conclusion:

The function $f(x) = x^3 + x^2 - 1$ is a polynomial function and is therefore continuous at every real number.

Given:

The function $f(x) = \frac{1}{x}$, with the domain $x \neq 0$. The domain is $\mathbb{R} \setminus \{0\}$.

To Discuss:

The continuity of the function $f(x)$ over its domain.

Solution:

A function is continuous if it is continuous at every point in its domain.

The domain of the function $f(x) = \frac{1}{x}$ is all real numbers except $x=0$. That is, the domain is $(-\infty, 0) \cup (0, \infty)$.

To discuss the continuity of $f(x)$, we need to check if it is continuous at every point $x=c$ within its domain.

Let $c$ be any arbitrary real number such that $c \neq 0$. We check the three conditions for continuity at $x=c$:

1. $f(c)$ must be defined.

2. $\lim\limits_{x \to c} f(x)$ must exist.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Let $c$ be any real number such that $c \neq 0$.

Step 1: Check if $f(c)$ is defined.

Substitute $x = c$ into the function $f(x)$: $f(c) = \frac{1}{c}$.

Since $c \neq 0$, $\frac{1}{c}$ is a finite real number. Thus, $f(c)$ is defined.

Step 2: Check if $\lim\limits_{x \to c} f(x)$ exists.

For the function $f(x) = \frac{1}{x}$, the limit as $x$ approaches $c$ (where $c \neq 0$) is:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} \frac{1}{x}$

Since $c \neq 0$, we can substitute $x=c$ directly into the expression:

$\lim\limits_{x \to c} \frac{1}{x} = \frac{1}{c}$

The limit exists and is equal to $\frac{1}{c}$.

Step 3: Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$.

From Step 1, we have $f(c) = \frac{1}{c}$.

From Step 2, we have $\lim\limits_{x \to c} f(x) = \frac{1}{c}$.

Comparing these two values, we see that $\lim\limits_{x \to c} f(x) = f(c)$, since both are equal to $\frac{1}{c}$.

Since all three conditions for continuity at $x=c$ are satisfied for any arbitrary real number $c \neq 0$, the function $f(x) = \frac{1}{x}$ is continuous at every point in its domain.

Conclusion:

The function $f(x) = \frac{1}{x}$ is continuous at every point in its domain $\mathbb{R} \setminus \{0\}$.

Given:

The function $f(x)$ defined as:

$f(x) = \begin{cases} x+2, & , & \text{if } x \leq 1 \\ x-2, & , & \text{if } x > 1 \end{cases}$

To Discuss:

The continuity of the function $f(x)$.

Solution:

The function $f(x)$ is defined piecewise. The definition changes at the point $x=1$.

We need to check the continuity at the point where the definition changes, which is $x=1$, and also at points where the definition does not change, i.e., for $x < 1$ and $x > 1$.

Case 1: Continuity for $x < 1$.

For any $c < 1$, $f(x) = x+2$ for all $x$ sufficiently close to $c$. The function $g(x) = x+2$ is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous for all $x < 1$.

Case 2: Continuity for $x > 1$.

For any $c > 1$, $f(x) = x-2$ for all $x$ sufficiently close to $c$. The function $h(x) = x-2$ is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous for all $x > 1$.

Case 3: Continuity at $x = 1$.

We need to check the three conditions for continuity at $x=1$:

1. $f(1)$ must be defined.

2. $\lim\limits_{x \to 1} f(x)$ must exist (i.e., $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x)$).

3. $\lim\limits_{x \to 1} f(x) = f(1)$.

Step 3.1: Evaluate $f(1)$.

According to the definition, if $x \leq 1$, $f(x) = x+2$. So, $f(1) = 1+2 = 3$. $f(1)$ is defined.

Step 3.2: Evaluate the limits as $x$ approaches 1.

Left-hand limit (LHL) at $x=1$:

For $x < 1$, $f(x) = x+2$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2) = 1+2 = 3$

Right-hand limit (RHL) at $x=1$:

For $x > 1$, $f(x) = x-2$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x-2) = 1-2 = -1$

Since LHL ($3$) $\neq$ RHL ($-1$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

Since the limit does not exist at $x=1$, the function $f(x)$ is not continuous at $x=1$.

Conclusion:

The function $f(x)$ is continuous for all real numbers except at $x=1$. The function is discontinuous at $x=1$.

So, the function is continuous on the intervals $(-\infty, 1)$ and $(1, \infty)$.

Given:

The function $f(x)$ defined as:

$f(x) = \begin{cases} x+2, & , & \text{if } x < 1 \\ 0, & , & \text{if } x = 1 \\ x-2, & , & \text{if } x > 1 \end{cases}$

To Find:

All the points of discontinuity of the function $f(x)$.

Solution:

A function can only be discontinuous at points where its definition changes or where the function is undefined. In this case, the definition of $f(x)$ changes at $x=1$. The function is defined for all real numbers.

We need to check the continuity at the point where the definition changes, which is $x=1$. For points where the definition does not change ($x < 1$ and $x > 1$), the function is defined by polynomial expressions, which are continuous.

Let's check the continuity at $x = 1$. We need to verify the three conditions for continuity:

1. $f(1)$ must be defined.

2. $\lim\limits_{x \to 1} f(x)$ must exist.

3. $\lim\limits_{x \to 1} f(x) = f(1)$.

Step 1: Evaluate $f(1)$.

According to the definition, if $x = 1$, $f(x) = 0$. So, $f(1) = 0$. $f(1)$ is defined.

Step 2: Evaluate the limits as $x$ approaches 1.

Left-hand limit (LHL) at $x=1$:

For $x < 1$, $f(x) = x+2$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2) = 1+2 = 3$

Right-hand limit (RHL) at $x=1$:

For $x > 1$, $f(x) = x-2$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x-2) = 1-2 = -1$

Since LHL ($3$) $\neq$ RHL ($-1$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

Since the limit does not exist at $x=1$, the function $f(x)$ is not continuous at $x=1$. Thus, $x=1$ is a point of discontinuity.

For any point $c < 1$, $f(x) = x+2$ in an interval around $c$. Since $y = x+2$ is a polynomial, it is continuous at $c$. Thus, $f(x)$ is continuous for all $x < 1$.

For any point $c > 1$, $f(x) = x-2$ in an interval around $c$. Since $y = x-2$ is a polynomial, it is continuous at $c$. Thus, $f(x)$ is continuous for all $x > 1$.

Therefore, the only point of discontinuity is $x=1$.

Answer:

The function $f(x)$ has a discontinuity at $x=1$. There are no other points of discontinuity.

Given:

The function $f(x)$ defined as:

$f(x) = \begin{cases} x+2, & , & \text{if } x < 0 \\ -x+2, & , & \text{if } x > 0 \end{cases}$

To Discuss:

The continuity of the function $f(x)$.

Solution:

The function $f(x)$ is defined piecewise. The definition changes at $x=0$. Note that the function is not defined at $x=0$ itself based on the given definition.

The domain of the function is $(-\infty, 0) \cup (0, \infty)$, i.e., all real numbers except 0.

We need to check the continuity at points where the definition does not change, i.e., for $x < 0$ and $x > 0$. We also need to consider if there's a discontinuity at the point where the definition changes, which is $x=0$. However, for a function to be continuous at a point, it must be defined at that point. Since $f(0)$ is not defined, the function is immediately discontinuous at $x=0$.

Case 1: Continuity for $x < 0$.

For any $c < 0$, $f(x) = x+2$ for all $x$ sufficiently close to $c$. The function $g(x) = x+2$ is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous for all $x < 0$.

Case 2: Continuity for $x > 0$.

For any $c > 0$, $f(x) = -x+2$ for all $x$ sufficiently close to $c$. The function $h(x) = -x+2$ is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous for all $x > 0$.

Case 3: Continuity at $x = 0$.

For a function to be continuous at $x=0$, $f(0)$ must be defined. Looking at the definition:

$f(x) = \begin{cases} x+2, & , & \text{if } x < 0 \\ -x+2, & , & \text{if } x > 0 \end{cases}$

There is no case for $x=0$. Therefore, $f(0)$ is not defined.

Since $f(0)$ is not defined, the function is discontinuous at $x=0$.

We can also check the limits at $x=0$ for completeness, although the function being undefined is sufficient to show discontinuity at that point.

Left-hand limit (LHL) at $x=0$:

For $x < 0$, $f(x) = x+2$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (x+2) = 0+2 = 2$

Right-hand limit (RHL) at $x=0$:

For $x > 0$, $f(x) = -x+2$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (-x+2) = -0+2 = 2$

The limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 2. However, since $f(0)$ is not defined, the function is discontinuous at $x=0$.

Therefore, the function is continuous for all real numbers except $x=0$.

Conclusion:

The function $f(x)$ is continuous on the intervals $(-\infty, 0)$ and $(0, \infty)$. It has a discontinuity at $x=0$ because the function is not defined at this point.

Given:

The function $f(x)$ defined as:

$f(x) = \begin{cases} x, & , & \text{if } x \geq 0 \\ x^2, & , & \text{if } x < 0 \end{cases}$

To Discuss:

The continuity of the function $f(x)$.

Solution:

The function $f(x)$ is defined piecewise. The definition changes at $x=0$. The domain of the function is all real numbers, $\mathbb{R}$.

We need to check the continuity at the point where the definition changes, which is $x=0$, and also at points where the definition does not change, i.e., for $x < 0$ and $x > 0$.

Case 1: Continuity for $x < 0$.

For any $c < 0$, $f(x) = x^2$ for all $x$ sufficiently close to $c$. The function $g(x) = x^2$ is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous for all $x < 0$.

Case 2: Continuity for $x > 0$.

For any $c > 0$, $f(x) = x$ for all $x$ sufficiently close to $c$. The function $h(x) = x$ is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous for all $x > 0$.

Case 3: Continuity at $x = 0$.

We need to check the three conditions for continuity at $x=0$:

1. $f(0)$ must be defined.

2. $\lim\limits_{x \to 0} f(x)$ must exist (i.e., $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x)$).

3. $\lim\limits_{x \to 0} f(x) = f(0)$.

Step 3.1: Evaluate $f(0)$.

According to the definition, if $x \geq 0$, $f(x) = x$. So, $f(0) = 0$. $f(0)$ is defined.

Step 3.2: Evaluate the limits as $x$ approaches 0.

Left-hand limit (LHL) at $x=0$:

For $x < 0$, $f(x) = x^2$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} x^2 = 0^2 = 0$

Right-hand limit (RHL) at $x=0$:

For $x \geq 0$, $f(x) = x$. We use this for the limit as $x \to 0^+$ because the definition for $x \geq 0$ applies to values just greater than 0.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} x = 0$

Since LHL ($0$) = RHL ($0$) = 0, the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 0.

Step 3.3: Compare $f(0)$ and $\lim\limits_{x \to 0} f(x)$.

From Step 3.1, we have $f(0) = 0$.

From Step 3.2, we have $\lim\limits_{x \to 0} f(x) = 0$.

Comparing these values, we see that $\lim\limits_{x \to 0} f(x) = f(0)$ (since $0 = 0$).

Since all three conditions for continuity are met at $x=0$, the function $f(x)$ is continuous at $x=0$.

Combining the cases, the function is continuous for $x < 0$, continuous for $x > 0$, and continuous at $x = 0$. Therefore, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number. It is a continuous function.

Given:

A general polynomial function $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $n$ is a non-negative integer and $a_0, a_1, \dots, a_n$ are real coefficients.

To Show:

That every polynomial function is continuous.

Proof:

A function is continuous if it is continuous at every point in its domain. The domain of any polynomial function is the set of all real numbers, $\mathbb{R}$. To show that a polynomial function $P(x)$ is continuous, we need to show that it is continuous at any arbitrary real number $c$.

For a function $P(x)$ to be continuous at $x=c$, the following three conditions must be satisfied:

1. $P(c)$ must be defined.

2. $\lim\limits_{x \to c} P(x)$ must exist.

3. $\lim\limits_{x \to c} P(x) = P(c)$.

Let $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ and let $c$ be any real number.

Step 1: Evaluate $P(c)$.

Substitute $x=c$ into the polynomial expression:

$P(c) = a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$

Since $c$ is a real number and $a_i$ are real coefficients, $c^i$ is a real number for any non-negative integer $i$. The sum and products of real numbers are real numbers. Therefore, $P(c)$ is a well-defined real number. Thus, the first condition is met.

Step 2: Evaluate the limit of $P(x)$ as $x$ approaches $c$, i.e., $\lim\limits_{x \to c} P(x)$.

We use the properties of limits:

(a) $\lim\limits_{x \to c} k = k$, where $k$ is a constant.

(b) $\lim\limits_{x \to c} x = c$.

(c) $\lim\limits_{x \to c} (f(x) + g(x)) = \lim\limits_{x \to c} f(x) + \lim\limits_{x \to c} g(x)$ (Sum rule).

(d) $\lim\limits_{x \to c} (k \cdot f(x)) = k \cdot \lim\limits_{x \to c} f(x)$ (Constant multiple rule).

(e) $\lim\limits_{x \to c} (f(x))^m = (\lim\limits_{x \to c} f(x))^m$ (Power rule), where $m$ is a positive integer.

Applying these properties to $\lim\limits_{x \to c} P(x)$:

$\lim\limits_{x \to c} P(x) = \lim\limits_{x \to c} (a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0)$

Using the sum rule:

$= \lim\limits_{x \to c} (a_n x^n) + \lim\limits_{x \to c} (a_{n-1} x^{n-1}) + \dots + \lim\limits_{x \to c} (a_1 x) + \lim\limits_{x \to c} a_0$

Using the constant multiple rule:

$= a_n \lim\limits_{x \to c} x^n + a_{n-1} \lim\limits_{x \to c} x^{n-1} + \dots + a_1 \lim\limits_{x \to c} x + \lim\limits_{x \to c} a_0$

Using the power rule and the limit of a constant:

$= a_n (\lim\limits_{x \to c} x)^n + a_{n-1} (\lim\limits_{x \to c} x)^{n-1} + \dots + a_1 (\lim\limits_{x \to c} x) + a_0$

Using $\lim\limits_{x \to c} x = c$:

$= a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$

The limit exists and is equal to $a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$. Thus, the second condition is met.

Step 3: Compare $P(c)$ and $\lim\limits_{x \to c} P(x)$.

From Step 1, we have $P(c) = a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$.

From Step 2, we have $\lim\limits_{x \to c} P(x) = a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$.

Comparing these two values, we see that $\lim\limits_{x \to c} P(x) = P(c)$. Thus, the third condition is met.

Since all three conditions for continuity at $x=c$ are satisfied for any arbitrary real number $c$, the polynomial function $P(x)$ is continuous at every real number.

Hence, every polynomial function is continuous.

Given:

The greatest integer function defined by $f(x) = [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$.

Examples: $[1.5] = 1$, $[2] = 2$, $[-0.7] = -1$, $[-3] = -3$.

The domain of this function is all real numbers, $\mathbb{R}$.

To Find:

All the points of discontinuity of the function $f(x) = [x]$.

Solution:

A function can be discontinuous at points where its definition or behavior changes abruptly. For the greatest integer function $f(x) = [x]$, the value of the function jumps at every integer value of $x$.

We need to check the continuity at two types of points: integer points and non-integer points.

Case 1: Continuity at a non-integer point $c$.

Let $c$ be a real number that is not an integer. This means $c$ lies strictly between two consecutive integers, say $n$ and $n+1$, where $n$ is an integer. So, $n < c < n+1$.

For any $x$ sufficiently close to $c$, $x$ will also be between $n$ and $n+1$. In such an interval around $c$, the value of $[x]$ is constant and equal to $n$. That is, for $x$ in an interval around $c$, $f(x) = [x] = n$.

Let's check the conditions for continuity at $x=c$:

1. $f(c) = [c] = n$. $f(c)$ is defined.

2. $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} [x]$. Since $f(x) = n$ for $x$ near $c$, $\lim\limits_{x \to c} [x] = n$. The limit exists.

3. $\lim\limits_{x \to c} f(x) = n$ and $f(c) = n$. Since $\lim\limits_{x \to c} f(x) = f(c)$, the function is continuous at any non-integer point $c$.

Case 2: Continuity at an integer point $c = n$.

Let $c$ be an integer, say $c = n$. We check the conditions for continuity at $x=n$:

1. $f(n) = [n]$. By definition, the greatest integer less than or equal to an integer $n$ is $n$. So, $f(n) = n$. $f(n)$ is defined.

2. Evaluate the limits as $x$ approaches $n$.

Left-hand limit (LHL) at $x=n$:

$\lim\limits_{x \to n^-} f(x) = \lim\limits_{x \to n^-} [x]$

As $x$ approaches $n$ from the left side (i.e., $x$ is slightly less than $n$, e.g., $n-0.001$), the greatest integer less than or equal to $x$ is the integer immediately below $n$, which is $n-1$.

LHL $= n-1$

Right-hand limit (RHL) at $x=n$:

$\lim\limits_{x \to n^+} f(x) = \lim\limits_{x \to n^+} [x]$

As $x$ approaches $n$ from the right side (i.e., $x$ is slightly greater than $n$, e.g., $n+0.001$), the greatest integer less than or equal to $x$ is $n$.

RHL $= n$

Since LHL ($n-1$) $\neq$ RHL ($n$), the limit $\lim\limits_{x \to n} f(x)$ does not exist at any integer point $n$.

Since the limit does not exist at any integer point, the function $f(x) = [x]$ is discontinuous at all integer points.

Combining both cases, the function is continuous at all non-integer points and discontinuous at all integer points.

Answer:

The points of discontinuity of the greatest integer function $f(x) = [x]$ are all integer points.

Given:

A rational function $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomial functions, and $Q(x)$ is not the zero polynomial.

To Prove:

That every rational function is continuous.

Proof:

A rational function $f(x) = \frac{P(x)}{Q(x)}$ is defined for all real numbers $x$ such that $Q(x) \neq 0$. The domain of a rational function is $\{x \in \mathbb{R} \mid Q(x) \neq 0\}$.

To prove that a rational function is continuous, we need to show that it is continuous at every point in its domain.

Let $f(x) = \frac{P(x)}{Q(x)}$ be a rational function, where $P(x)$ and $Q(x)$ are polynomials. Let $c$ be any arbitrary real number in the domain of $f(x)$. By definition of the domain, this means $Q(c) \neq 0$.

For $f(x)$ to be continuous at $x=c$, the following three conditions must be satisfied:

1. $f(c)$ must be defined.

2. $\lim\limits_{x \to c} f(x)$ must exist.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Step 1: Check if $f(c)$ is defined.

Since $c$ is in the domain of $f(x)$, $Q(c) \neq 0$.

$f(c) = \frac{P(c)}{Q(c)}$

Since $P(x)$ and $Q(x)$ are polynomials, $P(c)$ and $Q(c)$ are well-defined real numbers. Since $Q(c) \neq 0$, the fraction $\frac{P(c)}{Q(c)}$ is a well-defined real number. Thus, $f(c)$ is defined.

Step 2: Evaluate the limit of $f(x)$ as $x$ approaches $c$, i.e., $\lim\limits_{x \to c} f(x)$.

Using the property of limits that the limit of a quotient is the quotient of the limits (provided the limit of the denominator is not zero):

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} \frac{P(x)}{Q(x)}$

From Example 14, we know that every polynomial function is continuous. Therefore, $\lim\limits_{x \to c} P(x) = P(c)$ and $\lim\limits_{x \to c} Q(x) = Q(c)$ for any real number $c$.

Substitute these into the limit expression (i):

$\lim\limits_{x \to c} f(x) = \frac{P(c)}{Q(c)}$

Since $c$ is in the domain of $f(x)$, $Q(c) \neq 0$, so the denominator is not zero. The limit exists and is equal to $\frac{P(c)}{Q(c)}$.

Step 3: Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$.

From Step 1, we have $f(c) = \frac{P(c)}{Q(c)}$.

From Step 2, we have $\lim\limits_{x \to c} f(x) = \frac{P(c)}{Q(c)}$.

Comparing these two values, we see that $\lim\limits_{x \to c} f(x) = f(c)$. Thus, the third condition is met.

Since all three conditions for continuity at $x=c$ are satisfied for any arbitrary real number $c$ in the domain of $f(x)$, the rational function $f(x)$ is continuous at every point in its domain.

Hence, every rational function is continuous.

Given:

The sine function, $f(x) = \sin x$.

The domain of the sine function is the set of all real numbers, $\mathbb{R}$.

To Discuss:

The continuity of the sine function over its domain.

Solution:

To discuss the continuity of $f(x) = \sin x$, we need to check if it is continuous at every point in its domain $\mathbb{R}$. Let $c$ be an arbitrary real number. We need to verify the three conditions for continuity at $x=c$:

1. $f(c)$ must be defined.

2. $\lim\limits_{x \to c} f(x)$ must exist.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Step 1: Check if $f(c)$ is defined.

$f(c) = \sin c$. Since $\sin x$ is defined for all real numbers, $\sin c$ is a well-defined real number. Thus, $f(c)$ is defined.

Step 2: Evaluate the limit of $f(x)$ as $x$ approaches $c$, i.e., $\lim\limits_{x \to c} \sin x$.

Let $x = c + h$. As $x \to c$, $h \to 0$.

$\lim\limits_{x \to c} \sin x = \lim\limits_{h \to 0} \sin(c+h)$

Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

$= \lim\limits_{h \to 0} (\sin c \cos h + \cos c \sin h)$

Using the sum rule and constant multiple rule for limits:

$= \lim\limits_{h \to 0} (\sin c \cos h) + \lim\limits_{h \to 0} (\cos c \sin h)$

$= \sin c \lim\limits_{h \to 0} \cos h + \cos c \lim\limits_{h \to 0} \sin h$

We know the standard limits $\lim\limits_{h \to 0} \cos h = \cos 0 = 1$ and $\lim\limits_{h \to 0} \sin h = \sin 0 = 0$.

$= \sin c (1) + \cos c (0)$

$= \sin c + 0 = \sin c$

The limit $\lim\limits_{x \to c} \sin x$ exists and is equal to $\sin c$. Thus, the second condition is met.

Step 3: Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$.

From Step 1, we have $f(c) = \sin c$.

From Step 2, we have $\lim\limits_{x \to c} f(x) = \sin c$.

Comparing these two values, we see that $\lim\limits_{x \to c} f(x) = f(c)$. Thus, the third condition is met.

Since all three conditions for continuity at $x=c$ are satisfied for any arbitrary real number $c$, the sine function $f(x) = \sin x$ is continuous at every real number.

Conclusion:

The sine function is continuous on its entire domain, which is all real numbers.

Given:

The function $f(x) = \tan x$.

We know that $\tan x = \frac{\sin x}{\cos x}$.

The domain of $\tan x$ is all real numbers $x$ for which $\cos x \neq 0$. The values of $x$ for which $\cos x = 0$ are $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

So, the domain of $f(x) = \tan x$ is $\mathbb{R} \setminus \{x \in \mathbb{R} \mid x = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}\}$.

To Prove:

That the function $f(x) = \tan x$ is a continuous function over its domain.

Proof:

A function is continuous if it is continuous at every point in its domain. The domain of $f(x) = \tan x$ is the set of all real numbers except $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

To show that $f(x) = \tan x$ is continuous, we need to show that it is continuous at any arbitrary real number $c$ in its domain.

For a function $f(x)$ to be continuous at $x=c$, where $c$ is in the domain of $f(x)$, the following three conditions must be satisfied:

1. $f(c)$ must be defined.

2. $\lim\limits_{x \to c} f(x)$ must exist.

3. $\lim\limits_{x \to c} f(x) = f(c)$.

Let $c$ be any real number in the domain of $\tan x$. By definition of the domain, $\cos c \neq 0$.

Step 1: Check if $f(c)$ is defined.

$f(c) = \tan c = \frac{\sin c}{\cos c}$. Since $\cos c \neq 0$, $\tan c$ is a well-defined real number. Thus, $f(c)$ is defined.

Step 2: Evaluate the limit of $f(x)$ as $x$ approaches $c$, i.e., $\lim\limits_{x \to c} \tan x$.

$\lim\limits_{x \to c} \tan x = \lim\limits_{x \to c} \frac{\sin x}{\cos x}$

Using the property of limits that the limit of a quotient is the quotient of the limits (provided the limit of the denominator is not zero):

From Example 17 (continuity of sine function), we know that $\lim\limits_{x \to c} \sin x = \sin c$ for any real number $c$.

Similarly, we can show that the cosine function is continuous at every real number. Using a similar approach as Example 17 with the identity $\cos(c+h) = \cos c \cos h - \sin c \sin h$, we find that $\lim\limits_{x \to c} \cos x = \cos c$ for any real number $c$.

Since $c$ is in the domain of $\tan x$, we know that $\cos c \neq 0$. Therefore, the limit of the denominator $\lim\limits_{x \to c} \cos x = \cos c$ is not zero.

Substitute these limits into expression (i):

$\lim\limits_{x \to c} \tan x = \frac{\sin c}{\cos c} = \tan c$

The limit exists and is equal to $\tan c$. Thus, the second condition is met.

Step 3: Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$.

From Step 1, we have $f(c) = \tan c$.

From Step 2, we have $\lim\limits_{x \to c} f(x) = \tan c$.

Comparing these two values, we see that $\lim\limits_{x \to c} f(x) = f(c)$. Thus, the third condition is met.

Since all three conditions for continuity at $x=c$ are satisfied for any arbitrary real number $c$ in the domain of $\tan x$, the function $f(x) = \tan x$ is continuous at every point in its domain.

Hence, the function defined by $f(x) = \tan x$ is a continuous function.

Given:

The function $f(x) = \sin(x^2)$.

The domain of this function is all real numbers, $\mathbb{R}$, because $x^2$ is defined for all real $x$, and $\sin u$ is defined for all real $u$.

To Show:

That the function $f(x) = \sin(x^2)$ is a continuous function over its domain $\mathbb{R}$.

Proof:

We can show the continuity of the function $f(x) = \sin(x^2)$ by considering it as a composition of two functions.

Let $g(x) = x^2$ and $h(u) = \sin u$. Then $f(x) = h(g(x)) = \sin(x^2)$.

We know from previous examples that:

1. The function $g(x) = x^2$ is a polynomial function (specifically, a quadratic function). From Example 14, every polynomial function is continuous on its domain, which is $\mathbb{R}$. So, $g(x)$ is continuous everywhere on $\mathbb{R}$.

2. The function $h(u) = \sin u$ is the sine function. From Example 17, the sine function is continuous on its domain, which is all real numbers. So, $h(u)$ is continuous everywhere on $\mathbb{R}$.

We use the theorem which states that if a function $g$ is continuous at a point $c$, and a function $h$ is continuous at $g(c)$, then the composite function $h \circ g$ (defined by $(h \circ g)(x) = h(g(x))$) is continuous at $c$.

Let $c$ be an arbitrary real number in the domain of $f(x)$.

Since $g(x) = x^2$ is continuous at every real number, it is continuous at $x=c$.

The value of $g(c)$ is $c^2$. Since $c$ is a real number, $c^2$ is also a real number.

Since $h(u) = \sin u$ is continuous at every real number, it is continuous at $u = c^2 = g(c)$.

According to the theorem on the continuity of composite functions, since $g$ is continuous at $c$ and $h$ is continuous at $g(c)$, the composite function $(h \circ g)(x) = h(g(x)) = \sin(x^2) = f(x)$ is continuous at $c$.

Since $c$ was an arbitrary real number, the function $f(x) = \sin(x^2)$ is continuous at every real number in its domain $\mathbb{R}$.

Alternatively, we can directly check the conditions for continuity at an arbitrary point $c \in \mathbb{R}$.

1. $f(c) = \sin(c^2)$. Defined for all real $c$.

2. $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} \sin(x^2)$.

Let $u = x^2$. As $x \to c$, $u \to c^2$. Also, since the polynomial function $g(x) = x^2$ is continuous, the limit $\lim\limits_{x \to c} x^2 = c^2$.

Since the sine function is continuous, $\lim\limits_{u \to c^2} \sin u = \sin(c^2)$.

Therefore, $\lim\limits_{x \to c} \sin(x^2) = \sin(\lim\limits_{x \to c} x^2) = \sin(c^2)$. The limit exists.

3. Compare $f(c)$ and $\lim\limits_{x \to c} f(x)$. We have $f(c) = \sin(c^2)$ and $\lim\limits_{x \to c} f(x) = \sin(c^2)$. So, $\lim\limits_{x \to c} f(x) = f(c)$.

All conditions are met for any real number $c$.

Conclusion:

The function $f(x) = \sin(x^2)$ is a continuous function on $\mathbb{R}$.

Given:

The function $f(x) = |1 - x + |x||$, where $x$ is any real number.

The domain of the function is all real numbers, $\mathbb{R}$.

To Show:

That the function $f(x)$ is a continuous function over its domain $\mathbb{R}$.

Proof:

We can analyze the continuity of the function $f(x) = |1 - x + |x||$ by considering the properties of continuous functions and composition of functions.

We know the following:

1. Constant function $g(x) = k$ is continuous everywhere (Example 5).

2. Identity function $h(x) = x$ is continuous everywhere (Example 6).

3. Absolute value function $k(x) = |x|$ is continuous everywhere (Example 7).

4. If $u(x)$ and $v(x)$ are continuous functions, then $u(x) + v(x)$ and $u(x) - v(x)$ are continuous.

5. If $p(x)$ and $q(x)$ are continuous functions, the composite function $p(q(x))$ is continuous if $q$ is continuous at a point $c$ and $p$ is continuous at $q(c)$.

Let's break down the given function $f(x) = |1 - x + |x||$ into simpler functions:

Consider the function inside the outermost absolute value, let $g(x) = 1 - x + |x|$.

The term '1' is a constant function, which is continuous everywhere.

The term 'x' is the identity function, which is continuous everywhere. Thus, '-x' is also continuous (constant multiple -1 times continuous function).

The term '|x|' is the absolute value function, which is continuous everywhere.

The function $g(x) = 1 - x + |x|$ is the sum and difference of continuous functions (constant function, -1 times identity function, absolute value function). Therefore, $g(x)$ is continuous everywhere on $\mathbb{R}$.

Now, the function $f(x) = |g(x)| = |1 - x + |x||$. This is a composition of the absolute value function $k(u) = |u|$ and the function $g(x)$.

Let $k(u) = |u|$ and $u = g(x)$. Then $f(x) = k(g(x))$.

We know that $g(x) = 1 - x + |x|$ is continuous at every real number $c$.

We also know that the absolute value function $k(u) = |u|$ is continuous at every real number $u$. Since $g(c)$ is a real number for any real $c$, $k(u) = |u|$ is continuous at $u = g(c)$.

By the theorem on the continuity of composite functions, since $g$ is continuous at $c$ and $k$ is continuous at $g(c)$, the composite function $f(x) = k(g(x)) = |g(x)| = |1 - x + |x||$ is continuous at $c$.

Since $c$ was an arbitrary real number, the function $f(x) = |1 - x + |x||$ is continuous at every real number in its domain $\mathbb{R}$.

Conclusion:

The function $f(x) = |1 - x + |x||$ is continuous for all real numbers.

To prove that the function $f(x) = 5x - 3$ is continuous at a point $x=a$, we need to show that $\lim\limits_{x \to a} f(x) = f(a)$. This requires verifying three conditions:

1. $f(a)$ is defined.

2. $\lim\limits_{x \to a} f(x)$ exists.

3. $\lim\limits_{x \to a} f(x) = f(a)$.

Continuity at $x = 0$

Here, $a = 0$.

1. Evaluate $f(0)$:

$f(0) = 5(0) - 3 = 0 - 3 = -3$.

$f(0)$ is defined.

2. Evaluate $\lim\limits_{x \to 0} f(x)$:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (5x - 3)$.

Since $f(x) = 5x - 3$ is a polynomial function, the limit as $x$ approaches any value can be found by direct substitution.

$\lim\limits_{x \to 0} (5x - 3) = 5(0) - 3 = 0 - 3 = -3$.

The limit exists and is equal to $-3$.

3. Compare $\lim\limits_{x \to 0} f(x)$ and $f(0)$:

We have $\lim\limits_{x \to 0} f(x) = -3$ and $f(0) = -3$.

Thus, $\lim\limits_{x \to 0} f(x) = f(0)$.

Therefore, $f(x)$ is continuous at $x = 0$.

Continuity at $x = -3$

Here, $a = -3$.

1. Evaluate $f(-3)$:

$f(-3) = 5(-3) - 3 = -15 - 3 = -18$.

$f(-3)$ is defined.

2. Evaluate $\lim\limits_{x \to -3} f(x)$:

$\lim\limits_{x \to -3} f(x) = \lim\limits_{x \to -3} (5x - 3)$.

By direct substitution for the polynomial function:

$\lim\limits_{x \to -3} (5x - 3) = 5(-3) - 3 = -15 - 3 = -18$.

The limit exists and is equal to $-18$.

3. Compare $\lim\limits_{x \to -3} f(x)$ and $f(-3)$:

We have $\lim\limits_{x \to -3} f(x) = -18$ and $f(-3) = -18$.

Thus, $\lim\limits_{x \to -3} f(x) = f(-3)$.

Therefore, $f(x)$ is continuous at $x = -3$.

Continuity at $x = 5$

Here, $a = 5$.

1. Evaluate $f(5)$:

$f(5) = 5(5) - 3 = 25 - 3 = 22$.

$f(5)$ is defined.

2. Evaluate $\lim\limits_{x \to 5} f(x)$:

$\lim\limits_{x \to 5} f(x) = \lim\limits_{x \to 5} (5x - 3)$.

By direct substitution for the polynomial function:

$\lim\limits_{x \to 5} (5x - 3) = 5(5) - 3 = 25 - 3 = 22$.

The limit exists and is equal to $22$.

3. Compare $\lim\limits_{x \to 5} f(x)$ and $f(5)$:

We have $\lim\limits_{x \to 5} f(x) = 22$ and $f(5) = 22$.

Thus, $\lim\limits_{x \to 5} f(x) = f(5)$.

Therefore, $f(x)$ is continuous at $x = 5$.

Since the function $f(x) = 5x - 3$ satisfies the condition $\lim\limits_{x \to a} f(x) = f(a)$ for $a=0$, $a=-3$, and $a=5$, the function is proven to be continuous at these points.

To examine the continuity of the function $f(x)$ at a point $x=a$, we need to check if $\lim\limits_{x \to a} f(x) = f(a)$.

For the function $f(x) = 2x^2 - 1$ and the point $x = 3$, we need to verify if $\lim\limits_{x \to 3} f(x) = f(3)$.

Step 1: Evaluate $f(3)$

Substitute $x = 3$ into the function $f(x) = 2x^2 - 1$:

$f(3) = 2(3)^2 - 1$

$f(3) = 2(9) - 1$

$f(3) = 18 - 1$

$f(3) = 17$

$f(3)$ is defined and its value is $17$.

Step 2: Evaluate $\lim\limits_{x \to 3} f(x)$

We need to find the limit of $f(x) = 2x^2 - 1$ as $x$ approaches $3$:

$\lim\limits_{x \to 3} f(x) = \lim\limits_{x \to 3} (2x^2 - 1)$

Since $f(x) = 2x^2 - 1$ is a polynomial function, its limit as $x$ approaches any real number can be found by direct substitution.

$\lim\limits_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1$

$\lim\limits_{x \to 3} (2x^2 - 1) = 2(9) - 1$

$\lim\limits_{x \to 3} (2x^2 - 1) = 18 - 1$

$\lim\limits_{x \to 3} (2x^2 - 1) = 17$

The limit $\lim\limits_{x \to 3} f(x)$ exists and its value is $17$.

Step 3: Compare $f(3)$ and $\lim\limits_{x \to 3} f(x)$

From Step 1, we found $f(3) = 17$.

From Step 2, we found $\lim\limits_{x \to 3} f(x) = 17$.

Since $\lim\limits_{x \to 3} f(x) = f(3) = 17$, the condition for continuity at $x = 3$ is satisfied.

Conclusion

Therefore, the function $f(x) = 2x^2 - 1$ is continuous at $x = 3$.

To examine the continuity of a function, we typically look at its domain and check for points where the conditions for continuity might fail. The conditions for continuity at a point $x=a$ are:

1. $f(a)$ is defined.

2. $\lim\limits_{x \to a} f(x)$ exists.

3. $\lim\limits_{x \to a} f(x) = f(a)$.

We will examine each function individually.

(a) $f(x) = x - 5$

The function $f(x) = x - 5$ is a polynomial function.

The domain of this function is all real numbers, i.e., $(-\infty, \infty)$.

Polynomial functions are continuous at every point in their domain.

For any real number $a$, $f(a) = a - 5$, which is defined.

The limit as $x$ approaches $a$ is $\lim\limits_{x \to a} (x - 5) = a - 5$, which exists.

Since $\lim\limits_{x \to a} f(x) = f(a)$ for all $a \in \mathbb{R}$, the function is continuous everywhere.

Therefore, the function $f(x) = x - 5$ is continuous on $(-\infty, \infty)$.

(b) $f(x) = \frac{1}{x-5}$, $x \neq 5$

The function $f(x) = \frac{1}{x-5}$ is a rational function.

A rational function is defined as the ratio of two polynomial functions, $f(x) = \frac{P(x)}{Q(x)}$. This function is continuous at all points where the denominator $Q(x)$ is not equal to zero.

In this case, $P(x) = 1$ and $Q(x) = x - 5$. The denominator is zero when $x - 5 = 0$, which means $x = 5$.

The domain of the function is given as $x \neq 5$, which is $(-\infty, 5) \cup (5, \infty)$.

For any point $a$ in the domain (i.e., $a \neq 5$), $f(a) = \frac{1}{a-5}$ is defined.

The limit as $x$ approaches $a$ ($a \neq 5$) is $\lim\limits_{x \to a} \frac{1}{x-5} = \frac{1}{a-5}$, which exists.

Since $\lim\limits_{x \to a} f(x) = f(a)$ for all $a \neq 5$, the function is continuous on its domain.

The function is not continuous at $x = 5$ because the function is not defined at this point.

Therefore, the function $f(x) = \frac{1}{x-5}$ is continuous on $(-\infty, 5) \cup (5, \infty)$.

(c) $f(x) = \frac{x^2 - 25}{x + 5}$, $x \neq 5$

The function is given as $f(x) = \frac{x^2 - 25}{x + 5}$ with the restriction $x \neq 5$. Note that the denominator is zero when $x+5=0$, which means $x=-5$. The function is undefined at $x=-5$. The domain is given as $x \neq 5$, but the function definition itself implies $x \neq -5$ as well because of the denominator.

For $x \neq -5$, we can simplify the expression:

$f(x) = \frac{(x-5)(x+5)}{x+5}$

For $x \neq -5$, we can cancel the $(x+5)$ term:

$f(x) = x - 5$, for $x \neq -5$.

The function is effectively $g(x) = x-5$ for all $x$ except possibly at $x = -5$ and $x=5$ based on the given domain restriction.

Let's consider the function as defined:

Domain is $x \neq 5$ and implicitly $x \neq -5$. So the domain is $(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$.

For any point $a$ in the domain ($a \neq -5$ and $a \neq 5$), $f(a) = \frac{a^2 - 25}{a + 5} = \frac{(a-5)(a+5)}{a+5} = a-5$. $f(a)$ is defined.

The limit as $x$ approaches $a$ ($a \neq -5$) is $\lim\limits_{x \to a} \frac{x^2 - 25}{x + 5} = \lim\limits_{x \to a} (x-5) = a-5$. This limit exists for all $a \neq -5$.

For any $a$ in the domain ($a \neq -5$ and $a \neq 5$), $\lim\limits_{x \to a} f(x) = a-5$ and $f(a) = a-5$. So, $\lim\limits_{x \to a} f(x) = f(a)$. The function is continuous on its domain.

Let's examine the points excluded from the domain:

At $x = -5$: The function is not defined because the denominator $x+5$ is zero. Thus, the function is discontinuous at $x = -5$ (specifically, it's not defined there).

At $x = 5$: The problem statement gives the restriction $x \neq 5$. The function is not defined at $x=5$. Thus, the function is discontinuous at $x = 5$ (it's not in the domain). Even though the simplified expression $x-5$ is defined at $x=5$, the original function is not.

Therefore, the function $f(x) = \frac{x^2 - 25}{x + 5}$ is continuous on $(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$.

(d) $f(x) = |x - 5|$

The function $f(x) = |x - 5|$ is an absolute value function.

The absolute value function $|u|$ is continuous for all real numbers $u$.

The expression inside the absolute value is $g(x) = x - 5$, which is a polynomial function and thus continuous everywhere.

The function $f(x) = |g(x)| = |x-5|$ is the composition of the absolute value function and a polynomial function. The composition of continuous functions is continuous.

Alternatively, we can write the function as a piecewise function:

$f(x) = \begin{cases} (x - 5) & , & \text{if } x - 5 \geq 0 \Rightarrow x \geq 5 \\ -(x - 5) & , & \text{if } x - 5 < 0 \Rightarrow x < 5 \end{cases}$

$f(x) = \begin{cases} x - 5 & , & \text{if } x \geq 5 \\ 5 - x & , & \text{if } x < 5 \end{cases}$

For $x > 5$, $f(x) = x - 5$, which is a polynomial and thus continuous.

For $x < 5$, $f(x) = 5 - x$, which is a polynomial and thus continuous.

We need to check for continuity at the point where the definition changes, i.e., at $x = 5$.

1. Evaluate $f(5)$:

$f(5) = 5 - 5 = 0$ (using the definition for $x \geq 5$). $f(5)$ is defined.

2. Evaluate $\lim\limits_{x \to 5} f(x)$: We need to check the left-hand limit and the right-hand limit.

Left-hand limit: $\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (5 - x) = 5 - 5 = 0$.

Right-hand limit: $\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (x - 5) = 5 - 5 = 0$.

Since the left-hand limit equals the right-hand limit, the limit exists: $\lim\limits_{x \to 5} f(x) = 0$.

3. Compare $\lim\limits_{x \to 5} f(x)$ and $f(5)$:

We have $\lim\limits_{x \to 5} f(x) = 0$ and $f(5) = 0$.

Thus, $\lim\limits_{x \to 5} f(x) = f(5)$. The function is continuous at $x = 5$.

Since the function is continuous for $x < 5$, $x > 5$, and at $x = 5$, it is continuous for all real numbers.

Therefore, the function $f(x) = |x - 5|$ is continuous on $(-\infty, \infty)$.

To prove that the function $f(x) = x^n$ is continuous at $x = n$, where $n$ is a positive integer, we need to show that the following three conditions for continuity at a point are satisfied for $a=n$:

1. $f(n)$ is defined.

2. $\lim\limits_{x \to n} f(x)$ exists.

3. $\lim\limits_{x \to n} f(x) = f(n)$.

Let the function be $f(x) = x^n$, and the point be $x = n$, where $n$ is a positive integer.

Step 1: Check if $f(n)$ is defined

We evaluate the function at $x = n$:

$f(n) = n^n$

Since $n$ is a positive integer, $n^n$ is a well-defined real number. Thus, $f(n)$ is defined.

Step 2: Check if $\lim\limits_{x \to n} f(x)$ exists

We need to evaluate the limit of $f(x)$ as $x$ approaches $n$:

$\lim\limits_{x \to n} f(x) = \lim\limits_{x \to n} x^n$

The function $f(x) = x^n$, where $n$ is a positive integer, is a polynomial function. For any polynomial function $P(x)$, the limit as $x$ approaches any real number $a$ is equal to $P(a)$.

Using this property, we have:

$\lim\limits_{x \to n} x^n = n^n$

The limit exists and is equal to $n^n$.

Step 3: Compare $\lim\limits_{x \to n} f(x)$ and $f(n)$

From Step 1, we found that $f(n) = n^n$.

From Step 2, we found that $\lim\limits_{x \to n} f(x) = n^n$.

Comparing these two values, we see that $\lim\limits_{x \to n} f(x) = f(n)$.

Conclusion

Since all three conditions for continuity at $x=n$ are satisfied, the function $f(x) = x^n$ is proven to be continuous at $x = n$, where $n$ is a positive integer.

To examine the continuity of the function $f(x)$ at a point $x=a$, we need to check if the following conditions are met:

1. $f(a)$ is defined.

2. $\lim\limits_{x \to a} f(x)$ exists (i.e., $\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x)$).

3. $\lim\limits_{x \to a} f(x) = f(a)$.

We will examine the continuity at the three specified points: $x = 0$, $x = 1$, and $x = 2$.

Continuity at $x = 0$

For $x = 0$, the function is defined by the first case, $f(x) = x$, since $0 \leq 1$.

1. Evaluate $f(0)$:

$f(0) = 0$

$f(0)$ is defined.

2. Evaluate $\lim\limits_{x \to 0} f(x)$:

Since $x = 0$ is in the interval $x \leq 1$, the function is $f(x) = x$ near $x=0$.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} x = 0$

The limit exists and is $0$.

3. Compare $\lim\limits_{x \to 0} f(x)$ and $f(0)$:

$\lim\limits_{x \to 0} f(x) = 0$ and $f(0) = 0$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function is continuous at $x = 0$.

Conclusion: The function is continuous at $x = 0$.

Continuity at $x = 1$

For $x = 1$, this is the point where the function definition changes. We need to check the function value and the left and right-hand limits.

1. Evaluate $f(1)$:

According to the definition, if $x \leq 1$, $f(x) = x$. So for $x=1$, we use the first case.

$f(1) = 1$

$f(1)$ is defined.

2. Evaluate $\lim\limits_{x \to 1} f(x)$:

We need to find the left-hand limit and the right-hand limit at $x=1$.

Left-hand limit ($\lim\limits_{x \to 1^-} f(x)$): For $x < 1$, $f(x) = x$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} x = 1$

Right-hand limit ($\lim\limits_{x \to 1^+} f(x)$): For $x > 1$, $f(x) = 5$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 5 = 5$

Since the left-hand limit ($1$) and the right-hand limit ($5$) are not equal, $\lim\limits_{x \to 1} f(x)$ does not exist.

Conclusion: Since the limit does not exist at $x=1$, the function is not continuous at $x = 1$.

Continuity at $x = 2$

For $x = 2$, the function is defined by the second case, $f(x) = 5$, since $2 > 1$.

1. Evaluate $f(2)$:

$f(2) = 5$

$f(2)$ is defined.

2. Evaluate $\lim\limits_{x \to 2} f(x)$:

Since $x = 2$ is in the interval $x > 1$, the function is $f(x) = 5$ near $x=2$.

$\lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} 5 = 5$

The limit exists and is $5$.

3. Compare $\lim\limits_{x \to 2} f(x)$ and $f(2)$:

$\lim\limits_{x \to 2} f(x) = 5$ and $f(2) = 5$.

Since $\lim\limits_{x \to 2} f(x) = f(2)$, the function is continuous at $x = 2$.

Conclusion: The function is continuous at $x = 2$.

Summary:

The function $f(x)$ is continuous at $x = 0$ and at $x = 2$.

The function $f(x)$ is discontinuous at $x = 1$ because the limit of the function does not exist at this point.

To find the points of discontinuity of the function $f(x)$, we need to examine the behavior of the function at different points in its domain. The given function is a piecewise function.

$f(x) = \begin{cases} 2x+3,& if\; x ≤ 2 \\ 2x−3,& if\; x > 2 \end{cases}$

Case 1: For $x < 2$

For any point $c$ such that $c < 2$, $f(x) = 2x+3$ in the neighborhood of $c$.

This is a polynomial function, which is continuous everywhere.

So, $f(x)$ is continuous for all $x < 2$.

Case 2: For $x > 2$

For any point $c$ such that $c > 2$, $f(x) = 2x-3$ in the neighborhood of $c$.

This is a polynomial function, which is continuous everywhere.

So, $f(x)$ is continuous for all $x > 2$.

Case 3: At $x = 2$

This is the point where the definition of the function changes. We need to check the three conditions for continuity at $x=2$:

1. $f(2)$ must be defined.

According to the first part of the definition ($x \leq 2$), $f(2) = 2(2) + 3 = 4 + 3 = 7$.

$f(2)$ is defined and $f(2) = 7$.

2. $\lim\limits_{x \to 2} f(x)$ must exist.

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit: $\lim\limits_{x \to 2^-} f(x)$

For $x < 2$, $f(x) = 2x+3$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (2x+3) = 2(2) + 3 = 4 + 3 = 7$.

Right-hand limit: $\lim\limits_{x \to 2^+} f(x)$

For $x > 2$, $f(x) = 2x-3$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (2x-3) = 2(2) - 3 = 4 - 3 = 1$.

Since $\lim\limits_{x \to 2^-} f(x) = 7$ and $\lim\limits_{x \to 2^+} f(x) = 1$, the left-hand limit and the right-hand limit are not equal ($7 \neq 1$).

Therefore, $\lim\limits_{x \to 2} f(x)$ does not exist.

3. $\lim\limits_{x \to 2} f(x)$ must be equal to $f(2)$.

Since $\lim\limits_{x \to 2} f(x)$ does not exist, this condition is not met.

Thus, the function is discontinuous at $x = 2$.

From the analysis of the three cases, we find that the function is continuous for all $x < 2$ and for all $x > 2$. The function is discontinuous only at $x = 2$.

The only point of discontinuity is $x = 2$.

To find the points of discontinuity of the function $f(x)$, we need to examine the continuity of the function over the intervals and at the points where the definition of the function changes.

The function is defined as:

$f(x) = \begin{cases} |x|+3,& if\; x ≤ −3\\−2x,& if\; −3 < x < 3\\6x + 2,& if\; x ≥ 3 \end{cases}$

The potential points of discontinuity are the boundary points of the intervals, $x = -3$ and $x = 3$. We also need to check the continuity within each open interval.

Case 1: For $x < -3$

For any $x$ in this interval, $x$ is negative. Thus, $|x| = -x$.

So, for $x < -3$, $f(x) = -x + 3$.

This is a polynomial function, and polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $x < -3$.

Case 2: For $-3 < x < 3$

For any $x$ in this interval, $f(x) = -2x$.

This is a polynomial function, and polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $-3 < x < 3$.

Case 3: For $x > 3$

For any $x$ in this interval, $f(x) = 6x + 2$.

This is a polynomial function, and polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $x > 3$.

Case 4: At $x = -3$

We check the conditions for continuity at $x = -3$:

1. $f(-3)$:

Using the first case ($x \leq -3$), $f(-3) = |-3| + 3 = 3 + 3 = 6$. $f(-3)$ is defined.

2. $\lim\limits_{x \to -3} f(x)$:

Left-hand limit ($\lim\limits_{x \to -3^-} f(x)$): For $x < -3$, $f(x) = |x| + 3 = -x + 3$.

$\lim\limits_{x \to -3^-} (-x + 3) = -(-3) + 3 = 3 + 3 = 6$.

Right-hand limit ($\lim\limits_{x \to -3^+} f(x)$): For $x > -3$ and $x < 3$, $f(x) = -2x$.

$\lim\limits_{x \to -3^+} (-2x) = -2(-3) = 6$.

Since $\lim\limits_{x \to -3^-} f(x) = \lim\limits_{x \to -3^+} f(x) = 6$, the limit $\lim\limits_{x \to -3} f(x)$ exists and is $6$.

3. Compare $\lim\limits_{x \to -3} f(x)$ and $f(-3)$:

We have $\lim\limits_{x \to -3} f(x) = 6$ and $f(-3) = 6$.

Since $\lim\limits_{x \to -3} f(x) = f(-3)$, the function is continuous at $x = -3$.

Case 5: At $x = 3$

We check the conditions for continuity at $x = 3$:

1. $f(3)$:

Using the third case ($x \geq 3$), $f(3) = 6(3) + 2 = 18 + 2 = 20$. $f(3)$ is defined.

2. $\lim\limits_{x \to 3} f(x)$:

Left-hand limit ($\lim\limits_{x \to 3^-} f(x)$): For $x < 3$ and $x > -3$, $f(x) = -2x$.

$\lim\limits_{x \to 3^-} (-2x) = -2(3) = -6$.

Right-hand limit ($\lim\limits_{x \to 3^+} f(x)$): For $x > 3$, $f(x) = 6x + 2$.

$\lim\limits_{x \to 3^+} (6x + 2) = 6(3) + 2 = 18 + 2 = 20$.

Since $\lim\limits_{x \to 3^-} f(x) = -6$ and $\lim\limits_{x \to 3^+} f(x) = 20$, the left-hand limit and the right-hand limit are not equal ($-6 \neq 20$).

Therefore, $\lim\limits_{x \to 3} f(x)$ does not exist.

3. Compare $\lim\limits_{x \to 3} f(x)$ and $f(3)$:

Since $\lim\limits_{x \to 3} f(x)$ does not exist, the function is not continuous at $x = 3$.

From the analysis, the function is continuous in the intervals $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$. It is also continuous at $x = -3$. The only point where the function is discontinuous is $x = 3$.

The only point of discontinuity of $f$ is $x = 3$.

To find the points of discontinuity of the function $f(x)$, we need to examine its continuity across its domain. The function is defined as:

$f(x) = \begin{cases} \frac{|x|}{x},& if\; x ≠ 0\\0,& if\; x = 0 \end{cases}$

Let's rewrite the function $\frac{|x|}{x}$ for $x \neq 0$.

If $x > 0$, then $|x| = x$, so $\frac{|x|}{x} = \frac{x}{x} = 1$.

If $x < 0$, then $|x| = -x$, so $\frac{|x|}{x} = \frac{-x}{x} = -1$.

Thus, the function can be written in a piecewise form without the absolute value:

$f(x) = \begin{cases} -1,& if\; x < 0 \\ 0,& if\; x = 0 \\ 1,& if\; x > 0 \end{cases}$

We will examine the continuity of the function in the intervals and at the point where the definition changes, which is $x = 0$.

Case 1: For $x < 0$

For any point $c$ such that $c < 0$, the function is $f(x) = -1$ in the neighborhood of $c$.

This is a constant function, which is a polynomial of degree zero. Polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $x < 0$.

Case 2: For $x > 0$

For any point $c$ such that $c > 0$, the function is $f(x) = 1$ in the neighborhood of $c$.

This is a constant function, which is a polynomial. Polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $x > 0$.

Case 3: At $x = 0$

This is the point where the definition of the function changes. We need to check the three conditions for continuity at $x=0$:

1. $f(0)$ must be defined.

From the definition of the function, $f(0) = 0$. $f(0)$ is defined.

2. $\lim\limits_{x \to 0} f(x)$ must exist.

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit ($\lim\limits_{x \to 0^-} f(x)$): For $x < 0$, $f(x) = -1$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (-1) = -1$.

Right-hand limit ($\lim\limits_{x \to 0^+} f(x)$): For $x > 0$, $f(x) = 1$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (1) = 1$.

Since the left-hand limit ($-1$) and the right-hand limit ($1$) are not equal, the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

3. $\lim\limits_{x \to 0} f(x)$ must be equal to $f(0)$.

Since $\lim\limits_{x \to 0} f(x)$ does not exist, this condition cannot be met.

Thus, the function is discontinuous at $x = 0$.

From the analysis, we find that the function is continuous for all $x < 0$ and for all $x > 0$. The function is discontinuous only at $x = 0$.

The only point of discontinuity of $f$ is $x = 0$.

To find the points of discontinuity of the function $f(x)$, we examine its definition:

$f(x) = \begin{cases} \frac{x}{|x|},& if\; x < 0 \\ −1,& if\; x ≥ 0 \end{cases}$

Let's simplify the expression $\frac{x}{|x|}$ for the case when $x < 0$.

For $x < 0$, the absolute value of $x$ is $|x| = -x$.

So, for $x < 0$, $\frac{x}{|x|} = \frac{x}{-x}$.

We can cancel $x$ (since $x \neq 0$ in this case):

$\frac{\cancel{x}}{\cancel{-x}} = \frac{1}{-1} = -1$.

Now, we can rewrite the function $f(x)$ using this simplified form:

If $x < 0$, then $f(x) = -1$.

If $x \geq 0$, then $f(x) = -1$ (from the second case in the original definition).

Combining these two cases, we see that for all real numbers $x$, the function is $f(x) = -1$.

$f(x) = -1$ for all $x \in \mathbb{R}$.

The function $f(x) = -1$ is a constant function.

Constant functions are a type of polynomial function (specifically, a polynomial of degree 0).

Polynomial functions are known to be continuous at every point in their domain.

The domain of $f(x) = -1$ is $(-\infty, \infty)$, which includes all real numbers.

Since the function $f(x)$ simplifies to a constant function $f(x) = -1$ for all real numbers, and constant functions are continuous everywhere, the function $f(x)$ has no points of discontinuity.

The function is continuous on $(-\infty, \infty)$.

Therefore, there are no points of discontinuity for the function $f(x)$.

To find the points of discontinuity of the function $f(x)$, we need to examine its continuity across its domain. The function is defined as:

$f(x) = \begin{cases} x+1,& if\; x ≥1 \\ x^2+1,& if \;x < 1 \end{cases}$

The potential point of discontinuity is the boundary point where the definition changes, which is $x = 1$. We also need to check the continuity within each open interval defined by the piecewise definition.

Case 1: For $x < 1$

For any point $c$ such that $c < 1$, the function is $f(x) = x^2+1$ in the neighborhood of $c$.

This is a polynomial function, and polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

For any point $c$ such that $c > 1$, the function is $f(x) = x+1$ in the neighborhood of $c$.

This is a polynomial function, and polynomial functions are continuous everywhere.

Therefore, $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the point where the definition of the function changes. We need to check the three conditions for continuity at $x=1$:

1. $f(1)$ must be defined.

According to the first part of the definition ($x \geq 1$), $f(1) = 1 + 1 = 2$. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ must exist.

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit ($\lim\limits_{x \to 1^-} f(x)$): For $x < 1$, $f(x) = x^2+1$.

$\lim\limits_{x \to 1^-} (x^2+1) = (1)^2 + 1 = 1 + 1 = 2$.

Right-hand limit ($\lim\limits_{x \to 1^+} f(x)$): For $x > 1$, $f(x) = x+1$.

$\lim\limits_{x \to 1^+} (x+1) = 1 + 1 = 2$.

Since $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = 2$, the limit $\lim\limits_{x \to 1} f(x)$ exists and is $2$.

3. Compare $\lim\limits_{x \to 1} f(x)$ and $f(1)$:

We have $\lim\limits_{x \to 1} f(x) = 2$ and $f(1) = 2$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the function is continuous at $x = 1$.

From the analysis of the three cases, we find that the function is continuous for all $x < 1$, for all $x > 1$, and also at $x = 1$. This means the function is continuous for all real numbers.

Therefore, there are no points of discontinuity for the function $f(x)$.

To find the points of discontinuity of the function $f(x)$, we need to examine its continuity across its domain. The function is defined as:

$f(x) = \begin{cases} x^3−3,& if\; x ≤ 2 \\ x^2+1,& if\; x > 2 \end{cases}$

The potential point of discontinuity is the boundary point where the definition changes, which is $x = 2$. We also need to check the continuity within each open interval defined by the piecewise definition.

Case 1: For $x < 2$

For any point $c$ such that $c < 2$, the function is $f(x) = x^3-3$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x < 2$.

Case 2: For $x > 2$

For any point $c$ such that $c > 2$, the function is $f(x) = x^2+1$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x > 2$.

Case 3: At $x = 2$

This is the point where the definition of the function changes. We need to check the three conditions for continuity at $x=2$:

1. $f(2)$ must be defined.

According to the first part of the definition ($x \leq 2$), $f(2) = (2)^3 - 3 = 8 - 3 = 5$. $f(2)$ is defined.

2. $\lim\limits_{x \to 2} f(x)$ must exist.

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit ($\lim\limits_{x \to 2^-} f(x)$): For $x < 2$, $f(x) = x^3-3$.

$\lim\limits_{x \to 2^-} (x^3-3) = (2)^3 - 3 = 8 - 3 = 5$.

Right-hand limit ($\lim\limits_{x \to 2^+} f(x)$): For $x > 2$, $f(x) = x^2+1$.

$\lim\limits_{x \to 2^+} (x^2+1) = (2)^2 + 1 = 4 + 1 = 5$.

Since the left-hand limit ($5$) and the right-hand limit ($5$) are equal, the limit $\lim\limits_{x \to 2} f(x)$ exists and is $5$.

3. Compare $\lim\limits_{x \to 2} f(x)$ and $f(2)$:

We have $\lim\limits_{x \to 2} f(x) = 5$ and $f(2) = 5$.

Since $\lim\limits_{x \to 2} f(x) = f(2)$, the function is continuous at $x = 2$.

From the analysis of the three cases, we find that the function is continuous for all $x < 2$, for all $x > 2$, and also at $x = 2$. This means the function is continuous for all real numbers.

Therefore, there are no points of discontinuity for the function $f(x)$.

To find the points of discontinuity of the function $f(x)$, we need to examine its continuity across its domain. The function is defined as:

$f(x) = \begin{cases} x^{10}−1,& if\; x ≤ 1 \\ x^2,& if\; x > 1 \end{cases}$

The potential point of discontinuity is the boundary point where the definition changes, which is $x = 1$. We also need to check the continuity within each open interval defined by the piecewise definition.

Case 1: For $x < 1$

For any point $c$ such that $c < 1$, the function is $f(x) = x^{10}-1$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

For any point $c$ such that $c > 1$, the function is $f(x) = x^2$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the point where the definition of the function changes. We need to check the three conditions for continuity at $x=1$:

1. $f(1)$ must be defined.

According to the first part of the definition ($x \leq 1$), $f(1) = (1)^{10} - 1 = 1 - 1 = 0$. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ must exist.

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit ($\lim\limits_{x \to 1^-} f(x)$): For $x < 1$, $f(x) = x^{10}-1$.

$\lim\limits_{x \to 1^-} (x^{10}-1) = (1)^{10} - 1 = 1 - 1 = 0$.

Right-hand limit ($\lim\limits_{x \to 1^+} f(x)$): For $x > 1$, $f(x) = x^2$.

$\lim\limits_{x \to 1^+} (x^2) = (1)^2 = 1$.

Since the left-hand limit ($0$) and the right-hand limit ($1$) are not equal, the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

3. Compare $\lim\limits_{x \to 1} f(x)$ and $f(1)$:

Since $\lim\limits_{x \to 1} f(x)$ does not exist, this condition cannot be met.

Thus, the function is discontinuous at $x = 1$.

From the analysis of the three cases, we find that the function is continuous for all $x < 1$ and for all $x > 1$. The function is discontinuous only at $x = 1$.

The only point of discontinuity of $f$ is $x = 1$.

To determine if the function $f(x)$ is continuous, we need to examine its continuity over its entire domain, which is all real numbers. The function is defined as:

$f(x) = \begin{cases} x+5,& if\; x ≤ 1 \\ x−5,& if\; x > 1 \end{cases}$

We will examine the continuity of the function in the intervals and at the point where the definition changes, which is $x = 1$.

Case 1: For $x < 1$

For any point $c$ such that $c < 1$, the function is $f(x) = x+5$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

For any point $c$ such that $c > 1$, the function is $f(x) = x-5$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the point where the definition of the function changes. We need to check the three conditions for continuity at $x=1$:

1. $f(1)$ must be defined.

According to the first part of the definition ($x \leq 1$), $f(1) = 1 + 5 = 6$. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ must exist.

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit ($\lim\limits_{x \to 1^-} f(x)$): For $x < 1$, $f(x) = x+5$.

$\lim\limits_{x \to 1^-} (x+5) = 1 + 5 = 6$.

Right-hand limit ($\lim\limits_{x \to 1^+} f(x)$): For $x > 1$, $f(x) = x-5$.

$\lim\limits_{x \to 1^+} (x-5) = 1 - 5 = -4$.

Since the left-hand limit ($6$) and the right-hand limit ($-4$) are not equal, the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

3. $\lim\limits_{x \to 1} f(x)$ must be equal to $f(1)$.

Since $\lim\limits_{x \to 1} f(x)$ does not exist, this condition cannot be met.

Thus, the function is discontinuous at $x = 1$.

From the analysis, the function is continuous for all $x < 1$ and for all $x > 1$. However, it is discontinuous at $x = 1$.

For a function to be called a continuous function, it must be continuous at every point in its domain. In this case, the function is not continuous at $x=1$ (which is in its domain).

Therefore, the function $f(x)$ is not a continuous function.

Given Function:

The function $f(x)$ is defined piecewise as follows:

$f(x) = \begin{cases} 3, & \text{if}\; 0 \leq x \leq 1 \\ 4, & \text{if}\; 1 < x < 3 \\ 5, & \text{if}\; 3 \leq x \leq 10 \end{cases}$

The function $f(x)$ is defined on the closed interval $[0, 10]$. We need to examine the continuity of the function in the open intervals and at the transition points and endpoints.

Continuity in the open intervals:

For $0 < x < 1$, $f(x) = 3$. This is a constant function, which is continuous for all $x \in (0, 1)$.

For $1 < x < 3$, $f(x) = 4$. This is a constant function, which is continuous for all $x \in (1, 3)$.

For $3 < x < 10$, $f(x) = 5$. This is a constant function, which is continuous for all $x \in (3, 10)$.

Continuity at the transition point $x=1$:

Value of the function at $x=1$:

$f(1) = 3$ (from the first case, since $0 \leq 1 \leq 1$)

Left-hand limit at $x=1$:

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} 3 = 3$ (since for $x < 1$, $f(x)=3$)

Right-hand limit at $x=1$:

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 4 = 4$ (since for $x > 1$ and $x < 3$, $f(x)=4$)

Since the left-hand limit ($3$) is not equal to the right-hand limit ($4$) at $x=1$, the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

Therefore, the function $f(x)$ is discontinuous at $x=1$.

Continuity at the transition point $x=3$:

Value of the function at $x=3$:

$f(3) = 5$ (from the third case, since $3 \leq 3 \leq 10$)

Left-hand limit at $x=3$:

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} 4 = 4$ (since for $x < 3$ and $x > 1$, $f(x)=4$)

Right-hand limit at $x=3$:

$\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} 5 = 5$ (since for $x > 3$ and $x \leq 10$, $f(x)=5$)

Since the left-hand limit ($4$) is not equal to the right-hand limit ($5$) at $x=3$, the limit $\lim\limits_{x \to 3} f(x)$ does not exist.

Therefore, the function $f(x)$ is discontinuous at $x=3$.

Continuity at the endpoints of the domain $[0, 10]$:

At $x=0$:

Value of the function: $f(0) = 3$ (from the first case, since $0 \leq 0 \leq 1$).

Right-hand limit: $\lim\limits_{x \to 0^+} f(x)$. For $x$ slightly greater than $0$ (and $x \leq 1$), $f(x)=3$. So, $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 3 = 3$.

Since $f(0) = \lim\limits_{x \to 0^+} f(x)$, the function is continuous from the right at $x=0$.

At $x=10$:

Value of the function: $f(10) = 5$ (from the third case, since $3 \leq 10 \leq 10$).

Left-hand limit: $\lim\limits_{x \to 10^-} f(x)$. For $x$ slightly less than $10$ (and $x \geq 3$), $f(x)=5$. So, $\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^-} 5 = 5$.

Since $f(10) = \lim\limits_{x \to 10^-} f(x)$, the function is continuous from the left at $x=10$.

Conclusion:

The function $f(x)$ is continuous on the open intervals $(0, 1)$, $(1, 3)$, and $(3, 10)$.

The function is discontinuous at $x=1$ and $x=3$.

The function is continuous from the right at $x=0$ and continuous from the left at $x=10$.

Therefore, $f(x)$ is continuous on $[0, 1) \cup (1, 3) \cup (3, 10]$. The points of discontinuity are $x=1$ and $x=3$.

To find the points of discontinuity of the function $f(x)$, we need to examine its continuity across its domain. The function is defined as:

$f(x) = \begin{cases} 2x,& if\; x < 0 \\ 0,& if\; 0 ≤ x ≤ 1 \\ 4x,& if\; x > 1 \end{cases}$

The potential points of discontinuity are the boundary points of the intervals, $x = 0$ and $x = 1$. We also need to check the continuity within each open interval defined by the piecewise definition.

Case 1: For $x < 0$

For any point $c$ such that $c < 0$, the function is $f(x) = 2x$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x < 0$.

Case 2: For $0 < x < 1$

For any point $c$ such that $0 < c < 1$, the function is $f(x) = 0$ in the neighborhood of $c$.

This is a constant function. Constant functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $0 < x < 1$.

Case 3: For $x > 1$

For any point $c$ such that $c > 1$, the function is $f(x) = 4x$ in the neighborhood of $c$.

This is a polynomial function. Polynomial functions are continuous at every point in their domain.

Therefore, $f(x)$ is continuous for all $x > 1$.

Case 4: At $x = 0$

This is the first point where the definition of the function changes. We need to check the three conditions for continuity at $x=0$:

1. $f(0)$ must be defined.

According to the second part of the definition ($0 \leq x \leq 1$), for $x=0$, $f(0) = 0$. $f(0)$ is defined.

2. $\lim\limits_{x \to 0} f(x)$ must exist.

We need to find the left-hand limit and the right-hand limit at $x=0$.

Left-hand limit ($\lim\limits_{x \to 0^-} f(x)$): For $x < 0$, $f(x) = 2x$.

$\lim\limits_{x \to 0^-} (2x) = 2(0) = 0$.

Right-hand limit ($\lim\limits_{x \to 0^+} f(x)$): For $x > 0$ and $x \leq 1$, $f(x) = 0$.

$\lim\limits_{x \to 0^+} (0) = 0$.

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = 0$, the limit $\lim\limits_{x \to 0} f(x)$ exists and is $0$.

3. Compare $\lim\limits_{x \to 0} f(x)$ and $f(0)$:

We have $\lim\limits_{x \to 0} f(x) = 0$ and $f(0) = 0$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function is continuous at $x = 0$.

Case 5: At $x = 1$

This is the second point where the definition of the function changes. We need to check the three conditions for continuity at $x=1$:

1. $f(1)$ must be defined.

According to the second part of the definition ($0 \leq x \leq 1$), for $x=1$, $f(1) = 0$. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ must exist.

We need to find the left-hand limit and the right-hand limit at $x=1$.

Left-hand limit ($\lim\limits_{x \to 1^-} f(x)$): For $x < 1$ and $x \geq 0$, $f(x) = 0$.

$\lim\limits_{x \to 1^-} (0) = 0$.

Right-hand limit ($\lim\limits_{x \to 1^+} f(x)$): For $x > 1$, $f(x) = 4x$.

$\lim\limits_{x \to 1^+} (4x) = 4(1) = 4$.

Since the left-hand limit ($0$) and the right-hand limit ($4$) are not equal, the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

3. Compare $\lim\limits_{x \to 1} f(x)$ and $f(1)$:

Since $\lim\limits_{x \to 1} f(x)$ does not exist, this condition cannot be met.

Thus, the function is discontinuous at $x = 1$.

From the analysis of the cases, we find that the function is continuous for all $x < 0$, for all $0 < x < 1$, for all $x > 1$, and also at $x=0$. The function is discontinuous only at $x=1$.

The only point of discontinuity of $f$ is $x = 1$.

Given Function:

The function $f(x)$ is defined piecewise as follows:

$f(x) = \begin{cases} −2, & \text{if}\; x \leq −1 \\ 2x, & \text{if}\; −1 < x \leq 1 \\ 2, & \text{if}\; x > 1 \end{cases}$

We need to discuss the continuity of the function $f(x)$ by examining its continuity in the open intervals and at the transition points.

Continuity in the open intervals:

For $x < -1$, $f(x) = -2$. This is a constant function, which is continuous for all $x \in (-\infty, -1)$.

For $-1 < x < 1$, $f(x) = 2x$. This is a linear polynomial function, which is continuous for all $x \in (-1, 1)$.

For $x > 1$, $f(x) = 2$. This is a constant function, which is continuous for all $x \in (1, \infty)$.

Continuity at the transition point $x=-1$:

Value of the function at $x=-1$:

$f(-1) = -2$ (from the first case, since $x \leq -1$)

Left-hand limit at $x=-1$:

$\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^-} (-2) = -2$ (since for $x < -1$, $f(x)=-2$)

Right-hand limit at $x=-1$:

$\lim\limits_{x \to -1^+} f(x) = \lim\limits_{x \to -1^+} (2x)$ (since for $x > -1$ and $x \leq 1$, $f(x)=2x$)

Substitute $x=-1$: $\lim\limits_{x \to -1^+} (2x) = 2(-1) = -2$

Since $f(-1) = \lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^+} f(x) = -2$, the function $f(x)$ is continuous at $x=-1$.

Continuity at the transition point $x=1$:

Value of the function at $x=1$:

$f(1) = 2(1) = 2$ (from the second case, since $-1 < x \leq 1$)

Left-hand limit at $x=1$:

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (2x)$ (since for $x < 1$ and $x > -1$, $f(x)=2x$)

Substitute $x=1$: $\lim\limits_{x \to 1^-} (2x) = 2(1) = 2$

Right-hand limit at $x=1$:

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 2 = 2$ (since for $x > 1$, $f(x)=2$)

Since $f(1) = \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = 2$, the function $f(x)$ is continuous at $x=1$.

Conclusion:

The function $f(x)$ is continuous in the open intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

The function is also continuous at the transition points $x=-1$ and $x=1$.

Therefore, the function $f(x)$ is continuous for all real numbers. There are no points of discontinuity.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} ax+1, & \text{if}\; x \leq 3 \\ bx+3, & \text{if}\; x > 3 \end{cases}$

The function is continuous at $x=3$.

To Find:

The relationship between $a$ and $b$.

Solution:

For the function $f(x)$ to be continuous at $x=3$, the following condition must be satisfied:

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^+} f(x) = f(3)$

Let's evaluate each part:

1. Value of the function at $x=3$:

When $x \leq 3$, $f(x) = ax+1$.

$f(3) = a(3) + 1 = 3a + 1$

2. Left-hand limit at $x=3$:

$\lim\limits_{x \to 3^-} f(x)$

For $x < 3$, $f(x) = ax+1$.

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} (ax+1) = a(3) + 1 = 3a + 1$

3. Right-hand limit at $x=3$:

$\lim\limits_{x \to 3^+} f(x)$

For $x > 3$, $f(x) = bx+3$.

$\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} (bx+3) = b(3) + 3 = 3b + 3$

For continuity at $x=3$, we must have $\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^+} f(x)$.

Therefore, we set the left-hand limit equal to the right-hand limit:

$3a + 1 = 3b + 3$

Now, we find the relationship between $a$ and $b$ from this equation:

$3a - 3b = 3 - 1$

$3a - 3b = 2$

This can also be written as:

$3(a - b) = 2$

$a - b = \frac{2}{3}$

Or, solving for $a$: $a = b + \frac{2}{3}$

Or, solving for $b$: $b = a - \frac{2}{3}$

The relationship between $a$ and $b$ for the function to be continuous at $x=3$ is $\mathbf{3a - 3b = 2}$.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} \lambda(x^2-2x), & \text{if}\; x \leq 0 \\ 4x+1, & \text{if}\; x > 0 \end{cases}$

To Find:

1. The value of $\lambda$ for which $f(x)$ is continuous at $x=0$.

2. Discuss the continuity of $f(x)$ at $x=1$.

Solution:

Part 1: Continuity at $x=0$

For the function $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$

Let's evaluate each part:

1. Value of the function at $x=0$:

When $x \leq 0$, $f(x) = \lambda(x^2 - 2x)$.

$f(0) = \lambda(0^2 - 2(0)) = \lambda(0 - 0) = 0$

2. Left-hand limit at $x=0$:

$\lim\limits_{x \to 0^-} f(x)$

For $x < 0$, $f(x) = \lambda(x^2 - 2x)$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2(0)) = \lambda(0) = 0$

3. Right-hand limit at $x=0$:

$\lim\limits_{x \to 0^+} f(x)$

For $x > 0$, $f(x) = 4x+1$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (4x+1) = 4(0) + 1 = 0 + 1 = 1$

For continuity at $x=0$, we must have $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x)$.

From our calculations, we have:

$0 = 1$

This is a contradiction. The left-hand limit and the right-hand limit at $x=0$ are not equal, regardless of the value of $\lambda$.

Therefore, there is no value of $\lambda$ for which the function $f(x)$ is continuous at $x=0$.

Part 2: Continuity at $x=1$

We need to examine the continuity of $f(x)$ at $x=1$.

For $x > 0$, the function is defined as $f(x) = 4x+1$.

The point $x=1$ falls in the domain where $x > 0$.

So, for values of $x$ around $1$, $f(x) = 4x+1$.

The function $f(x) = 4x+1$ is a polynomial function (a linear function).

Polynomial functions are continuous for all real numbers.

Therefore, $f(x) = 4x+1$ is continuous at $x=1$.

To formally verify continuity at $x=1$:

Value of the function at $x=1$: $f(1) = 4(1) + 1 = 5$.

Limit at $x=1$: $\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (4x+1)$. Since $f(x)=4x+1$ for $x$ near $1$, the limit is $\lim\limits_{x \to 1} (4x+1) = 4(1) + 1 = 5$.

Since $f(1) = \lim\limits_{x \to 1} f(x) = 5$, the function is continuous at $x=1$.

Conclusion:

There is no value of $\lambda$ for which the function $f(x)$ is continuous at $x=0$.

The function $f(x)$ is continuous at $x=1$, regardless of the value of $\lambda$ (as long as $\lambda$ is a real number), because the definition of $f(x)$ for $x>0$ is a continuous polynomial.

Given:

The function is $g(x) = x - [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$.

To Show:

The function $g(x)$ is discontinuous at all integral points.

Proof:

Let $n$ be an arbitrary integral point. We need to examine the continuity of $g(x)$ at $x=n$.

For $g(x)$ to be continuous at $x=n$, the following condition must hold:

$\lim\limits_{x \to n^-} g(x) = \lim\limits_{x \to n^+} g(x) = g(n)$

Let's evaluate each part:

1. Value of the function at $x=n$:

$g(n) = n - [n]$

By the definition of the greatest integer function, for an integer $n$, $[n] = n$.

So, $g(n) = n - n = 0$.

2. Left-hand limit at $x=n$:

$\lim\limits_{x \to n^-} g(x) = \lim\limits_{x \to n^-} (x - [x])$

For $x$ approaching $n$ from the left side (i.e., $x < n$ but very close to $n$), the greatest integer less than or equal to $x$ is $n-1$.

So, $\lim\limits_{x \to n^-} [x] = n-1$.

Therefore, $\lim\limits_{x \to n^-} (x - [x]) = \lim\limits_{x \to n^-} x - \lim\limits_{x \to n^-} [x] = n - (n-1) = n - n + 1 = 1$.

3. Right-hand limit at $x=n$:

$\lim\limits_{x \to n^+} g(x) = \lim\limits_{x \to n^+} (x - [x])$

For $x$ approaching $n$ from the right side (i.e., $x > n$ but very close to $n$), the greatest integer less than or equal to $x$ is $n$.

So, $\lim\limits_{x \to n^+} [x] = n$.

Therefore, $\lim\limits_{x \to n^+} (x - [x]) = \lim\limits_{x \to n^+} x - \lim\limits_{x \to n^+} [x] = n - n = 0$.

Comparing the limits and the function value at $x=n$:

We have $\lim\limits_{x \to n^-} g(x) = 1$, $\lim\limits_{x \to n^+} g(x) = 0$, and $g(n) = 0$.

Since the left-hand limit (1) is not equal to the right-hand limit (0), the limit $\lim\limits_{x \to n} g(x)$ does not exist at $x=n$.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist, the function is discontinuous at $x=n$.

As $n$ was an arbitrary integer, this holds true for all integral points.

Conclusion:

Since $\lim\limits_{x \to n^-} g(x) \neq \lim\limits_{x \to n^+} g(x)$ for any integer $n$, the function $g(x) = x - [x]$ is discontinuous at all integral points.

Given:

The function is $f(x) = x^2 - \sin x + 5$.

We need to check its continuity at $x = \pi$.

To Determine:

Is $f(x)$ continuous at $x = \pi$?

Solution:

A function $f(x)$ is continuous at a point $x=c$ if and only if:

$\lim\limits_{x \to c} f(x) = f(c)$

In this case, $c = \pi$. So we need to check if $\lim\limits_{x \to \pi} f(x) = f(\pi)$.

Let's evaluate the function value at $x=\pi$:

$f(\pi) = (\pi)^2 - \sin(\pi) + 5$

We know that $\sin(\pi) = 0$.

So, $f(\pi) = \pi^2 - 0 + 5 = \pi^2 + 5$.

Now, let's evaluate the limit of the function as $x$ approaches $\pi$:

$\lim\limits_{x \to \pi} f(x) = \lim\limits_{x \to \pi} (x^2 - \sin x + 5)$

Using the properties of limits, the limit of a sum/difference is the sum/difference of the limits, and the limit of a polynomial is the function value. The sine function is also continuous everywhere.

$\lim\limits_{x \to \pi} (x^2 - \sin x + 5) = \lim\limits_{x \to \pi} x^2 - \lim\limits_{x \to \pi} \sin x + \lim\limits_{x \to \pi} 5$

$= (\pi)^2 - \sin(\pi) + 5$

$= \pi^2 - 0 + 5$

$= \pi^2 + 5$

Comparing the function value and the limit:

We found that $f(\pi) = \pi^2 + 5$ and $\lim\limits_{x \to \pi} f(x) = \pi^2 + 5$.

Since $\lim\limits_{x \to \pi} f(x) = f(\pi)$, the condition for continuity at $x=\pi$ is satisfied.

Alternatively, we can use the properties of continuous functions.

Let $g(x) = x^2$, $h(x) = \sin x$, and $k(x) = 5$.

The function $g(x) = x^2$ is a polynomial function, which is continuous everywhere.

The function $h(x) = \sin x$ is a trigonometric function, which is continuous everywhere.

The function $k(x) = 5$ is a constant function, which is continuous everywhere.

The given function $f(x) = g(x) - h(x) + k(x)$.

The difference and sum of continuous functions are continuous.

Since $g(x)$, $h(x)$, and $k(x)$ are continuous at $x=\pi$, their difference and sum, $f(x)$, must also be continuous at $x=\pi$.

Conclusion:

Yes, the function $f(x) = x^2 - \sin x + 5$ is continuous at $x = \pi$.

Given:

Three functions are given:

(a) $f(x) = \sin x + \cos x$

(b) $f(x) = \sin x - \cos x$

(c) $f(x) = \sin x \cdot \cos x$

To Discuss:

Discuss the continuity of each of the given functions.

Solution:

We know that the sine function, $g(x) = \sin x$, is continuous for all real numbers.

We also know that the cosine function, $h(x) = \cos x$, is continuous for all real numbers.

We can use the properties of continuous functions:

1. The sum of two continuous functions is continuous.

2. The difference of two continuous functions is continuous.

3. The product of two continuous functions is continuous.

(a) $f(x) = \sin x + \cos x$

This function is the sum of the sine function and the cosine function.

Since $\sin x$ is continuous for all real $x$ and $\cos x$ is continuous for all real $x$, their sum $f(x) = \sin x + \cos x$ is also continuous for all real numbers.

(b) $f(x) = \sin x - \cos x$

This function is the difference of the sine function and the cosine function.

Since $\sin x$ is continuous for all real $x$ and $\cos x$ is continuous for all real $x$, their difference $f(x) = \sin x - \cos x$ is also continuous for all real numbers.

(c) $f(x) = \sin x \cdot \cos x$

This function is the product of the sine function and the cosine function.

Since $\sin x$ is continuous for all real $x$ and $\cos x$ is continuous for all real $x$, their product $f(x) = \sin x \cdot \cos x$ is also continuous for all real numbers.

Conclusion:

All three functions (a), (b), and (c) are continuous everywhere, i.e., for all real numbers $x$.

Given:

The cosine, cosecant, secant, and cotangent functions.

To Discuss:

Discuss the continuity of each of these functions.

Solution:

We know that the fundamental trigonometric functions, $\sin x$ and $\cos x$, are continuous for all real numbers. The domain of both $\sin x$ and $\cos x$ is $(-\infty, \infty)$.

We also know that the quotient of two continuous functions is continuous everywhere the denominator is non-zero.

1. Cosine function ($f(x) = \cos x$)

The cosine function is a basic trigonometric function. Its graph is a smooth curve with no breaks or jumps over its entire domain.

The domain of $f(x) = \cos x$ is all real numbers, i.e., $x \in (-\infty, \infty)$.

The function $f(x) = \cos x$ is continuous for all real numbers.

2. Cosecant function ($f(x) = \text{cosec}\ x$)

The cosecant function is defined as $f(x) = \text{cosec}\ x = \frac{1}{\sin x}$.

This function is defined when the denominator is not zero, i.e., $\sin x \neq 0$.

The values of $x$ for which $\sin x = 0$ are $x = n\pi$, where $n$ is an integer ($n \in \mathbb{Z}$).

The domain of $f(x) = \text{cosec}\ x$ is $\{x \in \mathbb{R} \mid x \neq n\pi, n \in \mathbb{Z}\}$.

Since $\text{cosec}\ x$ is the quotient of the constant function $1$ (which is continuous) and the sine function $\sin x$ (which is continuous), $\text{cosec}\ x$ is continuous at all points where $\sin x \neq 0$.

Thus, the cosecant function is continuous on its domain, i.e., for all real numbers except integral multiples of $\pi$. The points of discontinuity are $x = n\pi$, $n \in \mathbb{Z}$.

3. Secant function ($f(x) = \sec x$)

The secant function is defined as $f(x) = \sec x = \frac{1}{\cos x}$.

This function is defined when the denominator is not zero, i.e., $\cos x \neq 0$.

The values of $x$ for which $\cos x = 0$ are $x = (n + \frac{1}{2})\pi = \frac{(2n+1)\pi}{2}$, where $n$ is an integer ($n \in \mathbb{Z}$).

The domain of $f(x) = \sec x$ is $\{x \in \mathbb{R} \mid x \neq \frac{(2n+1)\pi}{2}, n \in \mathbb{Z}\}$.

Since $\sec x$ is the quotient of the constant function $1$ (which is continuous) and the cosine function $\cos x$ (which is continuous), $\sec x$ is continuous at all points where $\cos x \neq 0$.

Thus, the secant function is continuous on its domain, i.e., for all real numbers except odd integral multiples of $\frac{\pi}{2}$. The points of discontinuity are $x = \frac{(2n+1)\pi}{2}$, $n \in \mathbb{Z}$.

4. Cotangent function ($f(x) = \cot x$)

The cotangent function is defined as $f(x) = \cot x = \frac{\cos x}{\sin x}$.

This function is defined when the denominator is not zero, i.e., $\sin x \neq 0$.

The values of $x$ for which $\sin x = 0$ are $x = n\pi$, where $n$ is an integer ($n \in \mathbb{Z}$).

The domain of $f(x) = \cot x$ is $\{x \in \mathbb{R} \mid x \neq n\pi, n \in \mathbb{Z}\}$.

Since $\cot x$ is the quotient of the cosine function $\cos x$ (which is continuous) and the sine function $\sin x$ (which is continuous), $\cot x$ is continuous at all points where $\sin x \neq 0$.

Thus, the cotangent function is continuous on its domain, i.e., for all real numbers except integral multiples of $\pi$. The points of discontinuity are $x = n\pi$, $n \in \mathbb{Z}$.

Summary:

- $\cos x$ is continuous everywhere.

- $\text{cosec}\ x$ is continuous on its domain, discontinuous at $x = n\pi$, $n \in \mathbb{Z}$.

- $\sec x$ is continuous on its domain, discontinuous at $x = \frac{(2n+1)\pi}{2}$, $n \in \mathbb{Z}$.

- $\cot x$ is continuous on its domain, discontinuous at $x = n\pi$, $n \in \mathbb{Z}$.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if}\; x < 0 \\ x+1, & \text{if}\; x \geq 0 \end{cases}$

To Find:

All points of discontinuity of $f(x)$.

Solution:

We need to examine the continuity of the function in the defined intervals and at the transition point $x=0$.

Continuity in the intervals:

For $x < 0$, $f(x) = \frac{\sin x}{x}$.

The function $\sin x$ is continuous for all real numbers. The function $g(x) = x$ is continuous for all real numbers. The quotient of two continuous functions is continuous wherever the denominator is non-zero. For $x < 0$, the denominator $x$ is non-zero.

Therefore, $f(x) = \frac{\sin x}{x}$ is continuous for all $x \in (-\infty, 0)$.

For $x > 0$, $f(x) = x+1$.

This is a polynomial function, which is continuous for all real numbers.

Therefore, $f(x) = x+1$ is continuous for all $x \in (0, \infty)$.

Continuity at the transition point $x=0$:

For $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$

Let's evaluate each part:

1. Value of the function at $x=0$:

When $x \geq 0$, $f(x) = x+1$.

$f(0) = 0 + 1 = 1$

2. Left-hand limit at $x=0$:

$\lim\limits_{x \to 0^-} f(x)$

For $x < 0$, $f(x) = \frac{\sin x}{x}$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{\sin x}{x}$

Using the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$:

$\lim\limits_{x \to 0^-} f(x) = 1$

3. Right-hand limit at $x=0$:

$\lim\limits_{x \to 0^+} f(x)$

For $x > 0$, $f(x) = x+1$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (x+1)$

By direct substitution:

$\lim\limits_{x \to 0^+} (x+1) = 0+1 = 1$

Comparing the limits and the function value at $x=0$:

We have $\lim\limits_{x \to 0^-} f(x) = 1$, $\lim\limits_{x \to 0^+} f(x) = 1$, and $f(0) = 1$.

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$, the function $f(x)$ is continuous at $x=0$.

Conclusion:

The function is continuous in the intervals $(-\infty, 0)$ and $(0, \infty)$, and it is also continuous at the transition point $x=0$.

Therefore, the function $f(x)$ is continuous for all real numbers. There are no points of discontinuity.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} x^2\sin \frac{1}{x}, & \text{if}\; x \neq 0 \\ 0, & \text{if}\; x = 0 \end{cases}$

To Determine:

Is the function $f(x)$ continuous? This means we need to check for continuity at all points in its domain, which is all real numbers.

Solution:

We need to check the continuity of $f(x)$ in the defined intervals ($x \neq 0$) and at the point where the definition changes ($x=0$).

Continuity for $x \neq 0$:

For any $x \neq 0$, the function is given by $f(x) = x^2 \sin \frac{1}{x}$.

Let's consider the functions $g(x) = x^2$ and $h(x) = \sin \frac{1}{x}$. The function $f(x)$ is the product of $g(x)$ and $h(x)$.

The function $g(x) = x^2$ is a polynomial function, which is continuous for all real numbers.

The function $h(x) = \sin \frac{1}{x}$ is a composite function. The inner function is $\frac{1}{x}$, which is continuous for all $x \neq 0$. The outer function is $\sin u$, which is continuous for all real numbers $u$. A composite of continuous functions is continuous where defined. Thus, $h(x) = \sin \frac{1}{x}$ is continuous for all $x \neq 0$.

The product of two continuous functions is continuous. Since $g(x)$ and $h(x)$ are continuous for all $x \neq 0$, their product $f(x) = x^2 \sin \frac{1}{x}$ is also continuous for all $x \neq 0$.

Continuity at $x=0$:

For $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim\limits_{x \to 0} f(x) = f(0)$

Let's evaluate the value of the function at $x=0$:

From the definition, $f(0) = 0$.

Now, let's evaluate the limit of the function as $x$ approaches $0$:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} x^2 \sin \frac{1}{x}$

We know that for any real value $u$, the value of $\sin u$ is bounded between -1 and 1:

$-1 \leq \sin u \leq 1$

Let $u = \frac{1}{x}$ for $x \neq 0$. Then:

$-1 \leq \sin \frac{1}{x} \leq 1$

Multiply the inequality by $x^2$. Since $x^2 \geq 0$ for all real $x$, the direction of the inequality remains the same:

$-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

Now, consider the limits of the bounding functions as $x \to 0$:

$\lim\limits_{x \to 0} (-x^2) = -(0)^2 = 0$

$\lim\limits_{x \to 0} (x^2) = (0)^2 = 0$

By the Squeeze Theorem (or Sandwich Theorem), since $\lim\limits_{x \to 0} (-x^2) = 0$ and $\lim\limits_{x \to 0} (x^2) = 0$, and $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$ for $x \neq 0$, the limit of the function in between must also be 0.

Therefore, $\lim\limits_{x \to 0} x^2 \sin \frac{1}{x} = 0$.

Comparing the limit and the function value at $x=0$:

We found that $\lim\limits_{x \to 0} f(x) = 0$ and $f(0) = 0$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function $f(x)$ is continuous at $x=0$.

Conclusion:

The function $f(x)$ is continuous for all $x \neq 0$ and is also continuous at $x=0$.

Therefore, the function $f(x)$ is continuous for all real numbers.

The answer to the question "Is the function defined by f(x) continuous?" is Yes.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} \sin x - \cos x, & \text{if}\; x \neq 0 \\ -1, & \text{if}\; x = 0 \end{cases}$

To Examine:

The continuity of the function $f(x)$ for all real numbers.

Solution:

We need to examine the continuity of the function in the defined interval ($x \neq 0$) and at the point where the definition changes ($x=0$).

Continuity for $x \neq 0$:

For any $x \neq 0$, the function is given by $f(x) = \sin x - \cos x$.

The function $\sin x$ is continuous for all real numbers.

The function $\cos x$ is continuous for all real numbers.

The difference of two continuous functions is continuous.

Therefore, $f(x) = \sin x - \cos x$ is continuous for all $x \neq 0$.

Continuity at $x=0$:

For $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim\limits_{x \to 0} f(x) = f(0)$

Let's evaluate the value of the function at $x=0$:

From the definition, $f(0) = -1$.

Now, let's evaluate the limit of the function as $x$ approaches $0$:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (\sin x - \cos x)$ (since for $x \neq 0$, $f(x) = \sin x - \cos x$)

Using the properties of limits:

$\lim\limits_{x \to 0} (\sin x - \cos x) = \lim\limits_{x \to 0} \sin x - \lim\limits_{x \to 0} \cos x$

We know the standard limits: $\lim\limits_{x \to 0} \sin x = \sin 0 = 0$ and $\lim\limits_{x \to 0} \cos x = \cos 0 = 1$.

So, $\lim\limits_{x \to 0} f(x) = 0 - 1 = -1$.

Comparing the limit and the function value at $x=0$:

We found that $\lim\limits_{x \to 0} f(x) = -1$ and $f(0) = -1$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function $f(x)$ is continuous at $x=0$.

Conclusion:

The function $f(x)$ is continuous for all $x \neq 0$ and is also continuous at $x=0$.

Therefore, the function $f(x)$ is continuous for all real numbers. There are no points of discontinuity.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} \frac{k\cos x}{\pi−2x}, & \text{if}\; x \neq \frac{\pi}{2} \\ 3, & \text{if}\; x = \frac{\pi}{2} \end{cases}$

The function is given to be continuous at $x = \frac{\pi}{2}$.

To Find:

The value of $k$.

Solution:

For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$, the following condition must be satisfied:

$\lim\limits_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$

Let's evaluate each part:

1. Value of the function at $x = \frac{\pi}{2}$:

From the definition, when $x = \frac{\pi}{2}$, $f(x) = 3$.

$f\left(\frac{\pi}{2}\right) = 3$

2. Limit of the function as $x$ approaches $\frac{\pi}{2}$:

$\lim\limits_{x \to \frac{\pi}{2}} f(x) = \lim\limits_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi−2x}$ (since for $x \neq \frac{\pi}{2}$, $f(x) = \frac{k\cos x}{\pi−2x}$)

If we substitute $x = \frac{\pi}{2}$, we get $\frac{k \cos(\frac{\pi}{2})}{\pi - 2(\frac{\pi}{2})} = \frac{k \cdot 0}{\pi - \pi} = \frac{0}{0}$, which is an indeterminate form.

We can use a substitution to evaluate this limit. Let $y = x - \frac{\pi}{2}$.

As $x \to \frac{\pi}{2}$, $y \to 0$.

Then $x = y + \frac{\pi}{2}$.

Substitute this into the expression for $f(x)$:

Numerator: $k\cos x = k\cos\left(y + \frac{\pi}{2}\right) = k(-\sin y) = -k\sin y$

Denominator: $\pi - 2x = \pi - 2\left(y + \frac{\pi}{2}\right) = \pi - 2y - \pi = -2y$

So the limit becomes:

$\lim\limits_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi−2x} = \lim\limits_{y \to 0} \frac{-k\sin y}{-2y}$

$= \lim\limits_{y \to 0} \frac{k\sin y}{2y}$

$= \frac{k}{2} \lim\limits_{y \to 0} \frac{\sin y}{y}$

Using the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$:

$\lim\limits_{x \to \frac{\pi}{2}} f(x) = \frac{k}{2} \cdot 1 = \frac{k}{2}$

For the function to be continuous at $x = \frac{\pi}{2}$, we must have $\lim\limits_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$.

Set the limit equal to the function value:

$\frac{k}{2} = 3$

Multiply both sides by 2:

$k = 3 \times 2$

$k = 6$

Conclusion:

The value of $k$ for which the function $f(x)$ is continuous at $x = \frac{\pi}{2}$ is $\mathbf{6}$.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} kx^2, & \text{if}\; x \leq 2 \\ 3, & \text{if}\; x > 2 \end{cases}$

The function is given to be continuous at $x = 2$.

To Find:

The value of $k$.

Solution:

For the function $f(x)$ to be continuous at $x=2$, the following condition must be satisfied:

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^+} f(x) = f(2)$

Let's evaluate each part:

1. Value of the function at $x=2$:

When $x \leq 2$, $f(x) = kx^2$.

$f(2) = k(2)^2 = 4k$

2. Left-hand limit at $x=2$:

$\lim\limits_{x \to 2^-} f(x)$

For $x < 2$, $f(x) = kx^2$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (kx^2) = k(2)^2 = 4k$

3. Right-hand limit at $x=2$:

$\lim\limits_{x \to 2^+} f(x)$

For $x > 2$, $f(x) = 3$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} 3 = 3$

For the function to be continuous at $x=2$, the left-hand limit and the right-hand limit must be equal.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^+} f(x)$

$4k = 3$

Solving for $k$:

$k = \frac{3}{4}$

Also, for continuity, $f(2)$ must be equal to the limit.

$f(2) = \lim\limits_{x \to 2} f(x)$

$4k = 3$

$k = \frac{3}{4}$

Both conditions yield the same value for $k$.

Conclusion:

The value of $k$ for which the function $f(x)$ is continuous at $x=2$ is $\mathbf{\frac{3}{4}}$.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} kx+1, & \text{if}\; x \leq \pi \\ \cos x, & \text{if}\; x > \pi \end{cases}$

The function is given to be continuous at $x = \pi$.

To Find:

The value of $k$.

Solution:

For the function $f(x)$ to be continuous at $x=\pi$, the following condition must be satisfied:

$\lim\limits_{x \to \pi^-} f(x) = \lim\limits_{x \to \pi^+} f(x) = f(\pi)$

Let's evaluate each part:

1. Value of the function at $x=\pi$:

When $x \leq \pi$, $f(x) = kx+1$.

$f(\pi) = k(\pi) + 1 = k\pi + 1$

2. Left-hand limit at $x=\pi$:

$\lim\limits_{x \to \pi^-} f(x)$

For $x < \pi$, $f(x) = kx+1$.

$\lim\limits_{x \to \pi^-} f(x) = \lim\limits_{x \to \pi^-} (kx+1) = k(\pi) + 1 = k\pi + 1$

3. Right-hand limit at $x=\pi$:

$\lim\limits_{x \to \pi^+} f(x)$

For $x > \pi$, $f(x) = \cos x$.

$\lim\limits_{x \to \pi^+} f(x) = \lim\limits_{x \to \pi^+} \cos x = \cos(\pi) = -1$

For the function to be continuous at $x=\pi$, we must have $\lim\limits_{x \to \pi^-} f(x) = \lim\limits_{x \to \pi^+} f(x)$.

Set the left-hand limit equal to the right-hand limit:

$k\pi + 1 = -1$

Now, solve for $k$:

$k\pi = -1 - 1$

$k\pi = -2$

$k = -\frac{2}{\pi}$

Also, we need to check if $f(\pi)$ is equal to this limit:

$f(\pi) = k\pi + 1 = \left(-\frac{2}{\pi}\right)\pi + 1 = -2 + 1 = -1$.

Since $\lim\limits_{x \to \pi^-} f(x) = \lim\limits_{x \to \pi^+} f(x) = f(\pi) = -1$ when $k = -\frac{2}{\pi}$, the function is continuous at $x=\pi$ for this value of $k$.

Conclusion:

The value of $k$ for which the function $f(x)$ is continuous at $x=\pi$ is $\mathbf{-\frac{2}{\pi}}$.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} kx+1, & \text{if}\; x \leq 5 \\ 3x-5, & \text{if}\; x > 5 \end{cases}$

The function is given to be continuous at $x = 5$.

To Find:

The value of $k$.

Solution:

For the function $f(x)$ to be continuous at $x=5$, the following condition must be satisfied:

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = f(5)$

Let's evaluate each part:

1. Value of the function at $x=5$:

When $x \leq 5$, $f(x) = kx+1$.

$f(5) = k(5) + 1 = 5k + 1$

2. Left-hand limit at $x=5$:

$\lim\limits_{x \to 5^-} f(x)$

For $x < 5$, $f(x) = kx+1$.

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (kx+1) = k(5) + 1 = 5k + 1$

3. Right-hand limit at $x=5$:

$\lim\limits_{x \to 5^+} f(x)$

For $x > 5$, $f(x) = 3x-5$.

$\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (3x-5) = 3(5) - 5 = 15 - 5 = 10$

For the function to be continuous at $x=5$, we must have $\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x)$.

Set the left-hand limit equal to the right-hand limit:

$5k + 1 = 10$

Now, solve for $k$:

$5k = 10 - 1$

$5k = 9$

$k = \frac{9}{5}$

Also, we need to check if $f(5)$ is equal to this limit:

$f(5) = 5k + 1 = 5\left(\frac{9}{5}\right) + 1 = 9 + 1 = 10$.

Since $\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = f(5) = 10$ when $k = \frac{9}{5}$, the function is continuous at $x=5$ for this value of $k$.

Conclusion:

The value of $k$ for which the function $f(x)$ is continuous at $x=5$ is $\mathbf{\frac{9}{5}}$.

Given:

The function $f(x)$ is defined piecewise as:

$f(x) = \begin{cases} 5, & \text{if}\; x \leq 2 \\ ax+b, & \text{if}\; 2 < x < 10 \\ 21, & \text{if}\; x \geq 10 \end{cases}$

The function is given to be continuous for all real numbers.

To Find:

The values of $a$ and $b$.

Solution:

For the function $f(x)$ to be continuous for all real numbers, it must be continuous at the points where the definition changes, which are $x=2$ and $x=10$.

Also, the function must be continuous within each defined interval.

For $x < 2$, $f(x) = 5$ (a constant function, continuous).

For $2 < x < 10$, $f(x) = ax+b$ (a linear polynomial, continuous).

For $x > 10$, $f(x) = 21$ (a constant function, continuous).

Thus, we only need to ensure continuity at the transition points $x=2$ and $x=10$.

Continuity at $x=2$:

For $f(x)$ to be continuous at $x=2$, the following condition must be satisfied:

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^+} f(x) = f(2)$

Value of the function at $x=2$:

$f(2) = 5$ (from the case $x \leq 2$)

Left-hand limit at $x=2$:

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} 5 = 5$ (for $x < 2$)

Right-hand limit at $x=2$:

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (ax+b) = a(2) + b = 2a+b$ (for $x > 2$)

For continuity at $x=2$, we must have $\lim\limits_{x \to 2^+} f(x) = f(2)$:

Continuity at $x=10$:

For $f(x)$ to be continuous at $x=10$, the following condition must be satisfied:

$\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^+} f(x) = f(10)$

Value of the function at $x=10$:

$f(10) = 21$ (from the case $x \geq 10$)

Left-hand limit at $x=10$:

$\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^-} (ax+b) = a(10) + b = 10a+b$ (for $x < 10$)

Right-hand limit at $x=10$:

$\lim\limits_{x \to 10^+} f(x) = \lim\limits_{x \to 10^+} 21 = 21$ (for $x > 10$)

For continuity at $x=10$, we must have $\lim\limits_{x \to 10^-} f(x) = f(10)$:

Solving the system of equations:

We have a system of two linear equations with two variables $a$ and $b$:

Equation (i): $2a + b = 5$

Equation (ii): $10a + b = 21$

Subtract Equation (i) from Equation (ii):

$(10a + b) - (2a + b) = 21 - 5$

$10a + b - 2a - b = 16$

$8a = 16$

$a = \frac{16}{8}$

$a = 2$

Substitute the value of $a=2$ into Equation (i):

$2(2) + b = 5$

$4 + b = 5$

$b = 5 - 4$

$b = 1$

Conclusion:

The values of $a$ and $b$ such that the function $f(x)$ is continuous are $\mathbf{a=2}$ and $\mathbf{b=1}$.

Given:

The function $f(x) = \cos (x^2)$.

To Show:

The function $f(x)$ is a continuous function for all real numbers.

Proof:

The given function $f(x) = \cos (x^2)$ is a composite function. We can express it as the composition of two functions.

Let the inner function be $g(x) = x^2$.

Let the outer function be $h(u) = \cos u$.

Then, the composite function is $h(g(x)) = h(x^2) = \cos(x^2)$, which is the given function $f(x)$.

Now, let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Consider the function $g(x) = x^2$.

This is a polynomial function. Polynomial functions are known to be continuous for all real numbers.

Thus, $g(x) = x^2$ is continuous on $\mathbb{R}$.

2. Consider the function $h(u) = \cos u$.

This is the cosine function. The cosine function is known to be continuous for all real numbers.

Thus, $h(u) = \cos u$ is continuous on $\mathbb{R}$.

We use the theorem regarding the continuity of composite functions:

Theorem: If $g$ is a continuous function at a point $c$, and $h$ is a continuous function at $g(c)$, then the composite function $h \circ g$ defined by $(h \circ g)(x) = h(g(x))$ is continuous at $c$.

Let $c$ be any arbitrary real number.

Since $g(x) = x^2$ is continuous for all real numbers, it is continuous at $x=c$.

The value of the inner function at $c$ is $g(c) = c^2$.

Since $h(u) = \cos u$ is continuous for all real numbers, it is continuous at $u = g(c) = c^2$.

Therefore, by the theorem on the continuity of composite functions, the function $f(x) = h(g(x)) = \cos(x^2)$ is continuous at $x=c$.

Since $c$ was an arbitrary real number, the function $f(x) = \cos(x^2)$ is continuous for all real numbers.

Conclusion:

The function $f(x) = \cos (x^2)$ is a continuous function because it is the composition of two functions ($g(x) = x^2$ and $h(u) = \cos u$) that are continuous everywhere.

Given:

The function is $f(x) = |\cos x|$.

To Show:

The function $f(x)$ is a continuous function for all real numbers.

Proof:

The given function $f(x) = |\cos x|$ is a composite function. We can express it as the composition of two functions.

Let the inner function be $g(x) = \cos x$.

Let the outer function be $h(u) = |u|$.

Then, the composite function is $h(g(x)) = h(\cos x) = |\cos x|$, which is the given function $f(x)$.

Now, let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Consider the function $g(x) = \cos x$.

This is the cosine function. The cosine function is known to be continuous for all real numbers.

Thus, $g(x) = \cos x$ is continuous on $\mathbb{R}$.

2. Consider the function $h(u) = |u|$.

This is the absolute value function. The absolute value function is known to be continuous for all real numbers.

Thus, $h(u) = |u|$ is continuous on $\mathbb{R}$.

We use the theorem regarding the continuity of composite functions:

Theorem: If $g$ is a continuous function at a point $c$, and $h$ is a continuous function at $g(c)$, then the composite function $h \circ g$ defined by $(h \circ g)(x) = h(g(x))$ is continuous at $c$.

Let $c$ be any arbitrary real number.

Since $g(x) = \cos x$ is continuous for all real numbers, it is continuous at $x=c$.

The value of the inner function at $c$ is $g(c) = \cos c$.

Since $h(u) = |u|$ is continuous for all real numbers, it is continuous at $u = g(c) = \cos c$ (because $\cos c$ is always a real number).

Therefore, by the theorem on the continuity of composite functions, the function $f(x) = h(g(x)) = |\cos x|$ is continuous at $x=c$.

Since $c$ was an arbitrary real number, the function $f(x) = |\cos x|$ is continuous for all real numbers.

Conclusion:

The function $f(x) = |\cos x|$ is a continuous function because it is the composition of two functions ($g(x) = \cos x$ and $h(u) = |u|$) that are continuous everywhere.

Given:

The function is $f(x) = \sin |x|$.

To Examine:

Examine the continuity of the function $f(x)$. We need to determine if it is continuous for all real numbers.

Solution:

The given function $f(x) = \sin |x|$ is a composite function. We can express it as the composition of two functions.

Let the inner function be $g(x) = |x|$.

Let the outer function be $h(u) = \sin u$.

Then, the composite function is $h(g(x)) = h(|x|) = \sin(|x|)$, which is the given function $f(x)$.

Now, let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Consider the function $g(x) = |x|$.

This is the absolute value function. The absolute value function is continuous for all real numbers.

For any real number $c$, $\lim\limits_{x \to c} |x| = |c|$. Since $\lim\limits_{x \to c} g(x) = g(c)$, $g(x)$ is continuous at every point $c \in \mathbb{R}$.

Thus, $g(x) = |x|$ is continuous on $\mathbb{R}$.

2. Consider the function $h(u) = \sin u$.

This is the sine function. The sine function is continuous for all real numbers.

For any real number $d$, $\lim\limits_{u \to d} \sin u = \sin d$. Since $\lim\limits_{u \to d} h(u) = h(d)$, $h(u)$ is continuous at every point $d \in \mathbb{R}$.

Thus, $h(u) = \sin u$ is continuous on $\mathbb{R}$.

We use the theorem regarding the continuity of composite functions:

Theorem: If $g$ is a continuous function at a point $c$, and $h$ is a continuous function at $g(c)$, then the composite function $h \circ g$ defined by $(h \circ g)(x) = h(g(x))$ is continuous at $c$.

Let $c$ be any arbitrary real number.

Since $g(x) = |x|$ is continuous for all real numbers, it is continuous at $x=c$.

The value of the inner function at $c$ is $g(c) = |c|$.

Since $h(u) = \sin u$ is continuous for all real numbers, it is continuous at $u = g(c) = |c|$ (because $|c|$ is always a real number).

Therefore, by the theorem on the continuity of composite functions, the function $f(x) = h(g(x)) = \sin(|x|)$ is continuous at $x=c$.

Since $c$ was an arbitrary real number, the function $f(x) = \sin(|x|)$ is continuous for all real numbers.

Conclusion:

The function $f(x) = \sin |x|$ is a continuous function because it is the composition of two functions ($g(x) = |x|$ and $h(u) = \sin u$) that are continuous everywhere.

Given:

The function is $f(x) = |x| - |x+1|$.

To Find:

All points of discontinuity of $f(x)$.

Solution:

The function $f(x)$ involves absolute values. We can rewrite the function by removing the absolute values based on the intervals where the expressions inside the absolute values change sign. The expressions are $x$ and $x+1$.

$|x|$ changes definition at $x=0$.

$|x+1|$ changes definition at $x+1=0$, i.e., $x=-1$.

The critical points are $x=-1$ and $x=0$. These points divide the real number line into three intervals: $x < -1$, $-1 \leq x < 0$, and $x \geq 0$.

Let's define $f(x)$ in each interval:

Case 1: $x < -1$

For $x < -1$, we have $x < 0$ and $x+1 < 0$.

So, $|x| = -x$ and $|x+1| = -(x+1)$.

$f(x) = |x| - |x+1| = (-x) - (-(x+1)) = -x + x + 1 = 1$.

Case 2: $-1 \leq x < 0$

For $-1 \leq x < 0$, we have $x < 0$ and $x+1 \geq 0$.

So, $|x| = -x$ and $|x+1| = x+1$.

$f(x) = |x| - |x+1| = (-x) - (x+1) = -x - x - 1 = -2x - 1$.

Case 3: $x \geq 0$

For $x \geq 0$, we have $x \geq 0$ and $x+1 > 0$.

So, $|x| = x$ and $|x+1| = x+1$.

$f(x) = |x| - |x+1| = x - (x+1) = x - x - 1 = -1$.

Combining these cases, the piecewise definition of $f(x)$ is:

$f(x) = \begin{cases} 1, & \text{if}\; x < -1 \\ -2x - 1, & \text{if}\; -1 \leq x < 0 \\ -1, & \text{if}\; x \geq 0 \end{cases}$

Now, we examine the continuity of $f(x)$ in the defined intervals and at the transition points $x=-1$ and $x=0$.

For $x \in (-\infty, -1)$, $f(x)=1$, which is a constant function and is continuous in this interval.

For $x \in (-1, 0)$, $f(x)=-2x-1$, which is a polynomial function and is continuous in this interval.

For $x \in (0, \infty)$, $f(x)=-1$, which is a constant function and is continuous in this interval.

We need to check for continuity at the transition points $x=-1$ and $x=0$.

Continuity at $x=-1$:

For continuity at $x=-1$, we check if $\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^+} f(x) = f(-1)$.

Value of the function at $x=-1$:

From the definition (case $-1 \leq x < 0$), $f(-1) = -2(-1) - 1 = 2 - 1 = 1$.

Left-hand limit at $x=-1$:

$\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^-} 1 = 1$ (from the case $x < -1$).

Right-hand limit at $x=-1$:

$\lim\limits_{x \to -1^+} f(x) = \lim\limits_{x \to -1^+} (-2x - 1) = -2(-1) - 1 = 2 - 1 = 1$ (from the case $-1 \leq x < 0$).

Since $\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^+} f(x) = f(-1) = 1$, the function is continuous at $x=-1$.

Continuity at $x=0$:

For continuity at $x=0$, we check if $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$.

Value of the function at $x=0$:

From the definition (case $x \geq 0$), $f(0) = -1$.

Left-hand limit at $x=0$:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (-2x - 1) = -2(0) - 1 = -1$ (from the case $-1 \leq x < 0$).

Right-hand limit at $x=0$:

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (-1) = -1$ (from the case $x \geq 0$).

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = -1$, the function is continuous at $x=0$.

Conclusion:

The function $f(x)$ is continuous in the open intervals $(-\infty, -1)$, $(-1, 0)$, and $(0, \infty)$.

The function is also continuous at the transition points $x=-1$ and $x=0$.

Therefore, the function $f(x) = |x| - |x+1|$ is continuous for all real numbers.

There are no points of discontinuity for the function $f(x)$.

Given:

The function $f(x) = \sin(x^2)$.

To Find:

The derivative of $f(x)$, i.e., $f'(x)$ or $\frac{d}{dx}(f(x))$.

Solution:

The given function is $f(x) = \sin(x^2)$. This is a composite function.

We can use the Chain Rule for differentiation.

Let $y = f(x) = \sin(x^2)$.

Let $u = x^2$.

Then $y = \sin u$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$.

Substitute back $u = x^2$:

$\frac{dy}{du} = \cos(x^2)$.

Next, find the derivative of $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$.

Now, apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = (\cos(x^2)) \cdot (2x)$

$\frac{dy}{dx} = 2x \cos(x^2)$

So, the derivative of $f(x) = \sin(x^2)$ is $f'(x) = 2x \cos(x^2)$.

Conclusion:

The derivative of the function $f(x) = \sin(x^2)$ is $\mathbf{f'(x) = 2x \cos(x^2)}$.

Given: The function to be differentiated is $f(x) = \sin(x^2 + 5)$.

Solution:

We need to find the derivative of the function $f(x) = \sin(x^2 + 5)$ with respect to $x$.

This function is a composition of two functions. We can use the chain rule for differentiation.

Let $y = \sin(u)$, where $u = x^2 + 5$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.

First, find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\sin(u)) = \cos(u)$

Substitute back $u = x^2 + 5$:

$\frac{dy}{du} = \cos(x^2 + 5)$

Next, find the derivative of $u$ with respect to $x$:

$u = x^2 + 5$

$\frac{du}{dx} = \frac{d}{dx}(x^2 + 5) = \frac{d}{dx}(x^2) + \frac{d}{dx}(5)$

$\frac{du}{dx} = 2x + 0 = 2x$

Now, apply the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Substitute the derivatives we found:

$\frac{dy}{dx} = \cos(x^2 + 5) \times (2x)$

Rearranging the terms:

$\frac{dy}{dx} = 2x \cos(x^2 + 5)$

Thus, the derivative of $\sin(x^2 + 5)$ with respect to $x$ is $2x \cos(x^2 + 5)$.

Given: The function to be differentiated is $f(x) = \cos(\sin x)$.

Solution:

We need to find the derivative of the function $f(x) = \cos(\sin x)$ with respect to $x$.

This function is a composite function, so we will use the chain rule.

Let $y = \cos(u)$, where $u = \sin x$.

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.

First, find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\cos(u)) = -\sin(u)$

Substitute back $u = \sin x$:

$\frac{dy}{du} = -\sin(\sin x)$

Next, find the derivative of $u$ with respect to $x$:

$u = \sin x$

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Substitute the derivatives we found:

$\frac{dy}{dx} = (-\sin(\sin x)) \times (\cos x)$

Rearranging the terms:

$\frac{dy}{dx} = -\cos x \sin(\sin x)$

Thus, the derivative of $\cos(\sin x)$ with respect to $x$ is $-\cos x \sin(\sin x)$.

Given: The function to be differentiated is $f(x) = \sin(ax + b)$.

Solution:

We need to find the derivative of the function $f(x) = \sin(ax + b)$ with respect to $x$.

This is a composite function, so we will use the chain rule for differentiation.

Let $y = \sin(u)$, where $u = ax + b$.

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.

First, find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$

Substitute back $u = ax + b$:

$\frac{dy}{du} = \cos(ax + b)$

Next, find the derivative of $u$ with respect to $x$:

$u = ax + b$

$\frac{du}{dx} = \frac{d}{dx}(ax + b)$

Using the sum rule and constant multiple rule:

$\frac{du}{dx} = a \frac{d}{dx}(x) + \frac{d}{dx}(b)$

We know that $\frac{d}{dx}(x) = 1$ and the derivative of a constant $b$ is 0.

$\frac{du}{dx} = a \times 1 + 0 = a$

Now, apply the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Substitute the derivatives we found:

$\frac{dy}{dx} = \cos(ax + b) \times (a)$

Rearranging the terms:

$\frac{dy}{dx} = a \cos(ax + b)$

Thus, the derivative of $\sin(ax + b)$ with respect to $x$ is $a \cos(ax + b)$.

Given: The function to be differentiated is $f(x) = \sec(\tan(\sqrt{x}))$.

Solution:

We need to find the derivative of the function $f(x) = \sec(\tan(\sqrt{x}))$ with respect to $x$.

This is a composite function with multiple layers, so we will apply the chain rule repeatedly.

Let $y = \sec(u)$, where $u = \tan(v)$, and $v = \sqrt{x}$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{dx}$.

Step 1: Differentiate the outermost function (secant) with respect to its argument ($u = \tan(\sqrt{x})$):

$\frac{dy}{du} = \frac{d}{du}(\sec u) = \sec u \tan u$

Substitute $u = \tan(\sqrt{x})$ back into the expression:

$\frac{dy}{du} = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))$

Step 2: Differentiate the middle function (tangent) with respect to its argument ($v = \sqrt{x}$):

$\frac{du}{dv} = \frac{d}{dv}(\tan v) = \sec^2 v$

Substitute $v = \sqrt{x}$ back into the expression:

$\frac{du}{dv} = \sec^2(\sqrt{x})$

Step 3: Differentiate the innermost function (square root) with respect to $x$:

$v = \sqrt{x} = x^{1/2}$

$\frac{dv}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{(1/2 - 1)} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Step 4: Apply the chain rule by multiplying the results from the steps above:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{dx}$

$\frac{dy}{dx} = [\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))] \times [\sec^2(\sqrt{x})] \times [\frac{1}{2\sqrt{x}}]$

Combine the terms:

$\frac{dy}{dx} = \frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$

Thus, the derivative of $\sec(\tan(\sqrt{x}))$ with respect to $x$ is $\frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$.

Given: The function to be differentiated is $f(x) = \frac{\sin (ax+b)}{\cos (cx+d)}$.

Solution:

We need to find the derivative of the function $f(x)$ with respect to $x$.

The function is in the form of a quotient, $\frac{u(x)}{v(x)}$, where $u(x) = \sin(ax+b)$ and $v(x) = \cos(cx+d)$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

First, find the derivative of $u(x)$ with respect to $x$, i.e., $u'(x)$.

$u(x) = \sin(ax+b)$. Using the chain rule, the derivative of $\sin(kx+c)$ is $k \cos(kx+c)$.

So, $u'(x) = a \cos(ax+b)$.

Next, find the derivative of $v(x)$ with respect to $x$, i.e., $v'(x)$.

$v(x) = \cos(cx+d)$. Using the chain rule, the derivative of $\cos(kx+c)$ is $-k \sin(kx+c)$.

So, $v'(x) = -c \sin(cx+d)$.

Now, apply the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Substitute the expressions for $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the formula:

$f'(x) = \frac{[a \cos(ax+b)][\cos(cx+d)] - [\sin(ax+b)][-c \sin(cx+d)]}{[\cos(cx+d)]^2}$

Simplify the numerator:

$f'(x) = \frac{a \cos(ax+b)\cos(cx+d) + c \sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

We can optionally rewrite the expression by splitting the fraction or using secant and tangent functions:

$f'(x) = a \frac{\cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)} + c \frac{\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

$f'(x) = a \cos(ax+b) \frac{1}{\cos(cx+d)} + c \sin(ax+b) \frac{\sin(cx+d)}{\cos(cx+d)} \frac{1}{\cos(cx+d)}$

$f'(x) = a \cos(ax+b) \sec(cx+d) + c \sin(ax+b) \tan(cx+d) \sec(cx+d)$

Both forms are valid.

Thus, the derivative of $\frac{\sin (ax+b)}{\cos (cx+d)}$ with respect to $x$ is $\frac{a \cos(ax+b)\cos(cx+d) + c \sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$ or $a \cos(ax+b) \sec(cx+d) + c \sin(ax+b) \tan(cx+d) \sec(cx+d)$.

Given: The function to be differentiated is $f(x) = \cos(x^3) \sin^2(x^5)$.

Solution:

We need to find the derivative of the function $f(x)$ with respect to $x$. The function is a product of two functions, $\cos(x^3)$ and $\sin^2(x^5)$. We will use the product rule and the chain rule.

Let $u(x) = \cos(x^3)$ and $v(x) = \sin^2(x^5)$. The product rule states that $\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)$.

First, find the derivative of $u(x) = \cos(x^3)$ using the chain rule.

Let $w = x^3$. Then $u = \cos(w)$.

$u'(x) = \frac{du}{dx} = \frac{du}{dw} \times \frac{dw}{dx}$

$\frac{du}{dw} = \frac{d}{dw}(\cos w) = -\sin w = -\sin(x^3)$

$\frac{dw}{dx} = \frac{d}{dx}(x^3) = 3x^2$

So, $u'(x) = (-\sin(x^3)) \times (3x^2) = -3x^2 \sin(x^3)$.

Next, find the derivative of $v(x) = \sin^2(x^5) = (\sin(x^5))^2$ using the chain rule.

Let $z = \sin(x^5)$. Then $v = z^2$.

$v'(x) = \frac{dv}{dx} = \frac{dv}{dz} \times \frac{dz}{dx}$

$\frac{dv}{dz} = \frac{d}{dz}(z^2) = 2z = 2\sin(x^5)$

Now, find $\frac{dz}{dx}$ for $z = \sin(x^5)$ using the chain rule again.

Let $y = x^5$. Then $z = \sin(y)$.

$\frac{dz}{dx} = \frac{dz}{dy} \times \frac{dy}{dx}$

$\frac{dz}{dy} = \frac{d}{dy}(\sin y) = \cos y = \cos(x^5)$

$\frac{dy}{dx} = \frac{d}{dx}(x^5) = 5x^4$

So, $\frac{dz}{dx} = \cos(x^5) \times (5x^4) = 5x^4 \cos(x^5)$.

Now, combine the parts for $v'(x)$:

$v'(x) = (2\sin(x^5)) \times (5x^4 \cos(x^5)) = 10x^4 \sin(x^5) \cos(x^5)$.

Finally, apply the product rule $\frac{d}{dx}(u v) = u'v + uv'$:

$f'(x) = (-3x^2 \sin(x^3)) (\sin^2(x^5)) + (\cos(x^3)) (10x^4 \sin(x^5) \cos(x^5))$

Rearranging the terms:

$f'(x) = -3x^2 \sin(x^3) \sin^2(x^5) + 10x^4 \cos(x^3) \sin(x^5) \cos(x^5)$

This can also be written as:

$f'(x) = 10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin(x^3) \sin^2(x^5)$

We can factor out a common term $\sin(x^5)$:

$f'(x) = \sin(x^5) [10x^4 \cos(x^3) \cos(x^5) - 3x^2 \sin(x^3) \sin(x^5)]$

Thus, the derivative of $\cos x^3 . \sin^2 (x^5)$ with respect to $x$ is $10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin(x^3) \sin^2(x^5)$.

Let the given function be $y$. We need to find the derivative of $y$ with respect to $x$, which is $\frac{dy}{dx}$.

Given Function:

$y = 2\sqrt{\cot (x^2)}$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We will find the derivative using the Chain Rule.

The function can be written as $y = 2 (\cot(x^2))^{1/2}$.

Applying the power rule and chain rule:

$\frac{dy}{dx} = \frac{d}{dx} [2 (\cot(x^2))^{1/2}]$

$\frac{dy}{dx} = 2 \cdot \frac{1}{2} (\cot(x^2))^{\frac{1}{2}-1} \cdot \frac{d}{dx}(\cot(x^2))$

$\frac{dy}{dx} = (\cot(x^2))^{-1/2} \cdot \frac{d}{dx}(\cot(x^2))$

$\frac{dy}{dx} = \frac{1}{\sqrt{\cot(x^2)}} \cdot \frac{d}{dx}(\cot(x^2))$

Now, we need to find the derivative of $\cot(x^2)$. We use the chain rule again.

Let $u = x^2$. Then we need to find $\frac{d}{dx}(\cot u)$. Using the chain rule, $\frac{d}{dx}(\cot u) = \frac{d}{du}(\cot u) \cdot \frac{du}{dx}$.

The derivative of $\cot u$ with respect to $u$ is $-\text{cosec}^2 u$. The derivative of $u = x^2$ with respect to $x$ is $2x$.

So, $\frac{d}{dx}(\cot(x^2)) = -\text{cosec}^2(x^2) \cdot 2x$

$\frac{d}{dx}(\cot(x^2)) = -2x \text{cosec}^2(x^2)$

Substitute this result back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\sqrt{\cot(x^2)}} \cdot (-2x \text{cosec}^2(x^2))$

$\frac{dy}{dx} = \frac{-2x \text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$

Now, let's simplify this expression using trigonometric identities: $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\text{cosec} \theta = \frac{1}{\sin \theta}$.

$\frac{dy}{dx} = \frac{-2x \left(\frac{1}{\sin^2(x^2)}\right)}{\sqrt{\frac{\cos(x^2)}{\sin(x^2)}}}$

$\frac{dy}{dx} = \frac{\frac{-2x}{\sin^2(x^2)}}{\frac{\sqrt{\cos(x^2)}}{\sqrt{\sin(x^2)}}}$

$\frac{dy}{dx} = \frac{-2x}{\sin^2(x^2)} \cdot \frac{\sqrt{\sin(x^2)}}{\sqrt{\cos(x^2)}}$

$\frac{dy}{dx} = \frac{-2x (\sin(x^2))^{1/2}}{(\sin(x^2))^2 (\cos(x^2))^{1/2}}$

$\frac{dy}{dx} = \frac{-2x}{(\sin(x^2))^{2 - 1/2} (\cos(x^2))^{1/2}}$

$\frac{dy}{dx} = \frac{-2x}{(\sin(x^2))^{3/2} \sqrt{\cos(x^2)}}$

We can rewrite the denominator using $\sin 2\theta = 2 \sin \theta \cos \theta$.

The denominator in the target answer is $\sin x^2 \sqrt{\sin 2x^2}$.

$\sin x^2 \sqrt{\sin 2x^2} = \sin x^2 \sqrt{2 \sin(x^2) \cos(x^2)}$

$= \sin x^2 \sqrt{2} \sqrt{\sin(x^2)} \sqrt{\cos(x^2)}$

$= \sqrt{2} (\sin x^2)^{1 + 1/2} \sqrt{\cos(x^2)}$

$= \sqrt{2} (\sin x^2)^{3/2} \sqrt{\cos(x^2)}$

Our calculated derivative has the denominator $(\sin(x^2))^{3/2} \sqrt{\cos(x^2)}$. To get the $\sqrt{2}$ term in the numerator as in the target answer, we multiply the numerator and denominator by $\sqrt{2}$.

$\frac{dy}{dx} = \frac{-2x}{(\sin(x^2))^{3/2} \sqrt{\cos(x^2)}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\frac{dy}{dx} = \frac{-2\sqrt{2}x}{\sqrt{2} (\sin(x^2))^{3/2} \sqrt{\cos(x^2)}}$

Using the identity $\sqrt{2} (\sin x^2)^{3/2} \sqrt{\cos(x^2)} = \sin x^2 \sqrt{\sin 2x^2}$ in the denominator:

$\frac{dy}{dx} = \frac{-2\sqrt{2}x}{\sin x^2 \sqrt{\sin 2x^2}}$

The derivative of $2\sqrt{\cot (x^2)}$ is $\frac{-2\sqrt{2}x}{\sin x^2 \sqrt{\sin 2x^2}}$.

Given: The function to be differentiated is $f(x) = \cos (\sqrt{x})$.

Solution:

We need to find the derivative of the function $f(x) = \cos (\sqrt{x})$ with respect to $x$.

This function is a composite function, so we will use the chain rule.

Let $y = \cos(u)$, where $u = \sqrt{x}$.

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.

First, find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u$

Substitute back $u = \sqrt{x}$:

$\frac{dy}{du} = -\sin(\sqrt{x})$

Next, find the derivative of $u$ with respect to $x$:

$u = \sqrt{x} = x^{1/2}$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2}$

$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$

Now, apply the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Substitute the derivatives we found:

$\frac{dy}{dx} = (-\sin(\sqrt{x})) \times (\frac{1}{2\sqrt{x}})$

Combine the terms:

$\frac{dy}{dx} = \frac{-\sin(\sqrt{x})}{2\sqrt{x}}$

Thus, the derivative of $\cos (\sqrt{x})$ with respect to $x$ is $\frac{-\sin(\sqrt{x})}{2\sqrt{x}}$.

Given: The function $f(x) = |x - 1|$, $x \in R$.

To Prove: The function $f(x)$ is not differentiable at $x = 1$.

Proof:

A function $f(x)$ is differentiable at a point $x=a$ if the limit of the difference quotient exists at that point. The derivative of $f(x)$ at $x=a$, denoted by $f'(a)$, is defined as:

$\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$

In this case, we need to check the differentiability of $f(x) = |x-1|$ at $x = 1$. So, we need to evaluate the limit:

$\lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$

First, let's find $f(1)$:

$f(1) = |1 - 1| = |0| = 0$

Next, let's find $f(1+h)$:

$f(1+h) = |(1+h) - 1| = |h|$

Now, substitute these values into the difference quotient:

$\frac{f(1+h) - f(1)}{h} = \frac{|h| - 0}{h} = \frac{|h|}{h}$

For the limit $\lim\limits_{h \to 0} \frac{|h|}{h}$ to exist, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal.

Let's find the LHL:

$\text{LHL} = \lim\limits_{h \to 0^-} \frac{|h|}{h}$

As $h \to 0^-$ (i.e., $h$ approaches 0 from the negative side), $h < 0$. By the definition of the absolute value function, for $h < 0$, $|h| = -h$.

$\text{LHL} = \lim\limits_{h \to 0^-} \frac{-h}{h} = \lim\limits_{h \to 0^-} (-1) = -1$

Now, let's find the RHL:

$\text{RHL} = \lim\limits_{h \to 0^+} \frac{|h|}{h}$

As $h \to 0^+$ (i.e., $h$ approaches 0 from the positive side), $h > 0$. By the definition of the absolute value function, for $h > 0$, $|h| = h$.

$\text{RHL} = \lim\limits_{h \to 0^+} \frac{h}{h} = \lim\limits_{h \to 0^+} (1) = 1$

We have found that the LHL is $-1$ and the RHL is $1$.

Since $\text{LHL} \neq \text{RHL}$, the limit $\lim\limits_{h \to 0} \frac{|h|}{h}$ does not exist.

Therefore, the derivative of $f(x)$ at $x = 1$ does not exist.

Hence, the function $f(x) = |x-1|$ is not differentiable at $x = 1$.

Alternate Solution (Using the definition of the derivative directly):

The derivative of $f(x)$ at $x=a$ is given by $f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$.

Here $a=1$, so we check $\lim\limits_{x \to 1} \frac{f(x) - f(1)}{x - 1}$.

$f(1) = |1-1| = 0$.

So, we evaluate $\lim\limits_{x \to 1} \frac{|x - 1| - 0}{x - 1} = \lim\limits_{x \to 1} \frac{|x - 1|}{x - 1}$.

Let's evaluate the left-hand limit as $x \to 1^-$:

As $x \to 1^-$ (i.e., $x$ approaches 1 from values less than 1), $x - 1 < 0$. So, $|x - 1| = -(x - 1)$.

$\lim\limits_{x \to 1^-} \frac{|x - 1|}{x - 1} = \lim\limits_{x \to 1^-} \frac{-(x - 1)}{x - 1} = \lim\limits_{x \to 1^-} (-1) = -1$

Now, let's evaluate the right-hand limit as $x \to 1^+$:

As $x \to 1^+$ (i.e., $x$ approaches 1 from values greater than 1), $x - 1 > 0$. So, $|x - 1| = x - 1$.

$\lim\limits_{x \to 1^+} \frac{|x - 1|}{x - 1} = \lim\limits_{x \to 1^+} \frac{x - 1}{x - 1} = \lim\limits_{x \to 1^+} (1) = 1$

Since the left-hand limit ($-1$) is not equal to the right-hand limit ($1$), the limit $\lim\limits_{x \to 1} \frac{|x - 1|}{x - 1}$ does not exist.

Therefore, the function $f(x) = |x-1|$ is not differentiable at $x = 1$.

Given: The greatest integer function $f(x) = [x]$ defined for $0 < x < 3$.

To Prove: The function $f(x)$ is not differentiable at $x = 1$ and $x = 2$.

Proof:

A function $f(x)$ is differentiable at a point $x=a$ if the limit of the difference quotient exists at that point and is finite. The derivative of $f(x)$ at $x=a$, denoted by $f'(a)$, is defined as:

$f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$

For this limit to exist, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal.

Let's check the differentiability at $x = 1$.

We need to evaluate $\lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$.

From the definition of $f(x)=[x]$, we have $f(1) = [1] = 1$ (since $0 < 1 < 3$).

Now, let's evaluate the LHL at $x=1$:

$\text{LHL} = \lim\limits_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$

As $h \to 0^-$, $h$ is a small negative number (i.e., $h < 0$). For $h$ values very close to 0 from the left (e.g., $-0.1, -0.001$), $1+h$ will be slightly less than 1 (e.g., $0.9, 0.999$).

For $-1 < h < 0$, we have $0 < 1+h < 1$.

So, $f(1+h) = [1+h] = 0$ for $-1 < h < 0$.

$\text{LHL} = \lim\limits_{h \to 0^-} \frac{0 - 1}{h} = \lim\limits_{h \to 0^-} \frac{-1}{h}$

As $h$ approaches 0 from the negative side, $\frac{1}{h}$ approaches $-\infty$, so $\frac{-1}{h}$ approaches $+\infty$.

$\text{LHL} = \infty$

Next, let's evaluate the RHL at $x=1$:

$\text{RHL} = \lim\limits_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$

As $h \to 0^+$, $h$ is a small positive number (i.e., $h > 0$). For $h$ values very close to 0 from the right (e.g., $0.1, 0.001$), $1+h$ will be slightly greater than 1 (e.g., $1.1, 1.001$).

For $0 < h < 1$, we have $1 \leq 1+h < 2$.

So, $f(1+h) = [1+h] = 1$ for $0 < h < 1$.

$\text{RHL} = \lim\limits_{h \to 0^+} \frac{1 - 1}{h} = \lim\limits_{h \to 0^+} \frac{0}{h} = \lim\limits_{h \to 0^+} 0 = 0$

Comparing the LHL and RHL at $x=1$, we have $\text{LHL} = \infty$ and $\text{RHL} = 0$.

Since $\text{LHL} \neq \text{RHL}$, the limit $\lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$ does not exist as a finite value.

Therefore, the function $f(x) = [x]$ is not differentiable at $x = 1$.

Now, let's check the differentiability at $x = 2$.

We need to evaluate $\lim\limits_{h \to 0} \frac{f(2+h) - f(2)}{h}$.

From the definition of $f(x)=[x]$, we have $f(2) = [2] = 2$ (since $0 < 2 < 3$).

Let's evaluate the LHL at $x=2$:

$\text{LHL} = \lim\limits_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$

As $h \to 0^-$, $h$ is a small negative number. For $h$ values very close to 0 from the left (e.g., $-0.1, -0.001$), $2+h$ will be slightly less than 2 (e.g., $1.9, 1.999$).

For $-1 < h < 0$, we have $1 < 2+h < 2$.

So, $f(2+h) = [2+h] = 1$ for $-1 < h < 0$.

$\text{LHL} = \lim\limits_{h \to 0^-} \frac{1 - 2}{h} = \lim\limits_{h \to 0^-} \frac{-1}{h}$

As $h$ approaches 0 from the negative side, $\frac{1}{h}$ approaches $-\infty$, so $\frac{-1}{h}$ approaches $+\infty$.

$\text{LHL} = \infty$

Next, let's evaluate the RHL at $x=2$:

$\text{RHL} = \lim\limits_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$

As $h \to 0^+$, $h$ is a small positive number. For $h$ values very close to 0 from the right (e.g., $0.1, 0.001$), $2+h$ will be slightly greater than 2 (e.g., $2.1, 2.001$).

For $0 < h < 1$, we have $2 \leq 2+h < 3$.

So, $f(2+h) = [2+h] = 2$ for $0 < h < 1$.

$\text{RHL} = \lim\limits_{h \to 0^+} \frac{2 - 2}{h} = \lim\limits_{h \to 0^+} \frac{0}{h} = \lim\limits_{h \to 0^+} 0 = 0$

Comparing the LHL and RHL at $x=2$, we have $\text{LHL} = \infty$ and $\text{RHL} = 0$.

Since $\text{LHL} \neq \text{RHL}$, the limit $\lim\limits_{h \to 0} \frac{f(2+h) - f(2)}{h}$ does not exist as a finite value.

Therefore, the function $f(x) = [x]$ is not differentiable at $x = 2$.

Thus, the greatest integer function $f(x) = [x]$ is not differentiable at $x=1$ and $x=2$ for the given domain $0 < x < 3$.

Given: The equation $x - y = \pi$.

To Find: $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y$ with respect to $x$ from the given equation $x - y = \pi$. We can do this by differentiating both sides of the equation with respect to $x$.

Differentiate each term with respect to $x$:

$\frac{d}{dx}(x - y) = \frac{d}{dx}(\pi)$

Using the linearity of differentiation (difference rule):

$\frac{d}{dx}(x) - \frac{d}{dx}(y) = \frac{d}{dx}(\pi)$

Now, we evaluate each derivative:

The derivative of $x$ with respect to $x$ is 1.

$\frac{d}{dx}(x) = 1$

The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$.

$\frac{d}{dx}(y) = \frac{dy}{dx}$

The derivative of a constant, like $\pi$, with respect to $x$ is 0.

$\frac{d}{dx}(\pi) = 0$

Substitute these derivatives back into the differentiated equation:

$1 - \frac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

Add $\frac{dy}{dx}$ to both sides of the equation:

$1 = \frac{dy}{dx}$

Thus, the derivative $\frac{dy}{dx}$ is 1.

Alternatively, we could first solve the given equation for $y$ in terms of $x$:

$x - y = \pi$

Subtract $x$ from both sides:

$-y = \pi - x$

Multiply by -1:

$y = x - \pi$

Now, differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x - \pi)$

Using the difference rule:

$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\pi)$

$\frac{dy}{dx} = 1 - 0$

$\frac{dy}{dx} = 1$

This confirms the previous result.

Given: The equation $y + \sin y = \cos x$.

To Find: $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y$ with respect to $x$ from the given implicit equation $y + \sin y = \cos x$. We will differentiate both sides of the equation with respect to $x$.

Differentiate each term with respect to $x$:

$\frac{d}{dx}(y + \sin y) = \frac{d}{dx}(\cos x)$

Using the sum rule on the left side:

$\frac{d}{dx}(y) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(\cos x)$

Now, evaluate each derivative:

The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

$\frac{d}{dx}(y) = \frac{dy}{dx}$

The derivative of $\sin y$ with respect to $x$ requires the chain rule, since $y$ is a function of $x$. We differentiate with respect to $y$ and multiply by $\frac{dy}{dx}$.

$\frac{d}{dx}(\sin y) = \frac{d}{dy}(\sin y) \times \frac{dy}{dx} = \cos y \times \frac{dy}{dx}$

The derivative of $\cos x$ with respect to $x$ is $-\sin x$.

$\frac{d}{dx}(\cos x) = -\sin x$

Substitute these derivatives back into the differentiated equation:

$\frac{dy}{dx} + \cos y \frac{dy}{dx} = -\sin x$

Now, we need to solve this equation for $\frac{dy}{dx}$. Notice that $\frac{dy}{dx}$ appears in two terms on the left side. Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx} (1 + \cos y) = -\sin x$

Finally, divide both sides by $(1 + \cos y)$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-\sin x}{1 + \cos y}$

This is the required derivative, provided that $1 + \cos y \neq 0$.

Thus, the derivative $\frac{dy}{dx}$ is $\frac{-\sin x}{1 + \cos y}$.

Given: The function $f(x) = \sin^{-1} x$.

To Find: The derivative $\frac{dy}{dx}$ of the function $f(x)$ with respect to $x$.

Solution:

Let $y = f(x)$, so $y = \sin^{-1} x$.

By the definition of the inverse sine function, the statement $y = \sin^{-1} x$ is equivalent to:

$x = \sin y$

The domain of $\sin^{-1} x$ is $[-1, 1]$, and its principal value range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, for $y = \sin^{-1} x$, we have $-1 \leq x \leq 1$ and $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

We differentiate the equation $x = \sin y$ implicitly with respect to $x$.

$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$

The derivative of $x$ with respect to $x$ is 1:

$\frac{d}{dx}(x) = 1$

The derivative of $\sin y$ with respect to $x$ requires the chain rule. We differentiate $\sin y$ with respect to $y$ and then multiply by $\frac{dy}{dx}$:

$\frac{d}{dx}(\sin y) = \frac{d}{dy}(\sin y) \times \frac{dy}{dx}$

We know that $\frac{d}{dy}(\sin y) = \cos y$.

So, $\frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx}$.

Substitute these derivatives back into the implicit differentiation equation:

$1 = \cos y \frac{dy}{dx}$

Now, we solve for $\frac{dy}{dx}$ by dividing both sides by $\cos y$ (assuming $\cos y \neq 0$):

$\frac{dy}{dx} = \frac{1}{\cos y}$

We need to express $\cos y$ in terms of $x$. We know that $x = \sin y$. Using the fundamental trigonometric identity $\sin^2 y + \cos^2 y = 1$, we can write:

$\cos^2 y = 1 - \sin^2 y$

Taking the square root of both sides:

$\cos y = \pm \sqrt{1 - \sin^2 y}$

Since the range of $y = \sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the value of $\cos y$ is non-negative in this interval (it is positive for $-\frac{\pi}{2} < y < \frac{\pi}{2}$ and zero at the endpoints). Therefore, we take the positive square root:

$\cos y = \sqrt{1 - \sin^2 y}$

Substitute $\sin y = x$ into this expression for $\cos y$:

$\cos y = \sqrt{1 - x^2}$

Now, substitute this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

The assumption that the derivative exists implies that $1 - x^2 > 0$, which means $-1 < x < 1$. At $x = \pm 1$, $\cos y = 0$, and the derivative is undefined (the tangent line is vertical).

Thus, the derivative of $f(x) = \sin^{-1} x$ with respect to $x$ is $\frac{1}{\sqrt{1 - x^2}}$ for $x \in (-1, 1)$.

Given:

$2x + 3y = \sin x$

Solution:

We are asked to find $\frac{dy}{dx}$. We will differentiate the given equation with respect to $x$ on both sides.

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin x)$

Using the sum rule of differentiation, we get:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$

Differentiating each term:

$2 + 3 \frac{dy}{dx} = \cos x$

Now, we need to isolate $\frac{dy}{dx}$. Subtract 2 from both sides:

$3 \frac{dy}{dx} = \cos x - 2$

Finally, divide by 3:

$\frac{dy}{dx} = \frac{\cos x - 2}{3}$

Or, this can also be written as:

$\frac{dy}{dx} = \frac{1}{3}(\cos x - 2)$

The derivative of the given equation is:

$\frac{dy}{dx} = \frac{\cos x - 2}{3}$

Given:

$2x + 3y = \sin y$

Solution:

We need to find $\frac{dy}{dx}$. We will differentiate the given equation with respect to $x$ on both sides.

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin y)$

Using the sum rule on the left side and the chain rule on the term involving $y$ on the right side:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx}$

Differentiating each term:

$2 + 3 \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

Now, we need to collect all terms containing $\frac{dy}{dx}$ on one side. Subtract $3 \frac{dy}{dx}$ from both sides:

$2 = \cos y \frac{dy}{dx} - 3 \frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the terms on the right side:

$2 = (\cos y - 3) \frac{dy}{dx}$

Finally, divide both sides by $(\cos y - 3)$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{2}{\cos y - 3}$

The derivative of the given equation is:

$\frac{dy}{dx} = \frac{2}{\cos y - 3}$

Given:

$ax + by^2 = \cos y$

Solution:

We need to find the derivative $\frac{dy}{dx}$. We will differentiate both sides of the given equation with respect to $x$.

$\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)$

Using the sum rule on the left side:

$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$

Differentiating $ax$ with respect to $x$ gives $a$.

For $by^2$, we use the chain rule: $\frac{d}{dx}(by^2) = b \frac{d}{dx}(y^2) = b \cdot 2y \cdot \frac{dy}{dx} = 2by \frac{dy}{dx}$.

For $\cos y$, we use the chain rule: $\frac{d}{dx}(\cos y) = \frac{d}{dy}(\cos y) \cdot \frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

Now, we want to collect all terms containing $\frac{dy}{dx}$ on one side. Add $\sin y \frac{dy}{dx}$ to both sides:

$a + 2by \frac{dy}{dx} + \sin y \frac{dy}{dx} = 0$

Subtract $a$ from both sides:

$2by \frac{dy}{dx} + \sin y \frac{dy}{dx} = -a$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$(2by + \sin y) \frac{dy}{dx} = -a$

Finally, divide both sides by $(2by + \sin y)$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-a}{2by + \sin y}$

This can also be written as:

$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

The derivative of the given equation is:

$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Given:

$xy + y^2 = \tan x + y$

Solution:

We need to find the derivative $\frac{dy}{dx}$. We will differentiate both sides of the given equation with respect to $x$.

$\frac{d}{dx}(xy + y^2) = \frac{d}{dx}(\tan x + y)$

Using the sum rule of differentiation on both sides:

$\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y)$

Now, we apply the rules of differentiation:

• For $\frac{d}{dx}(xy)$, we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$. Here, $u=x$ and $v=y$. So, $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y(1) = x \frac{dy}{dx} + y$.
• For $\frac{d}{dx}(y^2)$, we use the chain rule: $\frac{d}{dx}(f(y)^n) = n f(y)^{n-1} \frac{df}{dx}$. Here, $f(y)=y$ and $n=2$. So, $\frac{d}{dx}(y^2) = 2y^{2-1} \frac{dy}{dx} = 2y \frac{dy}{dx}$.
• For $\frac{d}{dx}(\tan x)$, this is a standard derivative: $\frac{d}{dx}(\tan x) = \sec^2 x$.
• For $\frac{d}{dx}(y)$, this is simply $\frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$\left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$

Rearrange the terms to group all $\frac{dy}{dx}$ terms on one side (e.g., the left side) and the other terms on the other side (the right side):

$x \frac{dy}{dx} + 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$(x + 2y - 1) \frac{dy}{dx} = \sec^2 x - y$

Finally, divide both sides by the term multiplying $\frac{dy}{dx}$ to solve for $\frac{dy}{dx}$, assuming $(x + 2y - 1) \neq 0$:

$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$

The derivative of the given equation is:

$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$

Given:

$x^2 + xy + y^2 = 100$

Solution:

We need to find the derivative $\frac{dy}{dx}$. We will differentiate both sides of the given equation with respect to $x$.

$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)$

Using the sum rule of differentiation on the left side:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(100)$

Now, we apply the rules of differentiation to each term:

• $\frac{d}{dx}(x^2) = 2x$
• For $\frac{d}{dx}(xy)$, we use the product rule, $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$, with $u=x$ and $v=y$. This gives: $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y(1) = x \frac{dy}{dx} + y$.
• For $\frac{d}{dx}(y^2)$, we use the chain rule, $\frac{d}{dx}(f(y)^n) = n f(y)^{n-1} \frac{df}{dx}$, with $f(y)=y$ and $n=2$. This gives: $\frac{d}{dx}(y^2) = 2y^{2-1} \frac{dy}{dx} = 2y \frac{dy}{dx}$.
• $\frac{d}{dx}(100) = 0$ (the derivative of a constant is zero).

Substituting these derivatives back into the equation:

$2x + \left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0$

Rearrange the terms to group all $\frac{dy}{dx}$ terms on one side and the other terms on the other side:

$x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - y$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$(x + 2y) \frac{dy}{dx} = -(2x + y)$

Finally, divide both sides by $(x + 2y)$ to solve for $\frac{dy}{dx}$, assuming $(x + 2y) \neq 0$:

$\frac{dy}{dx} = \frac{-(2x + y)}{x + 2y}$

This can also be written as:

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

The derivative of the given equation is:

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

Given:

$x^3 + x^2y + xy^2 + y^3 = 81$

Solution:

We need to find the derivative $\frac{dy}{dx}$. We will differentiate both sides of the given equation with respect to $x$.

$\frac{d}{dx}(x^3 + x^2y + xy^2 + y^3) = \frac{d}{dx}(81)$

Using the sum rule of differentiation on the left side:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81)$

Now, we differentiate each term:

• $\frac{d}{dx}(x^3) = 3x^2$
• For $\frac{d}{dx}(x^2y)$, we use the product rule, $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$, with $u=x^2$ and $v=y$:

  $\frac{d}{dx}(x^2y) = x^2 \frac{dy}{dx} + y \frac{d}{dx}(x^2) = x^2 \frac{dy}{dx} + y(2x) = x^2 \frac{dy}{dx} + 2xy$

• For $\frac{d}{dx}(xy^2)$, we use the product rule with $u=x$ and $v=y^2$. We also use the chain rule for $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$:

  $\frac{d}{dx}(xy^2) = x \frac{d}{dx}(y^2) + y^2 \frac{d}{dx}(x) = x (2y \frac{dy}{dx}) + y^2(1) = 2xy \frac{dy}{dx} + y^2$

• For $\frac{d}{dx}(y^3)$, we use the chain rule:

  $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$

• $\frac{d}{dx}(81) = 0$ (derivative of a constant)

Substituting these derivatives back into the equation:

$3x^2 + \left(x^2 \frac{dy}{dx} + 2xy\right) + \left(2xy \frac{dy}{dx} + y^2\right) + 3y^2 \frac{dy}{dx} = 0$

Rearrange the terms to group all $\frac{dy}{dx}$ terms on one side and the other terms on the other side:

$x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -3x^2 - 2xy - y^2$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$(x^2 + 2xy + 3y^2) \frac{dy}{dx} = -(3x^2 + 2xy + y^2)$

Finally, divide both sides by $(x^2 + 2xy + 3y^2)$ to solve for $\frac{dy}{dx}$, assuming $(x^2 + 2xy + 3y^2) \neq 0$:

$\frac{dy}{dx} = \frac{-(3x^2 + 2xy + y^2)}{x^2 + 2xy + 3y^2}$

This can also be written as:

$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

The derivative of the given equation is:

$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

Given:

$\sin^2 y + \cos(xy) = \kappa$

Here, $\kappa$ is a constant.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We will differentiate both sides of the given equation with respect to $x$.

$\frac{d}{dx}(\sin^2 y + \cos(xy)) = \frac{d}{dx}(\kappa)$

Using the sum rule on the left side:

$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)$

Now, we differentiate each term:

• For $\frac{d}{dx}(\sin^2 y)$, we use the chain rule. Let $u = \sin y$. Then $\frac{d}{dx}(\sin^2 y) = \frac{d}{dx}(u^2) = 2u \frac{du}{dx} = 2\sin y \frac{d}{dx}(\sin y)$. Applying the chain rule again for $\frac{d}{dx}(\sin y)$: $\frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx}$. So, $\frac{d}{dx}(\sin^2 y) = 2\sin y \cos y \frac{dy}{dx}$. Using the identity $2\sin y \cos y = \sin(2y)$, this term is $\sin(2y) \frac{dy}{dx}$.
• For $\frac{d}{dx}(\cos(xy))$, we use the chain rule and the product rule. Let $v = xy$. Then $\frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\cos v) = -\sin v \frac{dv}{dx} = -\sin(xy) \frac{d}{dx}(xy)$. Applying the product rule for $\frac{d}{dx}(xy)$: $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y(1) = x \frac{dy}{dx} + y$. So, $\frac{d}{dx}(\cos(xy)) = -\sin(xy) \left(x \frac{dy}{dx} + y\right) = -x \sin(xy) \frac{dy}{dx} - y \sin(xy)$.
• For $\frac{d}{dx}(\kappa)$, the derivative of a constant is 0. So, $\frac{d}{dx}(\kappa) = 0$.

Substituting these derivatives back into the equation:

$\sin(2y) \frac{dy}{dx} - x \sin(xy) \frac{dy}{dx} - y \sin(xy) = 0$

Rearrange the terms to group all $\frac{dy}{dx}$ terms on one side and the other term on the other side:

$\sin(2y) \frac{dy}{dx} - x \sin(xy) \frac{dy}{dx} = y \sin(xy)$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$(\sin(2y) - x \sin(xy)) \frac{dy}{dx} = y \sin(xy)$

Finally, divide both sides by $(\sin(2y) - x \sin(xy))$ to solve for $\frac{dy}{dx}$, assuming $(\sin(2y) - x \sin(xy)) \neq 0$:

$\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$

The derivative of the given equation is:

$\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$

Given:

$\sin^2 x + \cos^2 y = 1$

Solution:

We need to find the derivative $\frac{dy}{dx}$. We will differentiate both sides of the given equation with respect to $x$.

$\frac{d}{dx}(\sin^2 x + \cos^2 y) = \frac{d}{dx}(1)$

Using the sum rule of differentiation on the left side:

$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1)$

Now, we differentiate each term:

• For $\frac{d}{dx}(\sin^2 x)$, we use the chain rule. Let $u = \sin x$. Then $\frac{d}{dx}(\sin^2 x) = \frac{d}{dx}(u^2) = 2u \frac{du}{dx} = 2\sin x \frac{d}{dx}(\sin x) = 2\sin x \cos x$.
• For $\frac{d}{dx}(\cos^2 y)$, we use the chain rule. Let $v = \cos y$. Then $\frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(v^2) = 2v \frac{dv}{dx} = 2\cos y \frac{d}{dx}(\cos y)$. Applying the chain rule again for $\frac{d}{dx}(\cos y)$: $\frac{d}{dx}(\cos y) = -\sin y \frac{dy}{dx}$. So, $\frac{d}{dx}(\cos^2 y) = 2\cos y (-\sin y) \frac{dy}{dx} = -2\sin y \cos y \frac{dy}{dx}$.
• For $\frac{d}{dx}(1)$, the derivative of a constant is 0. So, $\frac{d}{dx}(1) = 0$.

Substituting these derivatives back into the equation:

$2\sin x \cos x - 2\sin y \cos y \frac{dy}{dx} = 0$

We know the double angle identities: $2\sin x \cos x = \sin(2x)$ and $2\sin y \cos y = \sin(2y)$.

So the equation becomes:

$\sin(2x) - \sin(2y) \frac{dy}{dx} = 0$

Rearrange the equation to isolate the term with $\frac{dy}{dx}$:

$-\sin(2y) \frac{dy}{dx} = -\sin(2x)$

Multiply both sides by -1:

$\sin(2y) \frac{dy}{dx} = \sin(2x)$

Finally, divide both sides by $\sin(2y)$ to solve for $\frac{dy}{dx}$, assuming $\sin(2y) \neq 0$:

$\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}$

The derivative of the given equation is:

$\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}$

Given:

$y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$

Solution (Using Substitution):

To find $\frac{dy}{dx}$, we can simplify the expression by using a trigonometric substitution.

Let $x = \tan \theta$. This substitution is valid for all real $x$, where $\theta = \tan^{-1} x$. We can consider the principal value branch of $\tan^{-1} x$, i.e., $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.

Substitute $x = \tan \theta$ into the expression for $y$:

$y = \sin^{-1} \left( \frac{2 \tan \theta}{1+\tan^2 \theta} \right)$

Using the trigonometric identity $\frac{2 \tan \theta}{1+\tan^2 \theta} = \sin(2\theta)$, the expression becomes:

$y = \sin^{-1} (\sin 2\theta)$

We know that $\sin^{-1}(\sin \alpha) = \alpha$ if $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$.

For the simplification $y = 2\theta$ to be valid, we need $2\theta$ to be in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. This means $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$.

Since $x = \tan \theta$, the condition $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$ corresponds to $-1 \le x \le 1$.

For the open interval $-1 < x < 1$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, which implies $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$.

Thus, for $-1 < x < 1$, we have:

$y = 2\theta$

Substituting back $\theta = \tan^{-1} x$:

$y = 2 \tan^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1} x)$

Using the standard derivative formula $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$:

$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = \frac{2}{1+x^2}$

This result is valid for $-1 < x < 1$.

The derivative of the given function for $-1 < x < 1$ is:

$\frac{dy}{dx} = \frac{2}{1+x^2}$

Alternate Solution (Using Direct Differentiation):

Given $y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$.

We use the chain rule for differentiation: $\frac{dy}{dx} = \frac{d}{du}(\sin^{-1} u) \cdot \frac{du}{dx}$, where $u = \frac{2x}{1+x^2}$.

The derivative of $\sin^{-1} u$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$, which is defined for $|u| < 1$.

First, we find the derivative of $u = \frac{2x}{1+x^2}$ with respect to $x$ using the quotient rule:

$\frac{du}{dx} = \frac{\frac{d}{dx}(2x) \cdot (1+x^2) - (2x) \cdot \frac{d}{dx}(1+x^2)}{(1+x^2)^2}$

$\frac{du}{dx} = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2}$

$\frac{du}{dx} = \frac{2+2x^2 - 4x^2}{(1+x^2)^2}$

$\frac{du}{dx} = \frac{2-2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2}$

Next, we calculate $\sqrt{1-u^2}$ where $u = \frac{2x}{1+x^2}$:

$\sqrt{1-u^2} = \sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2} = \sqrt{1 - \frac{4x^2}{(1+x^2)^2}} = \sqrt{\frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2}}$

$\sqrt{1-u^2} = \sqrt{\frac{1+2x^2+x^4 - 4x^2}{(1+x^2)^2}} = \sqrt{\frac{x^4 - 2x^2 + 1}{(1+x^2)^2}} = \sqrt{\frac{(x^2-1)^2}{(1+x^2)^2}}$

$\sqrt{1-u^2} = \frac{\sqrt{(x^2-1)^2}}{\sqrt{(1+x^2)^2}} = \frac{|x^2-1|}{1+x^2}$ (since $1+x^2 > 0$)

Now substitute $\frac{du}{dx}$ and $\sqrt{1-u^2}$ into the chain rule formula, for $|u| < 1$, which means $|x| \neq 1$:

$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} = \frac{1}{\frac{|x^2-1|}{1+x^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2}$

$\frac{dy}{dx} = \frac{1+x^2}{|x^2-1|} \cdot \frac{2(1-x^2)}{(1+x^2)^2}$

$\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)|x^2-1|}$

Since $|x^2-1| = |-(1-x^2)| = |1-x^2|$, we have:

$\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)|1-x^2|}$

This expression depends on the sign of $1-x^2$ (for $|x| \neq 1$).

If $-1 < x < 1$, then $1-x^2 > 0$, so $|1-x^2| = 1-x^2$.

$\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)(1-x^2)} = \frac{2}{1+x^2}$

If $x < -1$ or $x > 1$, then $1-x^2 < 0$, so $|1-x^2| = -(1-x^2) = x^2-1$.

$\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)(-(1-x^2))} = \frac{2}{-(1+x^2)} = -\frac{2}{1+x^2}$

Thus, the derivative is piecewise defined for $|x| \neq 1$:

$ \frac{dy}{dx} = \begin{cases} \frac{2}{1+x^2} & , & |x| < 1 \\ -\frac{2}{1+x^2} & , & |x| > 1 \end{cases} $

The derivative does not exist at $x = 1$ and $x = -1$.

Given:

$y = \tan^{-1} \left( \frac{3x − x^3}{1 −3x^2} \right)$

The condition given is $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We can simplify the expression using a trigonometric substitution.

Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.

Substitute $x = \tan \theta$ into the expression for $y$:

$y = \tan^{-1} \left( \frac{3\tan \theta − \tan^3 \theta}{1 −3\tan^2 \theta} \right)$

We use the trigonometric identity $\tan(3\theta) = \frac{3\tan \theta − \tan^3 \theta}{1 −3\tan^2 \theta}$.

So, the expression becomes:

$y = \tan^{-1} (\tan 3\theta)$

The given condition is $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$. Substituting $x = \tan \theta$, we get:

$-\frac{1}{\sqrt{3}} < \tan \theta < \frac{1}{\sqrt{3}}$

This inequality holds when $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$.

Multiplying the inequality by 3, we get the range for $3\theta$:

$3 \times (-\frac{\pi}{6}) < 3\theta < 3 \times (\frac{\pi}{6})$

$-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$

For any angle $\alpha$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have $\tan^{-1}(\tan \alpha) = \alpha$.

Since our $3\theta$ lies in this interval, we can simplify $y$ as:

$y = 3\theta$

Now, substitute back $\theta = \tan^{-1} x$:

$y = 3 \tan^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(3 \tan^{-1} x)$

Using the constant multiple rule and the standard derivative of $\tan^{-1} x$ ($\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$):

$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\tan^{-1} x)$

$\frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = \frac{3}{1+x^2}$

This result is valid for the given range $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

The derivative of the given function is:

$\frac{dy}{dx} = \frac{3}{1+x^2}$

Given:

$y = \cos^{-1} \left( \frac{1 − x^2}{1 + x^2} \right)$

The condition given is $0 < x < 1$.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We can simplify the expression using a trigonometric substitution.

Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.

Since $0 < x < 1$, taking the inverse tangent of the inequality, we get:

$\tan^{-1}(0) < \tan^{-1}(x) < \tan^{-1}(1)$

$0 < \theta < \frac{\pi}{4}$

Substitute $x = \tan \theta$ into the expression for $y$:

$y = \cos^{-1} \left( \frac{1 − \tan^2 \theta}{1 + \tan^2 \theta} \right)$

We use the trigonometric identity $\cos(2\theta) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$.

So, the expression becomes:

$y = \cos^{-1} (\cos 2\theta)$

We have the range for $\theta$ as $0 < \theta < \frac{\pi}{4}$. Multiplying the inequality by 2, we get the range for $2\theta$:

$2 \times 0 < 2\theta < 2 \times \frac{\pi}{4}$

$0 < 2\theta < \frac{\pi}{2}$

The principal value branch of $\cos^{-1} u$ is $[0, \pi]$. Since $0 < 2\theta < \frac{\pi}{2}$, the angle $2\theta$ is within this range. Therefore, $\cos^{-1}(\cos(2\theta)) = 2\theta$ for $0 < 2\theta < \pi$.

Thus, for the given range of $x$ (which implies $0 < 2\theta < \frac{\pi}{2}$), we can simplify $y$ as:

$y = 2\theta$

Now, substitute back $\theta = \tan^{-1} x$:

$y = 2 \tan^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1} x)$

Using the constant multiple rule and the standard derivative of $\tan^{-1} x$ ($\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$):

$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1} x)$

$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = \frac{2}{1+x^2}$

This result is valid for the given range $0 < x < 1$.

The derivative of the given function is:

$\frac{dy}{dx} = \frac{2}{1+x^2}$

Given:

$y = \sin^{-1} \left( \frac{1 − x^2}{1 + x^2} \right)$

The condition given is $0 < x < 1$.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We can simplify the expression using a trigonometric substitution.

Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.

Since the given condition is $0 < x < 1$, taking the inverse tangent of the inequality, we get:

$\tan^{-1}(0) < \tan^{-1}(x) < \tan^{-1}(1)$

$0 < \theta < \frac{\pi}{4}$

Substitute $x = \tan \theta$ into the expression for $y$:

$y = \sin^{-1} \left( \frac{1 − \tan^2 \theta}{1 + \tan^2 \theta} \right)$

We use the trigonometric identity $\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos(2\theta)$.

So, the expression becomes:

$y = \sin^{-1} (\cos 2\theta)$

We know that $\cos \alpha = \sin (\frac{\pi}{2} - \alpha)$. Using this, we can write $\cos 2\theta$ as $\sin (\frac{\pi}{2} - 2\theta)$.

$y = \sin^{-1} \left( \sin \left( \frac{\pi}{2} - 2\theta \right) \right)$

Now we need to check if the angle $\frac{\pi}{2} - 2\theta$ is in the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We have the range for $\theta$ as $0 < \theta < \frac{\pi}{4}$.

Multiplying by -2, we get: $0 \times (-2) > -2\theta > \frac{\pi}{4} \times (-2)$, which is $0 > -2\theta > -\frac{\pi}{2}$, or $-\frac{\pi}{2} < -2\theta < 0$.

Adding $\frac{\pi}{2}$ to the inequality: $\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - 2\theta < \frac{\pi}{2} + 0$.

$0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2}$

Since $0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2}$, the angle $\frac{\pi}{2} - 2\theta$ is within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, we can simplify $y$ as:

$y = \frac{\pi}{2} - 2\theta$

Now, substitute back $\theta = \tan^{-1} x$:

$y = \frac{\pi}{2} - 2\tan^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - 2\tan^{-1} x \right)$

Using the difference rule, constant multiple rule, and the standard derivative of $\tan^{-1} x$ ($\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$):

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - \frac{d}{dx}(2\tan^{-1} x)$

$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = -\frac{2}{1+x^2}$

This result is valid for the given range $0 < x < 1$.

The derivative of the given function is:

$\frac{dy}{dx} = -\frac{2}{1+x^2}$

Given:

$y = \cos^{-1} \left( \frac{2x}{1 + x^2} \right)$

The condition given is $-1 < x < 1$.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We can simplify the expression using a trigonometric substitution.

Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.

Since the given condition is $-1 < x < 1$, taking the inverse tangent of the inequality, we get:

$\tan^{-1}(-1) < \tan^{-1}(x) < \tan^{-1}(1)$

$-\frac{\pi}{4} < \theta < \frac{\pi}{4}$

Substitute $x = \tan \theta$ into the expression for $y$:

$y = \cos^{-1} \left( \frac{2\tan \theta}{1 + \tan^2 \theta} \right)$

We use the trigonometric identity $\frac{2\tan \theta}{1 + \tan^2 \theta} = \sin(2\theta)$.

So, the expression becomes:

$y = \cos^{-1} (\sin 2\theta)$

We know that $\sin \alpha = \cos (\frac{\pi}{2} - \alpha)$. Using this, we can write $\sin 2\theta$ as $\cos (\frac{\pi}{2} - 2\theta)$.

$y = \cos^{-1} \left( \cos \left( \frac{\pi}{2} - 2\theta \right) \right)$

Now we need to check if the angle $\frac{\pi}{2} - 2\theta$ is in the principal value branch of $\cos^{-1}$, which is $[0, \pi]$.

We have the range for $\theta$ as $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

Multiplying by -2, we get: $(-\frac{\pi}{4}) \times (-2) > -2\theta > (\frac{\pi}{4}) \times (-2)$, which is $\frac{\pi}{2} > -2\theta > -\frac{\pi}{2}$, or $-\frac{\pi}{2} < -2\theta < \frac{\pi}{2}$.

Adding $\frac{\pi}{2}$ to the inequality: $\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - 2\theta < \frac{\pi}{2} + \frac{\pi}{2}$.

$0 < \frac{\pi}{2} - 2\theta < \pi$

Since $0 < \frac{\pi}{2} - 2\theta < \pi$, the angle $\frac{\pi}{2} - 2\theta$ is within the interval $(0, \pi)$, which is a subset of $[0, \pi]$.

Therefore, for the given range of $x$, we can simplify $y$ as:

$y = \frac{\pi}{2} - 2\theta$

Now, substitute back $\theta = \tan^{-1} x$:

$y = \frac{\pi}{2} - 2\tan^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - 2\tan^{-1} x \right)$

Using the difference rule, constant multiple rule, and the standard derivative of $\tan^{-1} x$ ($\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$):

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - \frac{d}{dx}(2\tan^{-1} x)$

$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = -\frac{2}{1+x^2}$

This result is valid for the given range $-1 < x < 1$.

The derivative of the given function is:

$\frac{dy}{dx} = -\frac{2}{1+x^2}$

Given:

$y = \sin^{-1} \left( 2x \sqrt{1 − x^2} \right)$

The condition given is $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We can simplify the expression using a trigonometric substitution.

Let $x = \sin \theta$. Then $\theta = \sin^{-1} x$.

Since the given condition is $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, taking the inverse sine of the inequality (and considering the principal value branch of $\sin^{-1}$ which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ where $\sin \theta$ is increasing):

$\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) < \sin^{-1}(x) < \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$-\frac{\pi}{4} < \theta < \frac{\pi}{4}$

Substitute $x = \sin \theta$ into the expression for $y$:

$y = \sin^{-1} \left( 2 \sin \theta \sqrt{1 − \sin^2 \theta} \right)$

Using the trigonometric identity $1 - \sin^2 \theta = \cos^2 \theta$:

$y = \sin^{-1} \left( 2 \sin \theta \sqrt{\cos^2 \theta} \right)$

$y = \sin^{-1} \left( 2 \sin \theta |\cos \theta| \right)$

Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, $\cos \theta$ is positive (it lies in the first or fourth quadrant). Therefore, $|\cos \theta| = \cos \theta$.

$y = \sin^{-1} (2 \sin \theta \cos \theta)$

Using the trigonometric identity $2 \sin \theta \cos \theta = \sin(2\theta)$:

$y = \sin^{-1} (\sin 2\theta)$

Now we need to check if the angle $2\theta$ is in the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We have the range for $\theta$ as $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Multiplying the inequality by 2, we get the range for $2\theta$:

$2 \times \left(-\frac{\pi}{4}\right) < 2\theta < 2 \times \left(\frac{\pi}{4}\right)$

$-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$

Since $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$, the angle $2\theta$ is within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, we can simplify $y$ as:

$y = 2\theta$

Now, substitute back $\theta = \sin^{-1} x$:

$y = 2 \sin^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2 \sin^{-1} x)$

Using the constant multiple rule and the standard derivative of $\sin^{-1} x$ ($\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$ for $|x| < 1$):

$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\sin^{-1} x)$

$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$

This result is valid for the given range $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.

The derivative of the given function is:

$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$

Given:

$y = \sec^{-1} \left( \frac{1}{2x^2− 1} \right)$

The condition given is $0 < x < \frac{1}{\sqrt{2}}$.

Solution:

We need to find the derivative $\frac{dy}{dx}$. We can simplify the expression by using the property $\sec^{-1} u = \cos^{-1} \left(\frac{1}{u}\right)$.

$y = \cos^{-1} \left( \frac{1}{\frac{1}{2x^2− 1}} \right)$

$y = \cos^{-1} (2x^2 - 1)$

To further simplify this expression, we use a trigonometric substitution. Let $x = \cos \theta$. Then $\theta = \cos^{-1} x$.

Since the given condition is $0 < x < \frac{1}{\sqrt{2}}$, taking the inverse cosine of the inequality (note that $\cos^{-1} x$ is a decreasing function):

$\cos^{-1}(0) > \cos^{-1}(x) > \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$\frac{\pi}{2} > \theta > \frac{\pi}{4}$

So, the range for $\theta$ is $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.

Substitute $x = \cos \theta$ into the simplified expression for $y$:

$y = \cos^{-1} (2\cos^2 \theta - 1)$

Using the trigonometric identity $\cos(2\theta) = 2\cos^2 \theta - 1$, we get:

$y = \cos^{-1} (\cos 2\theta)$

Now we need to check if the angle $2\theta$ is in the principal value branch of $\cos^{-1}$, which is $[0, \pi]$.

We have the range for $\theta$ as $\frac{\pi}{4} < \theta < \frac{\pi}{2}$. Multiplying the inequality by 2, we get the range for $2\theta$:

$2 \times \frac{\pi}{4} < 2\theta < 2 \times \frac{\pi}{2}$

$\frac{\pi}{2} < 2\theta < \pi$

Since $\frac{\pi}{2} < 2\theta < \pi$, the angle $2\theta$ is within the principal value branch of $\cos^{-1}$.

Therefore, we can simplify $y$ as:

$y = 2\theta$

Now, substitute back $\theta = \cos^{-1} x$:

$y = 2 \cos^{-1} x$

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2 \cos^{-1} x)$

Using the constant multiple rule and the standard derivative of $\cos^{-1} x$ ($\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$ for $|x| < 1$):

$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\cos^{-1} x)$

$\frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$

$\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}$

This result is valid for the given range $0 < x < \frac{1}{\sqrt{2}}$, which is within the domain $|x| < 1$ for the derivative of $\cos^{-1} x$.

The derivative of the given function is:

$\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}$

Given:

The statement: $x = e^{\log x}$ for all real $x$.

Solution:

Let's analyze the properties of the functions involved in the statement $x = e^{\log x}$. We assume $\log x$ refers to the natural logarithm, often written as $\ln x$. The same logic applies if it refers to $\log_{10} x$ or $\log_b x$ for any base $b > 0, b \neq 1$.

The function $\log x$ (natural logarithm) is defined only for positive real numbers.

That is, the domain of $\log x$ is $(0, \infty)$, which means $x$ must be greater than 0 for $\log x$ to be a real number.

The expression $e^{\log x}$ involves the exponential function with base $e$. The base $e$ is a positive number ($e \approx 2.718$). An exponential function with a positive base, $e^z$, where $z$ is a real number, always results in a positive value.

Therefore, the right-hand side of the equation, $e^{\log x}$, is only defined for $x > 0$, and for these values of $x$, $e^{\log x}$ is always positive.

The left-hand side of the equation is $x$.

For the equality $x = e^{\log x}$ to hold, both sides must be defined and equal.

Since $e^{\log x}$ is only defined for $x > 0$, the equality can only potentially hold for $x > 0$.

For any $x > 0$, the definition of the logarithm states that $\log x$ is the power to which $e$ must be raised to get $x$. By definition, $e^{\log x} = x$ for all $x > 0$.

However, the original statement claims that the equality $x = e^{\log x}$ is true for all real $x$.

Consider real numbers $x \le 0$. For these values of $x$, $\log x$ is not defined in the set of real numbers. Consequently, $e^{\log x}$ is not defined for real $x \le 0$.

Since the equality does not hold (is not even defined on the right side) for $x \le 0$, the statement is not true for all real $x$.

The statement is only true for positive real numbers, i.e., for $x > 0$.

Therefore, the statement "x = e^{log x} for all real x" is false.

The given statement is not true for all real $x$. It is only true for $x > 0$.

Solution:

(i) Differentiate $e^{-x}$ with respect to $x$.

Let $y = e^{-x}$.

Using the chain rule, $\frac{d}{dx}(e^u) = e^u \frac{du}{dx}$, where $u = -x$.

We find the derivative of the inner function $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(-x)$

$\frac{du}{dx} = -1$

Now, substitute this into the chain rule formula:

$\frac{dy}{dx} = e^{-x} \cdot (-1)$

$\frac{dy}{dx} = -e^{-x}$

The derivative of $e^{-x}$ is $-e^{-x}$.

(ii) Differentiate $\sin(\log x)$ with respect to $x$, for $x > 0$.

Let $y = \sin(\log x)$.

Using the chain rule, $\frac{d}{dx}(\sin u) = \cos u \frac{du}{dx}$, where $u = \log x$. The condition $x > 0$ ensures that $\log x$ is defined.

We find the derivative of the inner function $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\log x)$

$\frac{du}{dx} = \frac{1}{x}$ (for $x > 0$)

Now, substitute this into the chain rule formula:

$\frac{dy}{dx} = \cos(\log x) \cdot \frac{1}{x}$

$\frac{dy}{dx} = \frac{\cos(\log x)}{x}$

The derivative of $\sin(\log x)$ for $x > 0$ is $\frac{\cos(\log x)}{x}$.

(iii) Differentiate $\cos^{-1}(e^x)$ with respect to $x$.

Let $y = \cos^{-1}(e^x)$.

Using the chain rule, $\frac{d}{dx}(\cos^{-1} u) = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx}$, where $u = e^x$.

The derivative of $\cos^{-1} u$ is defined for $|u| < 1$. Here, $u = e^x$, so we require $|e^x| < 1$. Since $e^x$ is always positive, this means $e^x < 1$. Taking the natural logarithm of both sides, $x < \log 1$, so $x < 0$. The derivative is defined for $x < 0$.

We find the derivative of the inner function $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(e^x)$

$\frac{du}{dx} = e^x$

Now, substitute this into the chain rule formula:

$\frac{dy}{dx} = \frac{-1}{\sqrt{1-(e^x)^2}} \cdot e^x$

$\frac{dy}{dx} = \frac{-e^x}{\sqrt{1-e^{2x}}}$

The derivative of $\cos^{-1}(e^x)$ for $x < 0$ is $\frac{-e^x}{\sqrt{1-e^{2x}}}$.

(iv) Differentiate $e^{\cos x}$ with respect to $x$.

Let $y = e^{\cos x}$.

Using the chain rule, $\frac{d}{dx}(e^u) = e^u \frac{du}{dx}$, where $u = \cos x$.

We find the derivative of the inner function $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\cos x)$

$\frac{du}{dx} = -\sin x$

Now, substitute this into the chain rule formula:

$\frac{dy}{dx} = e^{\cos x} \cdot (-\sin x)$

$\frac{dy}{dx} = -\sin x \cdot e^{\cos x}$

The derivative of $e^{\cos x}$ is $-\sin x \cdot e^{\cos x}$.

Let $y = \frac{e^x}{\sin x}$.

We need to find $\frac{dy}{dx}$. This function is in the form of a quotient, so we will use the quotient rule for differentiation.

The quotient rule states that if $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$, then the derivative is given by:

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

In this case, we have:

$u = e^x$

$v = \sin x$

Now, we find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x$

$\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Substitute these into the quotient rule formula:

$\frac{dy}{dx} = \frac{(\sin x) \frac{d}{dx}(e^x) - (e^x) \frac{d}{dx}(\sin x)}{(\sin x)^2}$

$\frac{dy}{dx} = \frac{(\sin x)(e^x) - (e^x)(\cos x)}{\sin^2 x}$

Factor out $e^x$ from the numerator:

$\frac{dy}{dx} = \frac{e^x (\sin x - \cos x)}{\sin^2 x}$

Alternatively, we can split the fraction:

$\frac{dy}{dx} = \frac{e^x \sin x}{\sin^2 x} - \frac{e^x \cos x}{\sin^2 x}$

$\frac{dy}{dx} = e^x \frac{\sin x}{\sin^2 x} - e^x \frac{\cos x}{\sin^2 x}$

$\frac{dy}{dx} = e^x \frac{1}{\sin x} - e^x \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$

$\frac{dy}{dx} = e^x \text{cosec } x - e^x \cot x \text{ cosec } x$

$\frac{dy}{dx} = e^x (\text{cosec } x - \cot x \text{ cosec } x)$

Both forms of the answer are acceptable. The simplest form is generally preferred.

The final answer is $\boxed{\frac{e^x(\sin x - \cos x)}{\sin^2 x}}$.

Let $y = e^{\sin^{-1} x}$.

We need to find $\frac{dy}{dx}$. This function is a composite function of the form $e^u$, where $u$ is a function of $x$. Therefore, we will use the chain rule for differentiation.

The chain rule states that if $y = f(u)$ and $u = g(x)$, then the derivative of $y$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

In this function, we can identify:

Outer function: $f(u) = e^u$

Inner function: $u = \sin^{-1} x$

First, we differentiate the outer function with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$

Next, we differentiate the inner function with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x)$

The derivative of $\sin^{-1} x$ is a standard derivative:

$\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$

Now, apply the chain rule by multiplying the results from the two differentiation steps:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Substitute the expressions for $\frac{dy}{du}$ and $\frac{du}{dx}$:

$\frac{dy}{dx} = (e^u) \cdot \left(\frac{1}{\sqrt{1-x^2}}\right)$

Substitute back the expression for $u$, which is $u = \sin^{-1} x$:

$\frac{dy}{dx} = e^{\sin^{-1} x} \cdot \left(\frac{1}{\sqrt{1-x^2}}\right)$

The derivative is:

$\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}$

The final answer is $\boxed{\frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}}$.

Let $y = e^{x^{3}}$.

We need to find $\frac{dy}{dx}$. This function is a composite function where the base is $e$ and the exponent is a function of $x$ ($x^3$). We will use the chain rule for differentiation.

The chain rule states that if $y = f(u)$ and $u = g(x)$, then the derivative of $y$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

In this case, we can identify the parts for the chain rule:

Let $u = x^{3}$.

Then $y = e^{u}$.

First, differentiate $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(e^{u})$

The derivative of $e^u$ with respect to $u$ is $e^u$:

$\frac{dy}{du} = e^{u}$

Next, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^{3})$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{du}{dx} = 3x^{3-1} = 3x^{2}$

Now, apply the chain rule by multiplying $\frac{dy}{du}$ and $\frac{du}{dx}$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Substitute the expressions we found:

$\frac{dy}{dx} = (e^{u}) \cdot (3x^{2})$

Substitute back the expression for $u$, which is $u = x^{3}$:

$\frac{dy}{dx} = e^{x^{3}} \cdot 3x^{2}$

Rearrange the terms for a standard form:

$\frac{dy}{dx} = 3x^{2} e^{x^{3}}$

The final answer is $\boxed{3x^{2} e^{x^{3}}}$.

Let $y = \sin (\tan^{-1} e^{-x})$.

We need to find $\frac{dy}{dx}$. This is a composition of multiple functions, so we will apply the chain rule repeatedly.

Let's break down the function into layers and apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$.

Let $u = \tan^{-1} e^{-x}$. Then $y = \sin u$.

Let $v = e^{-x}$. Then $u = \tan^{-1} v$.

Let $w = -x$. Then $v = e^w$.

Now, we differentiate each part:

1. Differentiate $y = \sin u$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$

2. Differentiate $u = \tan^{-1} v$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(\tan^{-1} v) = \frac{1}{1+v^2}$

3. Differentiate $v = e^w$ with respect to $w$:

$\frac{dv}{dw} = \frac{d}{dw}(e^w) = e^w$

4. Differentiate $w = -x$ with respect to $x$:

$\frac{dw}{dx} = \frac{d}{dx}(-x) = -1$

Now, multiply these derivatives together according to the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx} = (\cos u) \cdot \left(\frac{1}{1+v^2}\right) \cdot (e^w) \cdot (-1)$

Substitute back the expressions for $u$, $v$, and $w$ in terms of $x$:

$w = -x$

$v = e^w = e^{-x}$

$u = \tan^{-1} v = \tan^{-1} (e^{-x})$

So, $\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \left(\frac{1}{1+(e^{-x})^2}\right) \cdot (e^{-x}) \cdot (-1)$

Simplify the expression:

$(e^{-x})^2 = e^{-2x}$

$\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1+e^{-2x}} \cdot e^{-x} \cdot (-1)$

Rearrange the terms:

$\frac{dy}{dx} = - \frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}$

The final answer is $\boxed{- \frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}}$.

Let $y = \log(\cos e^x)$.

We need to find $\frac{dy}{dx}$. This is a composite function involving a logarithm, a cosine function, and an exponential function. We will use the chain rule repeatedly.

The chain rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. For multiple layers of composition, we extend this rule.

Let's define intermediate variables:

Let $u = \cos e^x$. Then $y = \log u$.

Let $v = e^x$. Then $u = \cos v$.

So, the structure is $y = \log(\cos(e^x))$. We differentiate from the outermost function inwards.

1. Differentiate $y = \log u$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}$

2. Differentiate $u = \cos v$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v$

3. Differentiate $v = e^x$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(e^x) = e^x$

Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

Substitute the derivatives we found:

$\frac{dy}{dx} = \left(\frac{1}{u}\right) \cdot (-\sin v) \cdot (e^x)$

Substitute back the expressions for $u$ and $v$ in terms of $x$:

$v = e^x$

$u = \cos v = \cos e^x$

So, $\frac{dy}{dx} = \left(\frac{1}{\cos e^x}\right) \cdot (-\sin e^x) \cdot (e^x)$

Simplify the expression:

$\frac{dy}{dx} = - \frac{\sin e^x}{\cos e^x} \cdot e^x$

Since $\frac{\sin \theta}{\cos \theta} = \tan \theta$, we have:

$\frac{dy}{dx} = - \tan e^x \cdot e^x$

Rearrange the terms for a standard form:

$\frac{dy}{dx} = -e^x \tan e^x$

The final answer is $\boxed{-e^x \tan e^x}$.

Let $y = e^{x} + e^{x^{2}} + e^{x^{3}} + e^{x^{4}} + e^{x^{5}}$.

We need to find $\frac{dy}{dx}$. We can differentiate each term separately because the derivative of a sum is the sum of the derivatives.

$\frac{dy}{dx} = \frac{d}{dx}(e^{x}) + \frac{d}{dx}(e^{x^{2}}) + \frac{d}{dx}(e^{x^{3}}) + \frac{d}{dx}(e^{x^{4}}) + \frac{d}{dx}(e^{x^{5}})$

We differentiate each term:

1. For the first term, $\frac{d}{dx}(e^{x})$:

The derivative of $e^x$ with respect to $x$ is $e^x$.

$\frac{d}{dx}(e^{x}) = e^{x}$

2. For the second term, $\frac{d}{dx}(e^{x^{2}})$:

This requires the chain rule. Let $u = x^2$. Then the derivative is $\frac{d}{du}(e^u) \cdot \frac{du}{dx}$.

$\frac{d}{du}(e^u) = e^u$

$\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$

So, $\frac{d}{dx}(e^{x^{2}}) = e^{x^2} \cdot 2x = 2xe^{x^2}$

3. For the third term, $\frac{d}{dx}(e^{x^{3}})$:

Using the chain rule, let $u = x^3$. Then the derivative is $\frac{d}{du}(e^u) \cdot \frac{du}{dx}$.

$\frac{d}{du}(e^u) = e^u$

$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$

So, $\frac{d}{dx}(e^{x^{3}}) = e^{x^3} \cdot 3x^2 = 3x^2e^{x^3}$

4. For the fourth term, $\frac{d}{dx}(e^{x^{4}})$:

Using the chain rule, let $u = x^4$. Then the derivative is $\frac{d}{du}(e^u) \cdot \frac{du}{dx}$.

$\frac{d}{du}(e^u) = e^u$

$\frac{du}{dx} = \frac{d}{dx}(x^4) = 4x^3$

So, $\frac{d}{dx}(e^{x^{4}}) = e^{x^4} \cdot 4x^3 = 4x^3e^{x^4}$

5. For the fifth term, $\frac{d}{dx}(e^{x^{5}})$:

Using the chain rule, let $u = x^5$. Then the derivative is $\frac{d}{du}(e^u) \cdot \frac{du}{dx}$.

$\frac{d}{du}(e^u) = e^u$

$\frac{du}{dx} = \frac{d}{dx}(x^5) = 5x^4$

So, $\frac{d}{dx}(e^{x^{5}}) = e^{x^5} \cdot 5x^4 = 5x^4e^{x^{5}}$

Now, sum the derivatives of each term:

$\frac{dy}{dx} = e^{x} + 2xe^{x^{2}} + 3x^2e^{x^{3}} + 4x^3e^{x^{4}} + 5x^4e^{x^{5}}$

The final answer is $\boxed{e^{x} + 2xe^{x^{2}} + 3x^2e^{x^{3}} + 4x^3e^{x^{4}} + 5x^4e^{x^{5}}}$.

Let $y = \sqrt{e^{\sqrt{x}}}$.

We need to find $\frac{dy}{dx}$. This is a composite function involving a square root, an exponential function, and another square root. We will use the chain rule multiple times.

We can rewrite the function as $y = (e^{\sqrt{x}})^{\frac{1}{2}}$.

Let's apply the chain rule step by step, differentiating from the outermost function inwards.

The outermost function is $(\cdot)^{\frac{1}{2}}$. Its derivative is $\frac{1}{2}(\cdot)^{-\frac{1}{2}}$.

Let $u = e^{\sqrt{x}}$. Then $y = u^{\frac{1}{2}}$.

$\frac{dy}{du} = \frac{d}{du}(u^{\frac{1}{2}}) = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}$

The next layer is $e^{(\cdot)}$. Its derivative is $e^{(\cdot)}$.

Let $v = \sqrt{x}$. Then $u = e^v$.

$\frac{du}{dv} = \frac{d}{dv}(e^v) = e^v$

The innermost function is $\sqrt{x}$. Its derivative is $\frac{1}{2\sqrt{x}}$.

$v = \sqrt{x} = x^{\frac{1}{2}}$.

$\frac{dv}{dx} = \frac{d}{dx}(x^{\frac{1}{2}}) = \frac{1}{2}x^{\frac{1}{2} - 1} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.

Substitute the derivatives we found:

$\frac{dy}{dx} = \left(\frac{1}{2\sqrt{u}}\right) \cdot (e^v) \cdot \left(\frac{1}{2\sqrt{x}}\right)$

Substitute back the expressions for $u$ and $v$ in terms of $x$:

$v = \sqrt{x}$

$u = e^v = e^{\sqrt{x}}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$

Combine the terms:

$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}}$

We can simplify the expression using $\sqrt{a} = a^{\frac{1}{2}}$ and $\frac{a}{a^{\frac{1}{2}}} = a^{\frac{1}{2}} = \sqrt{a}$:

$\frac{e^{\sqrt{x}}}{\sqrt{e^{\sqrt{x}}}} = \frac{(e^{\sqrt{x}})^1}{(e^{\sqrt{x}})^{\frac{1}{2}}} = (e^{\sqrt{x}})^{1 - \frac{1}{2}} = (e^{\sqrt{x}})^{\frac{1}{2}} = \sqrt{e^{\sqrt{x}}}$

So, $\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$

The final answer is $\boxed{\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}}$.

Let $y = \log (\log x)$.

We need to find $\frac{dy}{dx}$. This is a composite function where the argument of the outer logarithm function is another logarithm function. We will use the chain rule for differentiation.

The chain rule states that if $y = f(u)$ and $u = g(x)$, then the derivative is $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

In this case, we can identify the parts for the chain rule:

Let $u = \log x$.

Then $y = \log u$.

Assuming 'log' refers to the natural logarithm (ln), we have:

First, differentiate $y = \log u$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\log u)$

The derivative of $\log u$ with respect to $u$ is $\frac{1}{u}$.

$\frac{dy}{du} = \frac{1}{u}$

Next, differentiate $u = \log x$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\log x)$

The derivative of $\log x$ with respect to $x$ is $\frac{1}{x}$.

$\frac{du}{dx} = \frac{1}{x}$

Now, apply the chain rule by multiplying $\frac{dy}{du}$ and $\frac{du}{dx}$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Substitute the expressions we found:

$\frac{dy}{dx} = \left(\frac{1}{u}\right) \cdot \left(\frac{1}{x}\right)$

Substitute back the expression for $u$, which is $u = \log x$:

$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x}$

Combine the terms:

$\frac{dy}{dx} = \frac{1}{x \log x}$

The condition $x > 1$ is given because $\log x$ must be positive for $\log(\log x)$ to be defined in real numbers.

The final answer is $\boxed{\frac{1}{x \log x}}$.

Let $y = \frac{\cos x}{\log x}$.

We need to find $\frac{dy}{dx}$. This function is in the form of a quotient, so we will use the quotient rule for differentiation.

The quotient rule states that if $y = \frac{u}{v}$, where $u$ and $v$ are differentiable functions of $x$, then the derivative is given by:

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

In this function, we identify $u$ and $v$:

$u = \cos x$

$v = \log x$ (assuming natural logarithm, ln)

Now, we find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x$

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Substitute these into the quotient rule formula:

$\frac{dy}{dx} = \frac{(\log x) \frac{d}{dx}(\cos x) - (\cos x) \frac{d}{dx}(\log x)}{(\log x)^2}$

$\frac{dy}{dx} = \frac{(\log x)(-\sin x) - (\cos x)\left(\frac{1}{x}\right)}{(\log x)^2}$

Simplify the numerator:

$\frac{dy}{dx} = \frac{- \sin x \log x - \frac{\cos x}{x}}{(\log x)^2}$

To remove the fraction in the numerator, multiply the numerator and denominator by $x$:

$\frac{dy}{dx} = \frac{x(- \sin x \log x) - \cos x}{x(\log x)^2}$

$\frac{dy}{dx} = \frac{-x \sin x \log x - \cos x}{x(\log x)^2}$

The condition $x > 0$ ensures that $\log x$ is defined and in the real number system.

The final answer is $\boxed{\frac{-x \sin x \log x - \cos x}{x(\log x)^2}}$.

Let $y = \cos (\log x + e^x)$.

We need to find $\frac{dy}{dx}$. This is a composite function of the form $\cos(u)$, where $u$ is a function of $x$. We will use the chain rule for differentiation.

The chain rule states that if $y = f(u)$ and $u = g(x)$, then the derivative of $y$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

In this function, we identify:

Outer function: $y = \cos u$

Inner function: $u = \log x + e^x$

First, we differentiate the outer function with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\cos u)$

The derivative of $\cos u$ with respect to $u$ is $-\sin u$.

$\frac{dy}{du} = -\sin u$

Next, we differentiate the inner function with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\log x + e^x)$

Using the sum rule for differentiation, we differentiate each term separately:

$\frac{d}{dx}(\log x + e^x) = \frac{d}{dx}(\log x) + \frac{d}{dx}(e^x)$

Assuming 'log' refers to the natural logarithm (ln), the standard derivatives are:

$\frac{d}{dx}(\log x) = \frac{1}{x}$

$\frac{d}{dx}(e^x) = e^x$

So, $\frac{du}{dx} = \frac{1}{x} + e^x$

Now, apply the chain rule by multiplying $\frac{dy}{du}$ and $\frac{du}{dx}$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Substitute the expressions we found:

$\frac{dy}{dx} = (-\sin u) \cdot \left(\frac{1}{x} + e^x\right)$

Substitute back the expression for $u$, which is $u = \log x + e^x$:

$\frac{dy}{dx} = -\sin (\log x + e^x) \cdot \left(\frac{1}{x} + e^x\right)$

Rearrange the terms for a standard form:

$\frac{dy}{dx} = - \left(\frac{1}{x} + e^x\right) \sin (\log x + e^x)$

The condition $x > 0$ is specified because the natural logarithm function, $\log x$, is only defined for positive values of $x$.

The final answer is $\boxed{- \left(\frac{1}{x} + e^x\right) \sin (\log x + e^x)}$.

Let $y = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$.

We need to find $\frac{dy}{dx}$. Since the function is a complex expression involving products, quotients, and powers (square root), using logarithmic differentiation will simplify the process.

Rewrite the function using exponents:

$y = \left(\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}\right)^{1/2}$

Take the natural logarithm of both sides:

$\log y = \log \left(\left(\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}\right)^{1/2}\right)$

Use the logarithm property $\log(a^n) = n \log a$:

$\log y = \frac{1}{2} \log \left(\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}\right)$

Use the logarithm properties $\log(\frac{a}{b}) = \log a - \log b$ and $\log(ab) = \log a + \log b$:

$\log y = \frac{1}{2} [\log((x − 3) (x^2 + 4)) - \log(3x^2 + 4x + 5)]$

$\log y = \frac{1}{2} [\log(x − 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5)]$

Differentiate both sides with respect to $x$. On the left side, use the chain rule: $\frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx}$. On the right side, differentiate term by term:

$\frac{d}{dx}(\log y) = \frac{d}{dx} \left(\frac{1}{2} [\log(x − 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5)]\right)$

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left[\frac{d}{dx}(\log(x − 3)) + \frac{d}{dx}(\log(x^2 + 4)) - \frac{d}{dx}(\log(3x^2 + 4x + 5))\right]$

Now, differentiate each logarithmic term using the chain rule for $\log(f(x))$, which is $\frac{f'(x)}{f(x)}$:

$\frac{d}{dx}(\log(x − 3)) = \frac{\frac{d}{dx}(x-3)}{x-3} = \frac{1}{x-3}$

$\frac{d}{dx}(\log(x^2 + 4)) = \frac{\frac{d}{dx}(x^2+4)}{x^2+4} = \frac{2x}{x^2+4}$

$\frac{d}{dx}(\log(3x^2 + 4x + 5)) = \frac{\frac{d}{dx}(3x^2+4x+5)}{3x^2+4x+5} = \frac{6x+4}{3x^2+4x+5}$

Substitute these derivatives back into the equation for $\frac{1}{y}\frac{dy}{dx}$:

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left[\frac{1}{x-3} + \frac{2x}{x^2 + 4} - \frac{6x + 4}{3x^2 + 4x + 5}\right]$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \cdot \frac{1}{2} \left[\frac{1}{x-3} + \frac{2x}{x^2 + 4} - \frac{6x + 4}{3x^2 + 4x + 5}\right]$

Substitute the original expression for $y$:

$\frac{dy}{dx} = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \cdot \frac{1}{2} \left[\frac{1}{x-3} + \frac{2x}{x^2 + 4} - \frac{6x + 4}{3x^2 + 4x + 5}\right]$

The final answer is $\boxed{\frac{1}{2} \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \left[\frac{1}{x-3} + \frac{2x}{x^2 + 4} - \frac{6x + 4}{3x^2 + 4x + 5}\right]}$.

Let $y = a^x$, where $a$ is a positive constant.

We need to find $\frac{dy}{dx}$. We can use logarithmic differentiation to find the derivative of this exponential function with a constant base and variable exponent.

Take the natural logarithm of both sides of the equation:

$\log y = \log(a^x)$

Use the logarithm property $\log(b^n) = n \log b$ to bring the exponent $x$ down:

$\log y = x \log a$

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule, and on the right side, $\log a$ is a constant.

$\frac{d}{dx}(\log y) = \frac{d}{dx}(x \log a)$

Applying the chain rule on the left side ($\frac{d}{dx}(\log y) = \frac{d}{dy}(\log y) \cdot \frac{dy}{dx}$):

$\frac{d}{dy}(\log y) = \frac{1}{y}$

So, $\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$

On the right side, since $\log a$ is a constant, we use the constant multiple rule:

$\frac{d}{dx}(x \log a) = \log a \cdot \frac{d}{dx}(x)$

The derivative of $x$ with respect to $x$ is $1$:

$\frac{d}{dx}(x) = 1$

So, $\frac{d}{dx}(x \log a) = \log a \cdot 1 = \log a$

Equating the derivatives of both sides:

$\frac{1}{y} \frac{dy}{dx} = \log a$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \log a$

Substitute back the original expression for $y$, which is $a^x$:

$\frac{dy}{dx} = a^x \log a$

Thus, the derivative of $a^x$ with respect to $x$ is $a^x \log a$. Note that if $a=e$, $\log e = 1$, and the formula becomes $e^x \cdot 1 = e^x$, which is consistent with the known derivative of $e^x$.

The final answer is $\boxed{a^x \log a}$.

Let $y = x^{\sin x}$.

We need to find $\frac{dy}{dx}$. The function is in the form $u(x)^{v(x)}$, so we will use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log y = \log(x^{\sin x})$

Use the logarithm property $\log(a^b) = b \log a$:

$\log y = (\sin x) (\log x)$

Now, differentiate both sides with respect to $x$. The left side requires the chain rule, and the right side requires the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

$\frac{d}{dx}(\log y) = \frac{d}{dx}(\sin x \log x)$

Differentiating the left side (using chain rule):

$\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$

Differentiating the right side (using product rule with $u = \sin x$ and $v = \log x$):

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

$\frac{d}{dx}(\sin x \log x) = (\sin x) \cdot \left(\frac{1}{x}\right) + (\log x) \cdot (\cos x)$

$\frac{d}{dx}(\sin x \log x) = \frac{\sin x}{x} + \cos x \log x$

Equate the derivatives of both sides:

$\frac{1}{y} \frac{dy}{dx} = \frac{\sin x}{x} + \cos x \log x$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left(\frac{\sin x}{x} + \cos x \log x\right)$

Substitute back the original expression for $y$, which is $x^{\sin x}$:

$\frac{dy}{dx} = x^{\sin x} \left(\frac{\sin x}{x} + \cos x \log x\right)$

The condition $x > 0$ is necessary for $\log x$ to be defined in real numbers and for $x^{\sin x}$ to be generally defined for a real base.

The final answer is $\boxed{x^{\sin x} \left(\frac{\sin x}{x} + \cos x \log x\right)}$.

The given equation is $y^x + x^y + x^x = a^b$.

We need to find the derivative $\frac{dy}{dx}$. Here, $y$ is considered a function of $x$. The term $a^b$, where $a$ and $b$ are constants, is also a constant.

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{dx}(y^x + x^y + x^x) = \frac{d}{dx}(a^b)$

Using the sum rule for differentiation, we differentiate each term separately:

$\frac{d}{dx}(y^x) + \frac{d}{dx}(x^y) + \frac{d}{dx}(x^x) = \frac{d}{dx}(a^b)$

Since $a^b$ is a constant, its derivative is 0:

$\frac{d}{dx}(a^b) = 0$

We need to find the derivatives of the three terms on the left side. These are in the form of functions raised to the power of another function, so we use logarithmic differentiation for each term.

Let $u = y^x$, $v = x^y$, and $w = x^x$. We need to find $\frac{du}{dx}$, $\frac{dv}{dx}$, and $\frac{dw}{dx}$.

For $\frac{du}{dx}$ (where $u = y^x$):

Take logarithm on both sides: $\log u = \log(y^x) = x \log y$.

Differentiate implicitly with respect to $x$ using the product rule on the right side:

$\frac{1}{u}\frac{du}{dx} = \frac{d}{dx}(x \log y)$

$\frac{1}{u}\frac{du}{dx} = (1) \cdot \log y + x \cdot \left(\frac{1}{y}\frac{dy}{dx}\right)$

$\frac{1}{u}\frac{du}{dx} = \log y + \frac{x}{y}\frac{dy}{dx}$

Substitute $u = y^x$ back and solve for $\frac{du}{dx}$:

$\frac{du}{dx} = y^x \left(\log y + \frac{x}{y}\frac{dy}{dx}\right) = y^x \log y + y^x \cdot \frac{x}{y}\frac{dy}{dx} = y^x \log y + x y^{x-1}\frac{dy}{dx}$

For $\frac{dv}{dx}$ (where $v = x^y$):

Take logarithm on both sides: $\log v = \log(x^y) = y \log x$.

Differentiate implicitly with respect to $x$ using the product rule on the right side:

$\frac{1}{v}\frac{dv}{dx} = \frac{d}{dx}(y \log x)$

$\frac{1}{v}\frac{dv}{dx} = \left(\frac{dy}{dx}\right) \cdot \log x + y \cdot \left(\frac{1}{x}\right)$

$\frac{1}{v}\frac{dv}{dx} = \log x \frac{dy}{dx} + \frac{y}{x}$

Substitute $v = x^y$ back and solve for $\frac{dv}{dx}$:

$\frac{dv}{dx} = x^y \left(\log x \frac{dy}{dx} + \frac{y}{x}\right) = x^y \log x \frac{dy}{dx} + x^y \cdot \frac{y}{x} = x^y \log x \frac{dy}{dx} + y x^{y-1}$

For $\frac{dw}{dx}$ (where $w = x^x$):

Take logarithm on both sides: $\log w = \log(x^x) = x \log x$.

Differentiate with respect to $x$ using the product rule on the right side:

$\frac{1}{w}\frac{dw}{dx} = \frac{d}{dx}(x \log x)$

$\frac{1}{w}\frac{dw}{dx} = (1) \cdot \log x + x \cdot \left(\frac{1}{x}\right)$

$\frac{1}{w}\frac{dw}{dx} = \log x + 1$

Substitute $w = x^x$ back and solve for $\frac{dw}{dx}$:

$\frac{dw}{dx} = x^x (\log x + 1)$

Now, substitute these derivatives back into the differentiated equation $\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$:

$(y^x \log y + x y^{x-1}\frac{dy}{dx}) + (x^y \log x \frac{dy}{dx} + y x^{y-1}) + (x^x (\log x + 1)) = 0$

Group the terms containing $\frac{dy}{dx}$ and the terms that do not:

$(x y^{x-1}\frac{dy}{dx} + x^y \log x \frac{dy}{dx}) + (y^x \log y + y x^{y-1} + x^x (\log x + 1)) = 0$

Factor out $\frac{dy}{dx}$ from the first group:

$\frac{dy}{dx} (x y^{x-1} + x^y \log x) = - (y^x \log y + y x^{y-1} + x^x (\log x + 1))$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = - \frac{y^x \log y + y x^{y-1} + x^x (\log x + 1)}{x y^{x-1} + x^y \log x}$

The final answer is $\boxed{- \frac{y^x \log y + y x^{y-1} + x^x (\log x + 1)}{x y^{x-1} + x^y \log x}}$.

Given:

The function to differentiate is:

$y = \cos x \cdot \cos 2x \cdot \cos 3x$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = \cos x \cdot \cos 2x \cdot \cos 3x$.

Taking logarithm on both sides, we get:

Using the property of logarithm, $\log(abc) = \log a + \log b + \log c$:

Now, differentiate both sides of equation (ii) with respect to $x$:

Using the chain rule $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$:

Now, differentiate the cosine terms:

Substitute (v), (vi), and (vii) into (iv):

Simplify the terms using $\frac{\sin \theta}{\cos \theta} = \tan \theta$:

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

Substitute the original expression for $y$ back into the equation:

Alternate Form of the Answer:

We can distribute the $\cos x \cos 2x \cos 3x$ term:

$ \frac{dy}{dx} = - \cos x \cos 2x \cos 3x \tan x - 2 \cos x \cos 2x \cos 3x \tan 2x - 3 \cos x \cos 2x \cos 3x \tan 3x $

Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$ \frac{dy}{dx} = - \cos x \cos 2x \cos 3x \frac{\sin x}{\cos x} - 2 \cos x \cos 2x \cos 3x \frac{\sin 2x}{\cos 2x} - 3 \cos x \cos 2x \cos 3x \frac{\sin 3x}{\cos 3x} $

Cancel the common cosine terms:

$ \frac{dy}{dx} = - \cancel{\cos x} \cos 2x \cos 3x \frac{\sin x}{\cancel{\cos x}} - 2 \cos x \cancel{\cos 2x} \cos 3x \frac{\sin 2x}{\cancel{\cos 2x}} - 3 \cos x \cos 2x \cancel{\cos 3x} \frac{\sin 3x}{\cancel{\cos 3x}} $

$ \frac{dy}{dx} = - \sin x \cos 2x \cos 3x - 2 \cos x \sin 2x \cos 3x - 3 \cos x \cos 2x \sin 3x $

Final Answer:

The derivative of $\cos x \cdot \cos 2x \cdot \cos 3x$ with respect to $x$ is:

$ \frac{dy}{dx} = (\cos x \cos 2x \cos 3x) (-\tan x - 2\tan 2x - 3\tan 3x) $

or

$ \frac{dy}{dx} = -(\sin x \cos 2x \cos 3x + 2 \cos x \sin 2x \cos 3x + 3 \cos x \cos 2x \sin 3x) $

Given:

The function to differentiate is:

$y = \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = \left(\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}\right)^{1/2}$.

Taking the natural logarithm on both sides:

Using the property $\log(a^n) = n \log a$:

Using the property $\log(a/b) = \log a - \log b$:

Using the property $\log(ab) = \log a + \log b$:

Simplifying:

Now, differentiate both sides of equation (v) with respect to $x$:

Using the chain rule $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(x - a) = 1$:

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

Substitute the original expression for $y$ back into the equation:

Final Answer:

The derivative of $\sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}}$ with respect to $x$ is:

$ \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \left( \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5} \right) $

Given:

The function to differentiate is:

$y = (\log x)^{\cos x}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = (\log x)^{\cos x}$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(a^b) = b \log a$:

Now, differentiate both sides of equation (ii) with respect to $x$. On the left side, use the chain rule. On the right side, use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u = \cos x$ and $v = \log(\log x)$.

Differentiating the left side:

$ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} $

Differentiating the right side:

$ \frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x $

$ \frac{dv}{dx} = \frac{d}{dx}(\log(\log x)) $

Using the chain rule $\frac{d}{dx}(\log w) = \frac{1}{w} \frac{dw}{dx}$ with $w = \log x$, so $\frac{dw}{dx} = \frac{1}{x}$:

$ \frac{dv}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} $

Applying the product rule to the right side:

$ \frac{d}{dx}(\cos x \cdot \log(\log x)) = (-\sin x) \cdot \log(\log x) + (\cos x) \cdot \left(\frac{1}{x \log x}\right) $

$ \frac{d}{dx}(\cos x \cdot \log(\log x)) = -\sin x \log(\log x) + \frac{\cos x}{x \log x} $

Equating the derivatives of both sides:

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

Substitute the original expression for $y = (\log x)^{\cos x}$ back into equation (iv):

Final Answer:

The derivative of $(\log x)^{\cos x}$ with respect to $x$ is:

$ \frac{dy}{dx} = (\log x)^{\cos x} \left( -\sin x \log(\log x) + \frac{\cos x}{x \log x} \right) $

Given:

The function to differentiate is:

$y = x^x - 2^{\sin x}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = x^x - 2^{\sin x}$.

We can write this as $y = u - v$, where $u = x^x$ and $v = 2^{\sin x}$.

Then, $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.

First, let's find $\frac{du}{dx}$ for $u = x^x$.

Take the natural logarithm of both sides:

Using the property $\log(a^b) = b \log a$:

Differentiate both sides of equation (ii) with respect to $x$. Use the chain rule on the left and the product rule on the right:

$ \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} $

$ \frac{d}{dx}(x \log x) = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) $

$ \frac{d}{dx}(x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} $

$ \frac{d}{dx}(x \log x) = \log x + 1 $

So, equating the derivatives:

Multiply by $u$ to find $\frac{du}{dx}$:

Substitute $u = x^x$ back into equation (iv):

Next, let's find $\frac{dv}{dx}$ for $v = 2^{\sin x}$.

This is of the form $a^{f(x)}$, where $a=2$ and $f(x) = \sin x$.

The derivative of $a^{f(x)}$ is $a^{f(x)} \log a \cdot f'(x)$.

Here, $f'(x) = \frac{d}{dx}(\sin x) = \cos x$.

So, the derivative of $v$ is:

Now, combine the results from (v) and (vi) to find $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$:

Final Answer:

The derivative of $x^x - 2^{\sin x}$ with respect to $x$ is:

$ \frac{dy}{dx} = x^x (1 + \log x) - 2^{\sin x} \log 2 \cos x $

Given:

The function to differentiate is:

$y = (x + 3)^2 (x + 4)^3 (x + 5)^4$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = (x + 3)^2 (x + 4)^3 (x + 5)^4$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(abc) = \log a + \log b + \log c$:

Using the property of logarithm $\log(a^b) = b \log a$:

Now, differentiate both sides of equation (iii) with respect to $x$. On the left side, use the chain rule $\frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx}$. On the right side, use the sum rule and chain rule.

Applying the differentiation:

Since $\frac{d}{dx}(x + a) = 1$:

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

Substitute the original expression for $y$ back into equation (vii):

Final Answer:

The derivative of $(x + 3)^2 (x + 4)^3 (x + 5)^4$ with respect to $x$ is:

$ \frac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right) $

Given:

The function to differentiate is:

$y = \left( x + \frac{1}{x} \right)^x + x^{\left(1 + \frac{1}{x} \right)}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = \left( x + \frac{1}{x} \right)^x + x^{\left(1 + \frac{1}{x} \right)}$.

Let $u = \left( x + \frac{1}{x} \right)^x$ and $v = x^{\left(1 + \frac{1}{x} \right)}$.

Then $y = u + v$.

Differentiating with respect to $x$, we get:

First, consider $u = \left( x + \frac{1}{x} \right)^x$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiating both sides of equation (ii) with respect to $x$. Use the chain rule on the left and the product rule on the right:

Using the product rule $\frac{d}{dx}(fg) = f'g + fg'$ with $f=x$ and $g=\log \left( x + \frac{1}{x} \right)$:

$ f' = \frac{d}{dx}(x) = 1 $

$ g' = \frac{d}{dx}\left(\log \left( x + \frac{1}{x} \right)\right) = \frac{1}{x + \frac{1}{x}} \cdot \frac{d}{dx}\left( x + \frac{1}{x} \right) $

$ \frac{d}{dx}\left( x + \frac{1}{x} \right) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} $

$ g' = \frac{1}{\frac{x^2 + 1}{x}} \cdot \frac{x^2 - 1}{x^2} = \frac{x}{x^2 + 1} \cdot \frac{x^2 - 1}{x^2} = \frac{x^2 - 1}{x(x^2 + 1)} $

So, the right side of (iii) is:

$ 1 \cdot \log \left( x + \frac{1}{x} \right) + x \cdot \frac{x^2 - 1}{x(x^2 + 1)} = \log \left( x + \frac{1}{x} \right) + \frac{x^2 - 1}{x^2 + 1} $

Multiply by $u$:

Substitute $u = \left( x + \frac{1}{x} \right)^x$ back:

Next, consider $v = x^{\left(1 + \frac{1}{x} \right)}$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiating both sides of equation (vii) with respect to $x$. Use the chain rule on the left and the product rule on the right:

Using the product rule $\frac{d}{dx}(fg) = f'g + fg'$ with $f=1 + \frac{1}{x}$ and $g=\log x$:

$ f' = \frac{d}{dx}\left(1 + \frac{1}{x}\right) = \frac{d}{dx}(1 + x^{-1}) = 0 - x^{-2} = -\frac{1}{x^2} $

$ g' = \frac{d}{dx}(\log x) = \frac{1}{x} $

So, the right side of (viii) is:

$ \left(-\frac{1}{x^2}\right) \log x + \left(1 + \frac{1}{x}\right) \left(\frac{1}{x}\right) = -\frac{\log x}{x^2} + \frac{1}{x} + \frac{1}{x^2} = \frac{x + 1 - \log x}{x^2} $

Multiply by $v$:

Substitute $v = x^{\left(1 + \frac{1}{x} \right)}$ back:

Now, substitute the expressions for $\frac{du}{dx}$ from (vi) and $\frac{dv}{dx}$ from (xi) into equation (i):

Final Answer:

The derivative of $\left( x + \frac{1}{x} \right)^x + x^{\left(1 + \frac{1}{x} \right)}$ with respect to $x$ is:

$ \frac{dy}{dx} = \left( x + \frac{1}{x} \right)^x \left( \log \left( x + \frac{1}{x} \right) + \frac{x^2 - 1}{x^2 + 1} \right) + x^{\left(1 + \frac{1}{x} \right)} \left( \frac{x + 1 - \log x}{x^2} \right) $

Given:

The function to differentiate is:

$y = (\log x)^x + x^{\log x}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = (\log x)^x + x^{\log x}$.

Let $u = (\log x)^x$ and $v = x^{\log x}$.

Then $y = u + v$.

Differentiating with respect to $x$, we get:

First, consider $u = (\log x)^x$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiate both sides of equation (iii) with respect to $x$. Use the chain rule on the left and the product rule on the right $\frac{d}{dx}(fg) = f'g + fg'$ with $f=x$ and $g=\log(\log x)$.

$ f' = \frac{d}{dx}(x) = 1 $

$ g' = \frac{d}{dx}(\log(\log x)) = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} $

Applying the product rule to the right side of (iii):

Multiply both sides by $u$:

Substitute $u = (\log x)^x$ back into equation (vi):

Next, consider $v = x^{\log x}$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiate both sides of equation (ix) with respect to $x$. Use the chain rule on both sides:

Using the chain rule $\frac{d}{dx}(w^n) = nw^{n-1}\frac{dw}{dx}$ with $w = \log x$ and $n=2$, and $\frac{dw}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$:

Multiply both sides by $v$:

Substitute $v = x^{\log x}$ back into equation (xii):

Now, substitute the expressions for $\frac{du}{dx}$ from (vii) and $\frac{dv}{dx}$ from (xiii) into equation (i):

Final Answer:

The derivative of $(\log x)^x + x^{\log x}$ with respect to $x$ is:

$ \frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) + x^{\log x} \left( \frac{2 \log x}{x} \right) $

Given:

The function to differentiate is:

$y = (\sin x)^x + \sin^{-1} \sqrt{x}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = (\sin x)^x + \sin^{-1} \sqrt{x}$.

Let $u = (\sin x)^x$ and $v = \sin^{-1} \sqrt{x}$.

Then $y = u + v$.

Differentiating with respect to $x$, we get:

First, consider $u = (\sin x)^x$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiate both sides of equation (iii) with respect to $x$. Use the chain rule on the left and the product rule on the right $\frac{d}{dx}(fg) = f'g + fg'$ with $f=x$ and $g=\log(\sin x)$.

$ f' = \frac{d}{dx}(x) = 1 $

$ g' = \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x $

Applying the product rule to the right side of (iii):

Multiply both sides by $u$:

Substitute $u = (\sin x)^x$ back into equation (vi):

Next, consider $v = \sin^{-1} \sqrt{x}$.

Use the chain rule. The derivative of $\sin^{-1} w$ with respect to $w$ is $\frac{1}{\sqrt{1 - w^2}}$. Here $w = \sqrt{x}$.

Applying the chain rule:

We know that $(\sqrt{x})^2 = x$ and $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

Substitute these into (ix):

Now, substitute the expressions for $\frac{du}{dx}$ from (vii) and $\frac{dv}{dx}$ from (xi) into equation (i):

Final Answer:

The derivative of $(\sin x)^x + \sin^{-1} \sqrt{x}$ with respect to $x$ is:

$ \frac{dy}{dx} = (\sin x)^x (\log(\sin x) + x \cot x) + \frac{1}{2\sqrt{x(1 - x)}} $

Given:

The function to differentiate is:

$y = x^{\sin x} + (\sin x)^{\cos x}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = x^{\sin x} + (\sin x)^{\cos x}$.

Let $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$.

Then $y = u + v$.

Differentiating with respect to $x$, we get:

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

First, let's find $\frac{du}{dx}$ for $u = x^{\sin x}$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(a^b) = b \log a$:

Differentiate both sides of equation (2) with respect to $x$. Use the chain rule on the left and the product rule on the right $\frac{d}{dx}(fg) = f'g + fg'$ with $f=\sin x$ and $g=\log x$.

We know $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\log x) = \frac{1}{x}$. Substitute these into (3):

Multiply both sides by $u$:

Substitute $u = x^{\sin x}$ back into equation (5):

Next, consider $v = (\sin x)^{\cos x}$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(a^b) = b \log a$:

Differentiate both sides of equation (8) with respect to $x$. Use the chain rule on the left and the product rule on the right $\frac{d}{dx}(fg) = f'g + fg'$ with $f=\cos x$ and $g=\log(\sin x)$.

We know $\frac{d}{dx}(\cos x) = -\sin x$.

For $\frac{d}{dx}(\log(\sin x))$, use the chain rule: $\frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

Substitute these into (9):

Multiply both sides by $v$:

Substitute $v = (\sin x)^{\cos x}$ back into equation (11):

Now, substitute the expressions for $\frac{du}{dx}$ from (6) and $\frac{dv}{dx}$ from (12) into the equation $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$:

Final Answer:

The derivative of $x^{\sin x} + (\sin x)^{\cos x}$ with respect to $x$ is:

$ \frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} (-\sin x \log(\sin x) + \cos x \cot x) $

Given:

The function to differentiate is:

$y = x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}$.

Let $u = x^{x \cos x}$ and $v = \frac{x^2 + 1}{x^2 - 1}$.

Then $y = u + v$.

Differentiating with respect to $x$, we get:

First, let's find $\frac{du}{dx}$ for $u = x^{x \cos x}$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(a^b) = b \log a$:

Differentiate both sides of equation (iii) with respect to $x$. Use the chain rule on the left and the product rule for three functions on the right: $\frac{d}{dx}(fgh) = f'gh + fg'h + fgh'$, where $f=x$, $g=\cos x$, $h=\log x$.

We have $\frac{d}{dx}(x) = 1$, $\frac{d}{dx}(\cos x) = -\sin x$, and $\frac{d}{dx}(\log x) = \frac{1}{x}$. Substitute these into (iv):

Multiply both sides by $u$:

Substitute $u = x^{x \cos x}$ back into equation (vii):

Next, let's find $\frac{dv}{dx}$ for $v = \frac{x^2 + 1}{x^2 - 1}$.

Use the quotient rule: $\frac{d}{dx}\left(\frac{p}{q}\right) = \frac{p'q - pq'}{q^2}$. Here $p = x^2 + 1$ and $q = x^2 - 1$.

$ p' = \frac{d}{dx}(x^2 + 1) = 2x $

$ q' = \frac{d}{dx}(x^2 - 1) = 2x $

Simplify the numerator:

$ (2x)(x^2 - 1) - (x^2 + 1)(2x) = 2x^3 - 2x - (2x^3 + 2x) = 2x^3 - 2x - 2x^3 - 2x = -4x $

Now, substitute the expressions for $\frac{du}{dx}$ from (viii) and $\frac{dv}{dx}$ from (x) into equation (i):

Note: The term $\cos x \log x - x \sin x \log x + \cos x$ can be rewritten as $\cos x (1 + \log x) - x \sin x \log x$. The form in the final answer can vary.

Final Answer:

The derivative of $x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}$ with respect to $x$ is:

$ \frac{dy}{dx} = x^{x \cos x} (\cos x \log x - x \sin x \log x + \cos x) - \frac{4x}{(x^2 - 1)^2} $

Alternatively:

$ \frac{dy}{dx} = x^{x \cos x} (\cos x (1 + \log x) - x \sin x \log x) - \frac{4x}{(x^2 - 1)^2} $

Given:

The function to differentiate is:

$y = (x \cos x)^x + (x \sin x)^{\frac{1}{x}}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Given the function $y = (x \cos x)^x + (x \sin x)^{\frac{1}{x}}$.

Let $u = (x \cos x)^x$ and $v = (x \sin x)^{\frac{1}{x}}$.

Then $y = u + v$.

Differentiating with respect to $x$, we get:

First, consider $u = (x \cos x)^x$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Using the property $\log(ab) = \log a + \log b$:

Differentiate both sides of equation (iv) with respect to $x$. Use the chain rule on the left and the product rule on the right.

For $\frac{d}{dx}(x \log x)$, use the product rule: $\frac{d}{dx}(x \log x) = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.

For $\frac{d}{dx}(x \log(\cos x))$, use the product rule and chain rule: $\frac{d}{dx}(x \log(\cos x)) = \frac{d}{dx}(x) \cdot \log(\cos x) + x \cdot \frac{d}{dx}(\log(\cos x))$.

$ \frac{d}{dx}(\log(\cos x)) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x $

So, $\frac{d}{dx}(x \log(\cos x)) = 1 \cdot \log(\cos x) + x (-\tan x) = \log(\cos x) - x \tan x$.

Substituting these into equation (v):

Using the property $\log a + \log b = \log(ab)$:

Multiply both sides by $u$:

Substitute $u = (x \cos x)^x$ back into equation (ix):

Next, consider $v = (x \sin x)^{\frac{1}{x}}$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Using the property $\log(ab) = \log a + \log b$:

Differentiate both sides of equation (xiii) with respect to $x$. Use the chain rule on the left and the product rule on the right $\frac{d}{dx}(fg) = f'g + fg'$ with $f = \frac{1}{x} = x^{-1}$ and $g = \log x + \log(\sin x)$.

$ f' = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2} $

$ g' = \frac{d}{dx}(\log x + \log(\sin x)) = \frac{d}{dx}(\log x) + \frac{d}{dx}(\log(\sin x)) $

$ \frac{d}{dx}(\log x) = \frac{1}{x} $

$ \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x $

So, $g' = \frac{1}{x} + \cot x$.

Applying the product rule to the right side of (xiii):

Rewrite with a common denominator $x^2$:

Using the property $\log a + \log b = \log(ab)$:

Multiply both sides by $v$:

Substitute $v = (x \sin x)^{\frac{1}{x}}$ back into equation (xix):

Now, substitute the expressions for $\frac{du}{dx}$ from (x) and $\frac{dv}{dx}$ from (xx) into equation (i):

Final Answer:

The derivative of $(x \cos x)^x + (x \sin x)^{\frac{1}{x}}$ with respect to $x$ is:

$ \frac{dy}{dx} = (x \cos x)^x (\log(x \cos x) + 1 - x \tan x) + (x \sin x)^{\frac{1}{x}} \left( \frac{1 + x \cot x - \log(x \sin x)}{x^2} \right) $

Given:

The equation is:

$x^y + y^x = 1$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

Given the equation $x^y + y^x = 1$.

Let $u = x^y$ and $v = y^x$. The equation becomes $u + v = 1$.

Differentiating both sides with respect to $x$ implicitly:

This gives:

First, let's find $\frac{du}{dx}$ for $u = x^y$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiate both sides of equation (iii) with respect to $x$. Use the chain rule on the left and the product rule on the right:

Multiply by $u$ to find $\frac{du}{dx}$:

Substitute $u = x^y$ back:

Next, let's find $\frac{dv}{dx}$ for $v = y^x$.

Taking the natural logarithm on both sides:

Using the property $\log(a^b) = b \log a$:

Differentiate both sides of equation (v) with respect to $x$. Use the chain rule on the left and the product rule on the right:

Note that $\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$ by the chain rule.

Multiply by $v$ to find $\frac{dv}{dx}$:

Substitute $v = y^x$ back:

Substitute the expressions for $\frac{du}{dx}$ from (iv) and $\frac{dv}{dx}$ from (vi) into equation (ii):

Rearrange the terms to group $\frac{dy}{dx}$ on one side:

Solve for $\frac{dy}{dx}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} $

Given:

The equation is:

$y^x = x^y$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

Given the equation $y^x = x^y$.

Since both sides have variables in the exponent, we take the natural logarithm of both sides:

Using the property of logarithm $\log(a^b) = b \log a$, we bring the exponents down:

Now, differentiate both sides of equation (ii) implicitly with respect to $x$. We use the product rule on both sides.

Differentiating the left side, $x \log y$, with respect to $x$:

$ \frac{d}{dx}(x \log y) = \frac{d}{dx}(x) \cdot \log y + x \cdot \frac{d}{dx}(\log y) $

Using $\frac{d}{dx}(x) = 1$ and the chain rule $\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$:

$ \frac{d}{dx}(x \log y) = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} = \log y + \frac{x}{y} \frac{dy}{dx} $

Differentiating the right side, $y \log x$, with respect to $x$:

$ \frac{d}{dx}(y \log x) = \frac{d}{dx}(y) \cdot \log x + y \cdot \frac{d}{dx}(\log x) $

Using $\frac{d}{dx}(y) = \frac{dy}{dx}$ and $\frac{d}{dx}(\log x) = \frac{1}{x}$:

$ \frac{d}{dx}(y \log x) = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} = \log x \frac{dy}{dx} + \frac{y}{x} $

Equating the derivatives of both sides:

Group the terms containing $\frac{dy}{dx}$ on one side and the other terms on the other side:

Factor out $\frac{dy}{dx}$ from the left side:

Find a common denominator inside the parenthesis on the left and on the right side:

$ \frac{x}{y} - \log x = \frac{x}{y} - \frac{y \log x}{y} = \frac{x - y \log x}{y} $

$ \frac{y}{x} - \log y = \frac{y}{x} - \frac{x \log y}{x} = \frac{y - x \log y}{x} $

Substitute these back into equation (v):

Solve for $\frac{dy}{dx}$ by multiplying both sides by $\frac{y}{x - y \log x}$:

Alternate Form:

From equation (ii), $x \log y = y \log x$. This implies $\log y = \frac{y}{x} \log x$ and $\log x = \frac{x}{y} \log y$.

We can substitute these into the expression for $\frac{dy}{dx}$ in equation (viii).

Numerator: $y - x \log y = y - x \left(\frac{y}{x} \log x\right) = y - y \log x = y(1 - \log x)$.

Denominator: $x - y \log x = x - y \left(\frac{x}{y} \log y\right) = x - x \log y = x(1 - \log y)$.

Substituting these into equation (viii):

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \frac{y(y - x \log y)}{x(x - y \log x)} $

or

$ \frac{dy}{dx} = \frac{y(1 - \log x)}{x(1 - \log y)} $

Given:

The equation is:

$(\cos x)^y = (\cos y)^x$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

Given the equation $(\cos x)^y = (\cos y)^x$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(a^b) = b \log a$:

Now, differentiate both sides of equation (ii) implicitly with respect to $x$. Use the product rule on both sides: $\frac{d}{dx}(fg) = f'g + fg'$.

Differentiating the left side:

$ \frac{d}{dx}(y \log(\cos x)) = \frac{dy}{dx} \cdot \log(\cos x) + y \cdot \frac{d}{dx}(\log(\cos x)) $

Using the chain rule $\frac{d}{dx}(\log(\cos x)) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.

So, the left side derivative is $\frac{dy}{dx} \log(\cos x) - y \tan x$.

Differentiating the right side:

$ \frac{d}{dx}(x \log(\cos y)) = \frac{d}{dx}(x) \cdot \log(\cos y) + x \cdot \frac{d}{dx}(\log(\cos y)) $

Using $\frac{d}{dx}(x) = 1$ and the chain rule $\frac{d}{dx}(\log(\cos y)) = \frac{1}{\cos y} \cdot \frac{d}{dx}(\cos y) = \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx} = -\tan y \frac{dy}{dx}$.

So, the right side derivative is $1 \cdot \log(\cos y) + x (-\tan y \frac{dy}{dx}) = \log(\cos y) - x \tan y \frac{dy}{dx}$.

Equating the derivatives of both sides from equation (iii):

Rearrange the terms to group $\frac{dy}{dx}$ on one side and the other terms on the other side:

Factor out $\frac{dy}{dx}$ from the left side:

Solve for $\frac{dy}{dx}$ by dividing both sides by $(\log(\cos x) + x \tan y)$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y} $

Given:

The equation is:

$xy = e^{x - y}$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

Given the equation $xy = e^{x - y}$.

Differentiate both sides of the equation implicitly with respect to $x$:

On the left side, use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u=x$ and $v=y$:

$ \frac{d}{dx}(xy) = \frac{dx}{dx} \cdot y + x \cdot \frac{dy}{dx} = 1 \cdot y + x \frac{dy}{dx} = y + x \frac{dy}{dx} $

On the right side, use the chain rule $\frac{d}{dx}(e^u) = e^u \frac{du}{dx}$ with $u=x-y$:

$ \frac{d}{dx}(e^{x - y}) = e^{x - y} \cdot \frac{d}{dx}(x - y) $

Calculate the derivative of the exponent:

$ \frac{d}{dx}(x - y) = \frac{dx}{dx} - \frac{dy}{dx} = 1 - \frac{dy}{dx} $

Substitute this back into the right side derivative:

$ \frac{d}{dx}(e^{x - y}) = e^{x - y} (1 - \frac{dy}{dx}) $

Equate the derivatives of both sides from equation (i):

Distribute $e^{x - y}$ on the right side:

Group terms containing $\frac{dy}{dx}$ on the left side and other terms on the right side:

Factor out $\frac{dy}{dx}$ from the left side:

Solve for $\frac{dy}{dx}$:

Alternate Form:

From the original equation, we have $e^{x - y} = xy$. Substitute this into the expression for $\frac{dy}{dx}$ in equation (vi):

Factor out common terms in the numerator and denominator:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \frac{e^{x - y} - y}{x + e^{x - y}} $

or

$ \frac{dy}{dx} = \frac{y(x - 1)}{x(1 + y)} $

Given:

The function is:

$f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$

To Find:

The derivative $f'(x)$ and the value of $f'(1)$.

Solution:

Let $y = f(x)$. We have:

$y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$

To find the derivative, we can use logarithmic differentiation.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(abcd) = \log a + \log b + \log c + \log d$:

Now, differentiate both sides of equation (ii) with respect to $x$. Use the chain rule on both sides.

Applying the differentiation rules $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$:

Calculate the derivatives of the terms inside the logarithms:

$ \frac{d}{dx}(1 + x) = 1 $

$ \frac{d}{dx}(1 + x^2) = 2x $

$ \frac{d}{dx}(1 + x^4) = 4x^3 $

$ \frac{d}{dx}(1 + x^8) = 8x^7 $

Substitute these results back into equation (iv):

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

Substitute the original expression for $y = f(x)$ back into equation (vii):

Now, we need to find $f'(1)$. We substitute $x=1$ into the expression for $f'(x)$.

First, evaluate $f(1)$:

Now, substitute $x=1$ into the expression for $f'(x)$ derived in equation (vii):

Substitute $f(1)=16$ and simplify the terms in the parenthesis:

Substitute these values back into equation (ix):

Add the terms in the parenthesis:

$ \frac{1}{2} + 7 = \frac{1}{2} + \frac{14}{2} = \frac{1 + 14}{2} = \frac{15}{2} $

Substitute this back into equation (xi):

Final Answer:

The derivative of the function is:

$ f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left( \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right) $

And the value of $f'(1)$ is:

$ f'(1) = 120 $

Given:

The function is $f(x) = (x^2 - 5x + 8) (x^3 + 7x + 9)$. Let $y = f(x)$.

To Find:

The derivative $\frac{dy}{dx}$ using three different methods and determine if the results are the same.

Solution:

We will find the derivative of $y = (x^2 - 5x + 8) (x^3 + 7x + 9)$ using three methods.

Method (i): By using product rule

The product rule states that if $y = u \cdot v$, then $\frac{dy}{dx} = u'v + uv'$.

Let $u = x^2 - 5x + 8$ and $v = x^3 + 7x + 9$.

Find the derivatives of $u$ and $v$ with respect to $x$:

Apply the product rule:

Expand and simplify:

$ (2x - 5)(x^3 + 7x + 9) = 2x(x^3 + 7x + 9) - 5(x^3 + 7x + 9) $

$ = 2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45 $

$ = 2x^4 - 5x^3 + 14x^2 - 17x - 45 $

$ (x^2 - 5x + 8)(3x^2 + 7) = x^2(3x^2 + 7) - 5x(3x^2 + 7) + 8(3x^2 + 7) $

$ = 3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56 $

$ = 3x^4 - 15x^3 + 31x^2 - 35x + 56 $

Add the expanded terms:

$ \frac{dy}{dx} = (2x^4 - 5x^3 + 14x^2 - 17x - 45) + (3x^4 - 15x^3 + 31x^2 - 35x + 56) $

Method (ii): By expanding the product to obtain a single polynomial

Expand the given function $y = (x^2 - 5x + 8) (x^3 + 7x + 9)$:

$ y = x^2(x^3 + 7x + 9) - 5x(x^3 + 7x + 9) + 8(x^3 + 7x + 9) $

$ y = x^5 + 7x^3 + 9x^2 - 5x^4 - 35x^2 - 45x + 8x^3 + 56x + 72 $

Combine like terms:

$ y = x^5 - 5x^4 + (7x^3 + 8x^3) + (9x^2 - 35x^2) + (-45x + 56x) + 72 $

Now, differentiate the polynomial with respect to $x$:

$ \frac{dy}{dx} = 5x^{5-1} - 5(4x^{4-1}) + 15(3x^{3-1}) - 26(2x^{2-1}) + 11(1x^{1-1}) + 0 $

$ \frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11 $

Method (iii): By logarithmic differentiation

Given $y = (x^2 - 5x + 8) (x^3 + 7x + 9)$.

Take the natural logarithm of both sides:

Using the property $\log(ab) = \log a + \log b$:

Differentiate both sides with respect to $x$. Use the chain rule $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$.

We already found the derivatives of the terms: $\frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$ and $\frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$.

Substitute these into (3.3):

Multiply both sides by $y = (x^2 - 5x + 8) (x^3 + 7x + 9)$:

Distribute the term outside the parenthesis:

$ \frac{dy}{dx} = (x^2 - 5x + 8) (x^3 + 7x + 9) \cdot \frac{2x - 5}{x^2 - 5x + 8} + (x^2 - 5x + 8) (x^3 + 7x + 9) \cdot \frac{3x^2 + 7}{x^3 + 7x + 9} $

Cancel the common terms:

$ \frac{dy}{dx} = (x^3 + 7x + 9) (2x - 5) + (x^2 - 5x + 8) (3x^2 + 7) $

This expression is the same as the one obtained in equation (1.1) from the product rule method.

As shown in Method (i), expanding and simplifying this expression gives:

Comparison of Results:

From Method (i), we found $\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$ (equation 1.2).

From Method (ii), we found $\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$ (equation 2.2).

From Method (iii), we found $\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$ (equation 3.5).

All three methods yield the same derivative for the given function.

Final Answer:

The derivative of the function $f(x) = (x^2 – 5x + 8) (x^3 + 7x + 9)$ is $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

Yes, all three methods give the same answer.

Given:

u, v, and w are functions of x.

Let $y = u \cdot v \cdot w$.

To Prove:

$ \frac{d}{dx}(u \cdot v \cdot w) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx} $

Proof (Method 1: By repeated application of product rule):

Let $y = u \cdot (vw)$. Here we consider $u$ as the first function and the product $(vw)$ as the second function.

Applying the product rule $\frac{d}{dx}(f \cdot g) = \frac{df}{dx} g + f \frac{dg}{dx}$ with $f = u$ and $g = vw$:

Now, we need to find $\frac{d}{dx}(vw)$. We apply the product rule again, this time with $f = v$ and $g = w$:

Substitute the expression for $\frac{d}{dx}(vw)$ from equation (1.2) into equation (1.1):

Distribute the term $u$ in the second part of equation (1.3):

This proves the formula using the repeated application of the product rule.

Proof (Method 2: By logarithmic differentiation):

Let $y = u \cdot v \cdot w$.

Taking the natural logarithm on both sides:

Using the property of logarithm $\log(abc) = \log a + \log b + \log c$:

Now, differentiate both sides of equation (2.2) with respect to $x$ implicitly. Use the chain rule $\frac{d}{dx}(\log f(x)) = \frac{1}{f(x)} \frac{df}{dx}$ on the left side and the sum rule on the right side.

Applying the differentiation:

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

Substitute the original expression for $y = u \cdot v \cdot w$ back into equation (2.5):

Distribute the term $(u \cdot v \cdot w)$ inside the parenthesis:

$ \frac{dy}{dx} = (u \cdot v \cdot w) \cdot \frac{1}{u} \frac{du}{dx} + (u \cdot v \cdot w) \cdot \frac{1}{v} \frac{dv}{dx} + (u \cdot v \cdot w) \cdot \frac{1}{w} \frac{dw}{dx} $

Cancel the common terms in each part:

This can be written in the required format:

This also proves the formula.

Conclusion:

Both methods, repeated application of the product rule and logarithmic differentiation, lead to the same formula for the derivative of the product of three functions.

Given:

The parametric equations are:

$x = a \cos \theta$

$y = a \sin \theta$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given $x$ and $y$ as functions of the parameter $\theta$.

Differentiate $x$ with respect to $\theta$:

Since $a$ is a constant and $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:

Differentiate $y$ with respect to $\theta$:

Since $a$ is a constant and $\frac{d}{d\theta}(\sin \theta) = \cos \theta$:

Now, use the formula for the derivative of a parametric function:

Substitute the results from equations (ii) and (iv) into equation (v):

Simplify the expression:

Using the trigonometric identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = -\cot \theta $

Given:

The parametric equations are:

$x = at^2$

$y = 2at$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given $x$ and $y$ as functions of the parameter $t$.

Differentiate $x$ with respect to $t$:

Since $a$ is a constant and $\frac{d}{dt}(t^2) = 2t$:

Differentiate $y$ with respect to $t$:

Since $2a$ is a constant and $\frac{d}{dt}(t) = 1$:

Now, use the formula for the derivative of a parametric function:

Substitute the results from equations (ii) and (iv) into equation (v):

Simplify the expression by cancelling out the common term $2a$ (assuming $a \neq 0$ and $t \neq 0$):

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \frac{1}{t} $

Given:

The parametric equations are:

$x = a (\theta + \sin \theta)$

$y = a (1 - \cos \theta)$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given $x$ and $y$ as functions of the parameter $\theta$.

First, differentiate $x$ with respect to $\theta$:

Using the linearity of differentiation and known derivatives:

Next, differentiate $y$ with respect to $\theta$:

Using the linearity of differentiation and known derivatives:

Now, use the formula for the derivative of a parametric function:

Substitute the results from equations (i) and (ii) into equation (iii):

Simplify the expression by cancelling $a$ (assuming $a \neq 0$):

Use the half-angle trigonometric identities: $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$ and $1 + \cos \theta = 2 \cos^2(\frac{\theta}{2})$:

Cancel the common terms $2$ and $\cos(\frac{\theta}{2})$ (assuming $\cos(\frac{\theta}{2}) \neq 0$):

Using the trigonometric identity $\tan \phi = \frac{\sin \phi}{\cos \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \tan \left(\frac{\theta}{2}\right) $

Given:

The equation is:

$x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

Given the equation $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

Differentiate both sides of the equation implicitly with respect to $x$:

Using the sum rule on the left side and the rule for the derivative of a constant on the right side:

Apply the power rule $\frac{d}{dx}(u^n) = nu^{n-1} \frac{du}{dx}$. For the first term, $u=x$ and $\frac{du}{dx}=1$. For the second term, $u=y$ and $\frac{du}{dx}=\frac{dy}{dx}$ (by the chain rule):

Subtract $\frac{2}{3}x^{-\frac{1}{3}}$ from both sides:

Divide both sides by $\frac{2}{3}y^{-\frac{1}{3}}$ to solve for $\frac{dy}{dx}$:

Cancel the common term $\frac{2}{3}$:

Use the property $a^{-n} = \frac{1}{a^n}$ and $\frac{a^{-n}}{b^{-m}} = \frac{b^m}{a^n}$:

This can also be written as:

or

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = - \left(\frac{y}{x}\right)^{\frac{1}{3}} $

Solution:

Given parametric equations are:

$x = 2at^2$

$y = at^4$

We need to find $\frac{dy}{dx}$ without eliminating the parameter $t$.

The formula for parametric differentiation is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

First, find $\frac{dx}{dt}$ by differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(2at^2)$

Since $2a$ is a constant, we have:

$\frac{dx}{dt} = 2a \frac{d}{dt}(t^2)$

Using the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$:

$\frac{dx}{dt} = 2a(2t)$

$\frac{dx}{dt} = 4at$

Next, find $\frac{dy}{dt}$ by differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(at^4)$

Since $a$ is a constant, we have:

$\frac{dy}{dt} = a \frac{d}{dt}(t^4)$

Using the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$:

$\frac{dy}{dt} = a(4t^3)$

$\frac{dy}{dt} = 4at^3$

Now, substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the formula for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{4at^3}{4at}$

Assuming $a \neq 0$ and $t \neq 0$, we can cancel common terms:

$\frac{dy}{dx} = \frac{\cancel{4a}t^{\cancel{3}^2}}{\cancel{4a}\cancel{t}}$

$\frac{dy}{dx} = t^2$

Thus, the derivative $\frac{dy}{dx}$ is $t^2$.

Solution:

Given parametric equations are:

$x = a \cos \theta$

$y = b \cos \theta$

We need to find $\frac{dy}{dx}$ without eliminating the parameter $\theta$.

The formula for parametric differentiation is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta)$

Since $a$ is a constant, we have:

$\frac{dx}{d\theta} = a \frac{d}{d\theta}(\cos \theta)$

Using the derivative rule $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:

$\frac{dx}{d\theta} = a(-\sin \theta)$

$\frac{dx}{d\theta} = -a \sin \theta$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos \theta)$

Since $b$ is a constant, we have:

$\frac{dy}{d\theta} = b \frac{d}{d\theta}(\cos \theta)$

Using the derivative rule $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:

$\frac{dy}{d\theta} = b(-\sin \theta)$

$\frac{dy}{d\theta} = -b \sin \theta$

Now, substitute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ into the formula for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-b \sin \theta}{-a \sin \theta}$

Assuming $\sin \theta \neq 0$ and $a \neq 0$, we can cancel the common terms:

$\frac{dy}{dx} = \frac{\cancel{-b \sin \theta}}{\cancel{-a \sin \theta}}$

$\frac{dy}{dx} = \frac{b}{a}$

Thus, the derivative $\frac{dy}{dx}$ is $\frac{b}{a}$.

Solution:

Given parametric equations are:

$x = \sin t$

$y = \cos 2t$

We need to find $\frac{dy}{dx}$ without eliminating the parameter $t$.

The formula for parametric differentiation is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

First, find $\frac{dx}{dt}$ by differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(\sin t)$

Using the derivative rule $\frac{d}{dt}(\sin t) = \cos t$:

$\frac{dx}{dt} = \cos t$

Next, find $\frac{dy}{dt}$ by differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(\cos 2t)$

Using the chain rule, where $u=2t$, $\frac{d}{dt}(\cos u) = -\sin u \cdot \frac{du}{dt}$:

$\frac{dy}{dt} = -\sin(2t) \cdot \frac{d}{dt}(2t)$

$\frac{dy}{dt} = -\sin(2t) \cdot 2$

$\frac{dy}{dt} = -2 \sin(2t)$

Now, substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the formula for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2 \sin(2t)}{\cos t}$

Using the double angle identity $\sin(2t) = 2 \sin t \cos t$:

$\frac{dy}{dx} = \frac{-2 (2 \sin t \cos t)}{\cos t}$

Assuming $\cos t \neq 0$, we can cancel the common term $\cos t$:

$\frac{dy}{dx} = \frac{-4 \sin t \cancel{\cos t}}{\cancel{\cos t}}$

$\frac{dy}{dx} = -4 \sin t$

Thus, the derivative $\frac{dy}{dx}$ is $-4 \sin t$.

Solution:

Given parametric equations are:

$x = 4t$

$y = \frac{4}{t} = 4t^{-1}$

We need to find $\frac{dy}{dx}$ without eliminating the parameter $t$.

The formula for parametric differentiation is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

First, find $\frac{dx}{dt}$ by differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(4t)$

Since $4$ is a constant, we have:

$\frac{dx}{dt} = 4 \frac{d}{dt}(t)$

Using the derivative rule $\frac{d}{dt}(t) = 1$:

$\frac{dx}{dt} = 4(1)$

$\frac{dx}{dt} = 4$

Next, find $\frac{dy}{dt}$ by differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(4t^{-1})$

Since $4$ is a constant, we have:

$\frac{dy}{dt} = 4 \frac{d}{dt}(t^{-1})$

Using the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$:

$\frac{dy}{dt} = 4(-1 \cdot t^{-1-1})$

$\frac{dy}{dt} = 4(-t^{-2})$

$\frac{dy}{dt} = -4t^{-2} = -\frac{4}{t^2}$

Now, substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the formula for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-4/t^2}{4}$

$\frac{dy}{dx} = \frac{-4}{t^2} \cdot \frac{1}{4}$

Assuming $t \neq 0$, we can cancel the common term $4$:

$\frac{dy}{dx} = \frac{\cancel{-4}}{t^2} \cdot \frac{1}{\cancel{4}}$

$\frac{dy}{dx} = -\frac{1}{t^2}$

Thus, the derivative $\frac{dy}{dx}$ is $-\frac{1}{t^2}$.

Solution:

Given parametric equations are:

$x = \cos \theta - \cos 2\theta$

$y = \sin \theta - \sin 2\theta$

We need to find $\frac{dy}{dx}$ without eliminating the parameter $\theta$.

The formula for parametric differentiation is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(\cos \theta - \cos 2\theta)$

$\frac{dx}{d\theta} = \frac{d}{d\theta}(\cos \theta) - \frac{d}{d\theta}(\cos 2\theta)$

Using the derivative rules $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$ and $\frac{d}{d\theta}(\cos u) = -\sin u \frac{du}{d\theta}$ (with $u=2\theta$):

$\frac{dx}{d\theta} = -\sin \theta - (-\sin(2\theta) \cdot \frac{d}{d\theta}(2\theta))$

$\frac{dx}{d\theta} = -\sin \theta - (-\sin(2\theta) \cdot 2)$

$\frac{dx}{d\theta} = -\sin \theta + 2\sin 2\theta$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta - \sin 2\theta)$

$\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\sin 2\theta)$

Using the derivative rules $\frac{d}{d\theta}(\sin \theta) = \cos \theta$ and $\frac{d}{d\theta}(\sin u) = \cos u \frac{du}{d\theta}$ (with $u=2\theta$):

$\frac{dy}{d\theta} = \cos \theta - (\cos(2\theta) \cdot \frac{d}{d\theta}(2\theta))$

$\frac{dy}{d\theta} = \cos \theta - (\cos(2\theta) \cdot 2)$

$\frac{dy}{d\theta} = \cos \theta - 2\cos 2\theta$

Now, substitute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ into the formula for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\cos \theta - 2\cos 2\theta}{-\sin \theta + 2\sin 2\theta}$

Thus, the derivative $\frac{dy}{dx}$ is $\frac{\cos \theta - 2\cos 2\theta}{2\sin 2\theta - \sin \theta}$.

Solution:

Given parametric equations are:

We need to find $\frac{dy}{dx}$ without eliminating the parameter $\theta$.

The formula for parametric differentiation is:

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}[a (\theta - \sin \theta)]$

Since $a$ is a constant, we have:

$\frac{dx}{d\theta} = a \frac{d}{d\theta}(\theta - \sin \theta)$

Using the derivative rules $\frac{d}{d\theta}(\theta) = 1$ and $\frac{d}{d\theta}(\sin \theta) = \cos \theta$:

$\frac{dx}{d\theta} = a (1 - \cos \theta)$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}[a (1 + \cos \theta)]$

Since $a$ is a constant, we have:

$\frac{dy}{d\theta} = a \frac{d}{d\theta}(1 + \cos \theta)$

Using the derivative rules $\frac{d}{d\theta}(1) = 0$ and $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:

$\frac{dy}{d\theta} = a (0 - \sin \theta)$

$\frac{dy}{d\theta} = -a \sin \theta$

Now, substitute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ into the formula for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-a \sin \theta}{a (1 - \cos \theta)}$

Assuming $a \neq 0$, we can cancel the common term $a$:

$\frac{dy}{dx} = \frac{-\sin \theta}{1 - \cos \theta}$

We can simplify this expression further using trigonometric identities:

$\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$

$1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$

Substitute these identities into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})}{2 \sin^2(\frac{\theta}{2})}$

Assuming $\sin(\frac{\theta}{2}) \neq 0$, we can cancel common terms:

$\frac{dy}{dx} = \frac{-\cancel{2} \cancel{\sin(\frac{\theta}{2})} \cos(\frac{\theta}{2})}{\cancel{2} \sin^{\cancel{2}}(\frac{\theta}{2})}$

$\frac{dy}{dx} = -\frac{\cos(\frac{\theta}{2})}{\sin(\frac{\theta}{2})}$

Using the identity $\cot x = \frac{\cos x}{\sin x}$:

$\frac{dy}{dx} = -\cot(\frac{\theta}{2})$

Thus, the derivative $\frac{dy}{dx}$ is $-\cot(\frac{\theta}{2})$.

Solution:

Given parametric equations are:

$x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$

$y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$

To Find:

$\frac{dy}{dx}$

Solution:

The derivative $\frac{dy}{dx}$ for parametric equations is given by the formula:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

First, we find $\frac{dx}{dt}$ by differentiating $x$ with respect to $t$.

Using the quotient rule, $\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$, where $u = \sin^3 t$ and $v = \sqrt{\cos 2t} = (\cos 2t)^{\frac{1}{2}}$.

$\frac{du}{dt} = \frac{d}{dt}(\sin^3 t) = 3\sin^2 t \cdot \frac{d}{dt}(\sin t) = 3\sin^2 t \cos t$

$\frac{dv}{dt} = \frac{d}{dt}((\cos 2t)^{\frac{1}{2}}) = \frac{1}{2}(\cos 2t)^{-\frac{1}{2}} \cdot \frac{d}{dt}(\cos 2t) = \frac{1}{2\sqrt{\cos 2t}} \cdot (-\sin 2t \cdot 2) = -\frac{2\sin 2t}{2\sqrt{\cos 2t}} = -\frac{\sin 2t}{\sqrt{\cos 2t}}$

Applying the quotient rule:

$\frac{dx}{dt} = \frac{\sqrt{\cos 2t} (3\sin^2 t \cos t) - \sin^3 t \left(-\frac{\sin 2t}{\sqrt{\cos 2t}}\right)}{(\sqrt{\cos 2t})^2}$

$\frac{dx}{dt} = \frac{\frac{(\sqrt{\cos 2t})^2 (3\sin^2 t \cos t) + \sin^3 t \sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t \cos 2t + \sin^3 t (2\sin t \cos t)}{(\cos 2t)^{\frac{3}{2}}}$

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t \cos 2t + 2\sin^4 t \cos t}{(\cos 2t)^{\frac{3}{2}}}$

Factor out $\sin^2 t \cos t$ from the numerator:

$\frac{dx}{dt} = \frac{\sin^2 t \cos t (3\cos 2t + 2\sin^2 t)}{(\cos 2t)^{\frac{3}{2}}}$

Using the identity $3\cos 2t + 2\sin^2 t = 3(1-2\sin^2 t) + 2\sin^2 t = 3-6\sin^2 t + 2\sin^2 t = 3-4\sin^2 t$:

$\frac{dx}{dt} = \frac{\sin^2 t \cos t (3 - 4\sin^2 t)}{(\cos 2t)^{\frac{3}{2}}}$

Next, we find $\frac{dy}{dt}$ by differentiating $y$ with respect to $t$.

Using the quotient rule, $\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$, where $u = \cos^3 t$ and $v = \sqrt{\cos 2t}$.

$\frac{du}{dt} = \frac{d}{dt}(\cos^3 t) = 3\cos^2 t \cdot \frac{d}{dt}(\cos t) = 3\cos^2 t (-\sin t) = -3\cos^2 t \sin t$

$\frac{dv}{dt} = -\frac{\sin 2t}{\sqrt{\cos 2t}}$ (calculated previously)

Applying the quotient rule:

$\frac{dy}{dt} = \frac{\sqrt{\cos 2t} (-3\cos^2 t \sin t) - \cos^3 t \left(-\frac{\sin 2t}{\sqrt{\cos 2t}}\right)}{(\sqrt{\cos 2t})^2}$

$\frac{dy}{dt} = \frac{\frac{(\sqrt{\cos 2t})^2 (-3\cos^2 t \sin t) + \cos^3 t \sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t \cos 2t + \cos^3 t (2\sin t \cos t)}{(\cos 2t)^{\frac{3}{2}}}$

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t \cos 2t + 2\cos^4 t \sin t}{(\cos 2t)^{\frac{3}{2}}}$

Factor out $\cos^2 t \sin t$ from the numerator:

$\frac{dy}{dt} = \frac{\cos^2 t \sin t (-3\cos 2t + 2\cos^2 t)}{(\cos 2t)^{\frac{3}{2}}}$

Using the identity $-3\cos 2t + 2\cos^2 t = -3(2\cos^2 t - 1) + 2\cos^2 t = -6\cos^2 t + 3 + 2\cos^2 t = 3-4\cos^2 t$:

$\frac{dy}{dt} = \frac{\cos^2 t \sin t (3 - 4\cos^2 t)}{(\cos 2t)^{\frac{3}{2}}}$

Now, we compute $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

$\frac{dy}{dx} = \frac{\frac{\cos^2 t \sin t (3 - 4\cos^2 t)}{(\cos 2t)^{\frac{3}{2}}}}{\frac{\sin^2 t \cos t (3 - 4\sin^2 t)}{(\cos 2t)^{\frac{3}{2}}}}$

Assuming $\cos 2t \neq 0$, we cancel the term $(\cos 2t)^{\frac{3}{2}}$ from the numerator and denominator:

$\frac{dy}{dx} = \frac{\cos^2 t \sin t (3 - 4\cos^2 t)}{\sin^2 t \cos t (3 - 4\sin^2 t)}$

Assuming $\sin t \neq 0$ and $\cos t \neq 0$, we cancel $\sin t$ and $\cos t$ terms:

$\frac{dy}{dx} = \frac{\cos t (3 - 4\cos^2 t)}{\sin t (3 - 4\sin^2 t)}$

Using the triple angle identities:

$\sin 3t = 3\sin t - 4\sin^3 t = \sin t (3 - 4\sin^2 t)$

$\cos 3t = 4\cos^3 t - 3\cos t$, so $3\cos t - 4\cos^3 t = -(4\cos^3 t - 3\cos t) = -\cos 3t$. Thus, $\cos t (3 - 4\cos^2 t) = -\cos 3t$.

Substitute these into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-\cos 3t}{\sin 3t}$

$\frac{dy}{dx} = -\cot 3t$

The final answer is $\boxed{-\cot 3t}$.

Solution:

Given parametric equations are:

To Find:

$\frac{dy}{dx}$

Solution:

The derivative $\frac{dy}{dx}$ for parametric equations is given by the formula:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

First, find $\frac{dx}{dt}$ by differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt} \left[ a \left( \cos t + \log \tan \frac{t}{2} \right) \right]$

$\frac{dx}{dt} = a \frac{d}{dt} \left( \cos t + \log \tan \frac{t}{2} \right)$

$\frac{dx}{dt} = a \left( \frac{d}{dt}(\cos t) + \frac{d}{dt}\left(\log \tan \frac{t}{2}\right) \right)$

We know $\frac{d}{dt}(\cos t) = -\sin t$.

For the second term, use the chain rule:

$\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{dt}\left(\tan \frac{t}{2}\right)$

$\frac{d}{dt}\left(\tan \frac{t}{2}\right) = \sec^2 \frac{t}{2} \cdot \frac{d}{dt}\left(\frac{t}{2}\right) = \sec^2 \frac{t}{2} \cdot \frac{1}{2}$

So, $\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}$

Substitute $\tan \frac{t}{2} = \frac{\sin(t/2)}{\cos(t/2)}$ and $\sec^2 \frac{t}{2} = \frac{1}{\cos^2(t/2)}$:

$\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{2\cos^2(t/2)} = \frac{1}{2\sin(t/2)\cos(t/2)}$

Using the identity $\sin t = 2\sin(\frac{t}{2})\cos(\frac{t}{2})$, we get:

$\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{1}{\sin t}$

Now, substitute back into $\frac{dx}{dt}$:

$\frac{dx}{dt} = a \left( -\sin t + \frac{1}{\sin t} \right)$

Combine the terms:

$\frac{dx}{dt} = a \left( \frac{-\sin^2 t + 1}{\sin t} \right)$

Using the identity $1 - \sin^2 t = \cos^2 t$:

$\frac{dx}{dt} = a \left( \frac{\cos^2 t}{\sin t} \right)$

Next, find $\frac{dy}{dt}$ by differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(a \sin t)$

$\frac{dy}{dt} = a \frac{d}{dt}(\sin t)$

Using the derivative rule $\frac{d}{dt}(\sin t) = \cos t$:

$\frac{dy}{dt} = a \cos t$

Now, compute $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

$\frac{dy}{dx} = \frac{a \cos t}{a \left( \frac{\cos^2 t}{\sin t} \right)}$

Assuming $a \neq 0$ and $\cos t \neq 0$ and $\sin t \neq 0$, we can simplify:

$\frac{dy}{dx} = \frac{\cancel{a} \cos t}{\cancel{a}} \cdot \frac{\sin t}{\cos^2 t}$

$\frac{dy}{dx} = \frac{\cos t \sin t}{\cos^2 t}$

$\frac{dy}{dx} = \frac{\cancel{\cos t} \sin t}{\cos^{\cancel{2}} t}$

$\frac{dy}{dx} = \frac{\sin t}{\cos t}$

Using the identity $\frac{\sin t}{\cos t} = \tan t$:

$\frac{dy}{dx} = \tan t$

Thus, the derivative $\frac{dy}{dx}$ is $\tan t$. The final answer is $\boxed{\tan t}$.

Solution:

Given parametric equations are:

$x = a \sec \theta$

$y = b \tan \theta$

To Find:

$\frac{dy}{dx}$

Solution:

The derivative $\frac{dy}{dx}$ for parametric equations is given by the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \sec \theta)$

Since $a$ is a constant, we have:

$\frac{dx}{d\theta} = a \frac{d}{d\theta}(\sec \theta)$

Using the derivative rule $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$:

$\frac{dx}{d\theta} = a \sec \theta \tan \theta$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \tan \theta)$

Since $b$ is a constant, we have:

$\frac{dy}{d\theta} = b \frac{d}{d\theta}(\tan \theta)$

Using the derivative rule $\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta$:

$\frac{dy}{d\theta} = b \sec^2 \theta$

Now, compute $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:

$\frac{dy}{dx} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta}$

Assuming $a \neq 0$, $\sec \theta \neq 0$ and $\tan \theta \neq 0$, we can simplify:

$\frac{dy}{dx} = \frac{b}{a} \frac{\sec^2 \theta}{\sec \theta \tan \theta}$

$\frac{dy}{dx} = \frac{b}{a} \frac{\cancel{\sec \theta} \sec \theta}{\cancel{\sec \theta} \tan \theta}$

$\frac{dy}{dx} = \frac{b}{a} \frac{\sec \theta}{\tan \theta}$

Substitute $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$\frac{dy}{dx} = \frac{b}{a} \frac{1/\cos \theta}{\sin \theta/\cos \theta}$

$\frac{dy}{dx} = \frac{b}{a} \left( \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} \right)$

$\frac{dy}{dx} = \frac{b}{a} \left( \frac{1}{\cancel{\cos \theta}} \cdot \frac{\cancel{\cos \theta}}{\sin \theta} \right)$

$\frac{dy}{dx} = \frac{b}{a \sin \theta}$

Using the identity $\frac{1}{\sin \theta} = \text{cosec} \theta$:

$\frac{dy}{dx} = \frac{b}{a} \text{ cosec } \theta$

Thus, the derivative $\frac{dy}{dx}$ is $\frac{b}{a \sin \theta}$ or $\frac{b}{a} \text{ cosec } \theta$.

Solution:

Given parametric equations are:

$x = a (\cos \theta + \theta \sin \theta)$

$y = a (\sin \theta – \theta \cos \theta)$

To Find:

$\frac{dy}{dx}$

Solution:

The derivative $\frac{dy}{dx}$ for parametric equations is given by the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}[a (\cos \theta + \theta \sin \theta)]$

$\frac{dx}{d\theta} = a \frac{d}{d\theta}(\cos \theta + \theta \sin \theta)$

$\frac{dx}{d\theta} = a \left( \frac{d}{d\theta}(\cos \theta) + \frac{d}{d\theta}(\theta \sin \theta) \right)$

We know $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$.

Using the product rule for $\frac{d}{d\theta}(\theta \sin \theta)$: $\frac{d}{d\theta}(uv) = u \frac{dv}{d\theta} + v \frac{du}{d\theta}$ with $u = \theta$ and $v = \sin \theta$:

$\frac{d}{d\theta}(\theta \sin \theta) = \theta \frac{d}{d\theta}(\sin \theta) + \sin \theta \frac{d}{d\theta}(\theta)$

$\frac{d}{d\theta}(\theta \sin \theta) = \theta (\cos \theta) + \sin \theta (1)$

$\frac{d}{d\theta}(\theta \sin \theta) = \theta \cos \theta + \sin \theta$

Substitute these derivatives back into the expression for $\frac{dx}{d\theta}$:

$\frac{dx}{d\theta} = a (-\sin \theta + \theta \cos \theta + \sin \theta)$

$\frac{dx}{d\theta} = a (\theta \cos \theta)$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}[a (\sin \theta – \theta \cos \theta)]$

$\frac{dy}{d\theta} = a \frac{d}{d\theta}(\sin \theta – \theta \cos \theta)$

$\frac{dy}{d\theta} = a \left( \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\theta \cos \theta) \right)$

We know $\frac{d}{d\theta}(\sin \theta) = \cos \theta$.

Using the product rule for $\frac{d}{d\theta}(\theta \cos \theta)$: $\frac{d}{d\theta}(uv) = u \frac{dv}{d\theta} + v \frac{du}{d\theta}$ with $u = \theta$ and $v = \cos \theta$:

$\frac{d}{d\theta}(\theta \cos \theta) = \theta \frac{d}{d\theta}(\cos \theta) + \cos \theta \frac{d}{d\theta}(\theta)$

$\frac{d}{d\theta}(\theta \cos \theta) = \theta (-\sin \theta) + \cos \theta (1)$

$\frac{d}{d\theta}(\theta \cos \theta) = -\theta \sin \theta + \cos \theta$

Substitute these derivatives back into the expression for $\frac{dy}{d\theta}$:

$\frac{dy}{d\theta} = a [\cos \theta - (-\theta \sin \theta + \cos \theta)]$

$\frac{dy}{d\theta} = a [\cos \theta + \theta \sin \theta - \cos \theta]$

$\frac{dy}{d\theta} = a (\theta \sin \theta)$

Now, compute $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:

$\frac{dy}{dx} = \frac{a (\theta \sin \theta)}{a (\theta \cos \theta)}$

Assuming $a \neq 0$ and $\theta \neq 0$, we can cancel the common terms $a$ and $\theta$:

$\frac{dy}{dx} = \frac{\cancel{a} \cancel{\theta} \sin \theta}{\cancel{a} \cancel{\theta} \cos \theta}$

$\frac{dy}{dx} = \frac{\sin \theta}{\cos \theta}$

Using the identity $\frac{\sin \theta}{\cos \theta} = \tan \theta$:

$\frac{dy}{dx} = \tan \theta$

Thus, the derivative $\frac{dy}{dx}$ is $\tan \theta$.

Given:

To Show:

$\frac{dy}{dx} = -\frac{y}{x}$

Solution:

The given parametric equations are:

$x = (a^{\sin^{-1} t})^{\frac{1}{2}} = a^{\frac{1}{2} \sin^{-1} t}$

$y = (a^{\cos^{-1} t})^{\frac{1}{2}} = a^{\frac{1}{2} \cos^{-1} t}$

We need to find $\frac{dy}{dx}$ without eliminating the parameter $t$. The formula for parametric differentiation is $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

First, differentiate $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(a^{\frac{1}{2} \sin^{-1} t})$

Using the derivative rule $\frac{d}{dt}(a^u) = a^u \log_e a \frac{du}{dt}$, where $u = \frac{1}{2} \sin^{-1} t$:

$\frac{du}{dt} = \frac{d}{dt}\left(\frac{1}{2} \sin^{-1} t\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} = \frac{1}{2\sqrt{1-t^2}}$

So, $\frac{dx}{dt} = a^{\frac{1}{2} \sin^{-1} t} \log_e a \cdot \left(\frac{1}{2\sqrt{1-t^2}}\right)$

Next, differentiate $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(a^{\frac{1}{2} \cos^{-1} t})$

Using the derivative rule $\frac{d}{dt}(a^u) = a^u \log_e a \frac{du}{dt}$, where $u = \frac{1}{2} \cos^{-1} t$:

$\frac{du}{dt} = \frac{d}{dt}\left(\frac{1}{2} \cos^{-1} t\right) = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-t^2}}\right) = -\frac{1}{2\sqrt{1-t^2}}$

So, $\frac{dy}{dt} = a^{\frac{1}{2} \cos^{-1} t} \log_e a \cdot \left(-\frac{1}{2\sqrt{1-t^2}}\right)$

$\frac{dy}{dt} = -a^{\frac{1}{2} \cos^{-1} t} \log_e a \cdot \frac{1}{2\sqrt{1-t^2}}$

Now, compute $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

$\frac{dy}{dx} = \frac{-a^{\frac{1}{2} \cos^{-1} t} \log_e a \cdot \frac{1}{2\sqrt{1-t^2}}}{a^{\frac{1}{2} \sin^{-1} t} \log_e a \cdot \frac{1}{2\sqrt{1-t^2}}}$

Assuming $\log_e a \neq 0$ (i.e., $a > 0, a \neq 1$) and $1-t^2 > 0$ (i.e., $t \in (-1, 1)$), we can cancel the common terms $\log_e a$ and $\frac{1}{2\sqrt{1-t^2}}$:

$\frac{dy}{dx} = \frac{-a^{\frac{1}{2} \cos^{-1} t} \cancel{\log_e a} \cdot \cancel{\frac{1}{2\sqrt{1-t^2}}}}{a^{\frac{1}{2} \sin^{-1} t} \cancel{\log_e a} \cdot \cancel{\frac{1}{2\sqrt{1-t^2}}}}$

$\frac{dy}{dx} = \frac{-a^{\frac{1}{2} \cos^{-1} t}}{a^{\frac{1}{2} \sin^{-1} t}}$

From the given equations, we have:

$x = a^{\frac{1}{2} \sin^{-1} t}$

$y = a^{\frac{1}{2} \cos^{-1} t}$

Substituting these back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-y}{x}$

Hence, shown.

Solution:

Given the function:

$y = x^3 + \tan x$

We need to find the second derivative $\frac{d^2y}{dx^2}$.

First, we find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^3 + \tan x)$

$\frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(\tan x)$

Using the power rule for $x^3$ and the standard derivative for $\tan x$:

$\frac{dy}{dx} = 3x^{3-1} + \sec^2 x$

$\frac{dy}{dx} = 3x^2 + \sec^2 x$

Next, we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(3x^2 + \sec^2 x)$

$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(\sec^2 x)$

For the first term:

$\frac{d}{dx}(3x^2) = 3 \frac{d}{dx}(x^2) = 3(2x) = 6x$

For the second term, we use the chain rule. Let $u = \sec x$. Then $\frac{d}{dx}(\sec^2 x) = \frac{d}{dx}(u^2)$.

$\frac{d}{dx}(u^2) = 2u \frac{du}{dx}$

We know $\frac{du}{dx} = \frac{d}{dx}(\sec x) = \sec x \tan x$.

So, $\frac{d}{dx}(\sec^2 x) = 2(\sec x)(\sec x \tan x) = 2\sec^2 x \tan x$

Combining the derivatives of both terms:

$\frac{d^2y}{dx^2} = 6x + 2\sec^2 x \tan x$

Thus, the second derivative is $6x + 2\sec^2 x \tan x$.

The final answer is $\boxed{6x + 2\sec^2 x \tan x}$.

Given:

To Prove:

$\frac{d^2y}{dx^2} + y = 0$

Proof:

We are given the function:

$y = A \sin x + B \cos x$

First, we find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(A \sin x + B \cos x)$

Using the sum rule and constant multiple rule:

$\frac{dy}{dx} = A \frac{d}{dx}(\sin x) + B \frac{d}{dx}(\cos x)$

Using the standard derivative rules $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$:

$\frac{dy}{dx} = A (\cos x) + B (-\sin x)$

$\frac{dy}{dx} = A \cos x - B \sin x$

Next, we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(A \cos x - B \sin x)$

Using the difference rule and constant multiple rule:

$\frac{d^2y}{dx^2} = A \frac{d}{dx}(\cos x) - B \frac{d}{dx}(\sin x)$

Using the standard derivative rules $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\sin x) = \cos x$:

$\frac{d^2y}{dx^2} = A (-\sin x) - B (\cos x)$

$\frac{d^2y}{dx^2} = -A \sin x - B \cos x$

Now, consider the expression $\frac{d^2y}{dx^2} + y$. Substitute the expressions for $\frac{d^2y}{dx^2}$ and $y$:

$\frac{d^2y}{dx^2} + y = (-A \sin x - B \cos x) + (A \sin x + B \cos x)$

Rearrange the terms:

$\frac{d^2y}{dx^2} + y = -A \sin x + A \sin x - B \cos x + B \cos x$

Combine like terms:

$\frac{d^2y}{dx^2} + y = (-A + A) \sin x + (-B + B) \cos x$

$\frac{d^2y}{dx^2} + y = 0 \cdot \sin x + 0 \cdot \cos x$

$\frac{d^2y}{dx^2} + y = 0 + 0$

$\frac{d^2y}{dx^2} + y = 0$

Since the left-hand side equals $0$, which is the right-hand side of the equation to be proven, the equation is satisfied.

Hence, proven.

Given:

To Prove:

$\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$

Proof:

The given function is $y = 3e^{2x} + 2e^{3x}$.

First, we find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(3e^{2x} + 2e^{3x})$

Using the linearity of differentiation and the rule $\frac{d}{dx}(e^{ax}) = ae^{ax}$:

$\frac{dy}{dx} = 3 \frac{d}{dx}(e^{2x}) + 2 \frac{d}{dx}(e^{3x})$

$\frac{dy}{dx} = 3(2e^{2x}) + 2(3e^{3x})$

$\frac{dy}{dx} = 6e^{2x} + 6e^{3x}$

Next, we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(6e^{2x} + 6e^{3x})$

Using the linearity of differentiation and the rule $\frac{d}{dx}(e^{ax}) = ae^{ax}$ again:

$\frac{d^2y}{dx^2} = 6 \frac{d}{dx}(e^{2x}) + 6 \frac{d}{dx}(e^{3x})$

$\frac{d^2y}{dx^2} = 6(2e^{2x}) + 6(3e^{3x})$

$\frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x}$

Now, we substitute the expressions for $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the left-hand side of the equation we need to prove:

L.H.S. = $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$

L.H.S. = $(12e^{2x} + 18e^{3x}) - 5(6e^{2x} + 6e^{3x}) + 6(3e^{2x} + 2e^{3x})$

Distribute the constants $-5$ and $6$:

L.H.S. = $12e^{2x} + 18e^{3x} - 30e^{2x} - 30e^{3x} + 18e^{2x} + 12e^{3x}$

Group the terms with $e^{2x}$ and $e^{3x}$:

L.H.S. = $(12e^{2x} - 30e^{2x} + 18e^{2x}) + (18e^{3x} - 30e^{3x} + 12e^{3x})$

Combine the coefficients:

L.H.S. = $(12 - 30 + 18)e^{2x} + (18 - 30 + 12)e^{3x}$

L.H.S. = $(0)e^{2x} + (0)e^{3x}$

L.H.S. = $0 + 0$

L.H.S. = $0$

Since the left-hand side equals $0$, which is the right-hand side of the equation to be proven, the equation is satisfied.

Hence, proven.

Given:

To Show:

$(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0$

Proof:

The given function is:

$y = \sin^{-1} x$

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x)$

Using the standard derivative rule for inverse sine function:

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

To make the differentiation of the second derivative easier, we can rewrite the first derivative as:

$\frac{dy}{dx} = (1 - x^2)^{-\frac{1}{2}}$

Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$. Use the chain rule.

$\frac{d^2y}{dx^2} = \frac{d}{dx}((1 - x^2)^{-\frac{1}{2}})$

Let $u = 1 - x^2$. Then $\frac{d}{dx}(u^{-\frac{1}{2}}) = -\frac{1}{2} u^{-\frac{3}{2}} \frac{du}{dx}$.

$\frac{du}{dx} = \frac{d}{dx}(1 - x^2) = 0 - 2x = -2x$

So, $\frac{d^2y}{dx^2} = -\frac{1}{2} (1 - x^2)^{-\frac{3}{2}} (-2x)$

$\frac{d^2y}{dx^2} = x (1 - x^2)^{-\frac{3}{2}}$

Rewrite the expression using square roots:

$\frac{d^2y}{dx^2} = \frac{x}{(1 - x^2)^{\frac{3}{2}}} = \frac{x}{(1 - x^2)\sqrt{1 - x^2}}$

Now, substitute the expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the left-hand side of the equation we need to show:

L.H.S. = $(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx}$

L.H.S. = $(1 – x^2) \left( \frac{x}{(1 - x^2)\sqrt{1 - x^2}} \right) - x \left( \frac{1}{\sqrt{1 - x^2}} \right)$

In the first term, assuming $1 - x^2 \neq 0$, we can cancel $(1 - x^2)$ in the numerator and denominator:

L.H.S. = $\cancel{(1 – x^2)} \left( \frac{x}{\cancel{(1 - x^2)}\sqrt{1 - x^2}} \right) - x \left( \frac{1}{\sqrt{1 - x^2}} \right)$

L.H.S. = $\frac{x}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}}$

L.H.S. = $0$

Since the left-hand side equals $0$, which is the right-hand side of the equation to be proven, the equation is satisfied.

Hence, shown.

Let the given function be denoted by $y$.

$y = x^2 + 3x + 2$

To find the second order derivative, we first find the first order derivative.

Differentiating $y$ with respect to $x$, we get:

$\frac{dy}{dx} = \frac{d}{dx}(x^2 + 3x + 2)$

Using the sum rule and power rule of differentiation:

$\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(2)$

$\frac{dy}{dx} = 2x^{2-1} + 3 \cdot 1 + 0$

$\frac{dy}{dx} = 2x + 3$

Now, we find the second order derivative by differentiating the first order derivative with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(2x + 3)$

Using the sum rule and constant multiple rule:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x) + \frac{d}{dx}(3)$

$\frac{d^2y}{dx^2} = 2 \cdot 1 + 0$

$\frac{d^2y}{dx^2} = 2$

Thus, the second order derivative of the function $x^2 + 3x + 2$ is 2.

The final answer is $\boxed{2}$.

Let the given function be denoted by $y$.

$y = x^{20}$

To find the second order derivative, we first find the first order derivative.

Differentiating $y$ with respect to $x$, we use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

$\frac{dy}{dx} = \frac{d}{dx}(x^{20})$

$\frac{dy}{dx} = 20x^{20-1}$

$\frac{dy}{dx} = 20x^{19}$

Now, we find the second order derivative by differentiating the first order derivative with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(20x^{19})$

Using the constant multiple rule and the power rule:

$\frac{d^2y}{dx^2} = 20 \cdot \frac{d}{dx}(x^{19})$

$\frac{d^2y}{dx^2} = 20 \cdot 19x^{19-1}$

$\frac{d^2y}{dx^2} = 380x^{18}$

Thus, the second order derivative of the function $x^{20}$ is $380x^{18}$.

The final answer is $\boxed{380x^{18}}$.

Let the given function be denoted by $y$.

$y = x \cos x$

To find the second order derivative, we first find the first order derivative.

We use the product rule for differentiation, which states that $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Here, let $u=x$ and $v=\cos x$.

We find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x) = 1$

$\frac{dv}{dx} = \frac{d}{dx}(\cos x) = -\sin x$

Applying the product rule to $y = x \cos x$:

$\frac{dy}{dx} = (x)(-\sin x) + (\cos x)(1)$

$\frac{dy}{dx} = -x \sin x + \cos x$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-x \sin x + \cos x)$

Using the sum/difference rule, we differentiate each term separately:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-x \sin x) + \frac{d}{dx}(\cos x)$

For the term $\frac{d}{dx}(-x \sin x)$, we again use the product rule. Let $u' = -x$ and $v' = \sin x$.

We find the derivatives of $u'$ and $v'$ with respect to $x$:

$\frac{du'}{dx} = \frac{d}{dx}(-x) = -1$

$\frac{dv'}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Applying the product rule to $-x \sin x$:

$\frac{d}{dx}(-x \sin x) = (-x)(\cos x) + (\sin x)(-1) = -x \cos x - \sin x$

The derivative of the second term is:

$\frac{d}{dx}(\cos x) = -\sin x$

Combining the results for the second derivative:

$\frac{d^2y}{dx^2} = (-x \cos x - \sin x) + (-\sin x)$

$\frac{d^2y}{dx^2} = -x \cos x - 2 \sin x$

Thus, the second order derivative of the function $x \cos x$ is $-x \cos x - 2 \sin x$.

The final answer is $\boxed{-x \cos x - 2 \sin x}$.

Let the given function be denoted by $y$.

$y = \log x$

(Assuming $\log x$ represents the natural logarithm, often denoted as $\log_e x$)

$y = \log_e x$

To find the second order derivative, we first find the first order derivative.

Differentiating $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\log_e x)$

$\frac{dy}{dx} = \frac{1}{x}$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

We can rewrite $\frac{1}{x}$ as $x^{-1}$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x}\right) = \frac{d}{dx}(x^{-1})$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ with $n=-1$:

$\frac{d^2y}{dx^2} = (-1)x^{-1-1}$

$\frac{d^2y}{dx^2} = -x^{-2}$

$\frac{d^2y}{dx^2} = -\frac{1}{x^2}$

Thus, the second order derivative of the function $\log x$ (natural logarithm) is $-\frac{1}{x^2}$.

The final answer is $\boxed{-\frac{1}{x^2}}$.

Let the given function be denoted by $y$.

$y = x^3 \log x$

(Assuming $\log x$ represents the natural logarithm, often denoted as $\log_e x$)

$y = x^3 \log_e x$

To find the second order derivative, we first find the first order derivative.

We use the product rule for differentiation, which states that $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Here, let $u=x^3$ and $v=\log_e x$.

We find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$

$\frac{dv}{dx} = \frac{d}{dx}(\log_e x) = \frac{1}{x}$

Applying the product rule to $y = x^3 \log_e x$:

$\frac{dy}{dx} = (x^3)\left(\frac{1}{x}\right) + (\log_e x)(3x^2)$

$\frac{dy}{dx} = x^{3-1} + 3x^2 \log_e x$

$\frac{dy}{dx} = x^2 + 3x^2 \log_e x$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(x^2 + 3x^2 \log_e x)$

Using the sum rule, we differentiate each term separately:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x^2 \log_e x)$

For the first term: $\frac{d}{dx}(x^2) = 2x$.

For the second term $\frac{d}{dx}(3x^2 \log_e x)$, we again use the product rule. Let $u' = 3x^2$ and $v' = \log_e x$.

We find the derivatives of $u'$ and $v'$ with respect to $x$:

$\frac{du'}{dx} = \frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x$

$\frac{dv'}{dx} = \frac{d}{dx}(\log_e x) = \frac{1}{x}$

Applying the product rule to $3x^2 \log_e x$:

$\frac{d}{dx}(3x^2 \log_e x) = (3x^2)\left(\frac{1}{x}\right) + (\log_e x)(6x)$

$\frac{d}{dx}(3x^2 \log_e x) = 3x + 6x \log_e x$

Combining the derivatives of both terms for the second derivative:

$\frac{d^2y}{dx^2} = (2x) + (3x + 6x \log_e x)$

$\frac{d^2y}{dx^2} = 2x + 3x + 6x \log_e x$

$\frac{d^2y}{dx^2} = 5x + 6x \log_e x$

We can factor out $x$ from the expression:

$\frac{d^2y}{dx^2} = x(5 + 6 \log_e x)$

If $\log x$ refers to the base-10 logarithm, the derivative $\frac{d}{dx}(\log_{10} x) = \frac{1}{x \log_e 10}$. In that case, the first derivative would be $\frac{dy}{dx} = x^3 \left(\frac{1}{x \log_e 10}\right) + (\log_{10} x)(3x^2) = \frac{x^2}{\log_e 10} + 3x^2 \log_{10} x$. The second derivative would be more complex. However, in calculus context, $\log x$ usually means $\log_e x$ (natural logarithm).

Thus, assuming $\log x = \log_e x$, the second order derivative of the function $x^3 \log x$ is $5x + 6x \log_e x$ or $x(5 + 6 \log_e x)$.

The final answer is $\boxed{x(5 + 6 \log_e x)}$.

Let the given function be denoted by $y$.

$y = e^x \sin 5x$

To find the second order derivative, we first find the first order derivative.

We use the product rule for differentiation: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Here, let $u=e^x$ and $v=\sin 5x$.

We find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x$

$\frac{dv}{dx} = \frac{d}{dx}(\sin 5x)$

Using the chain rule for $\sin 5x$, where the outer function is $\sin u'$ and the inner function is $u' = 5x$.

$\frac{d}{dx}(\sin 5x) = \cos(5x) \cdot \frac{d}{dx}(5x) = \cos(5x) \cdot 5 = 5 \cos 5x$

Applying the product rule to $y = e^x \sin 5x$:

$\frac{dy}{dx} = (e^x)(5 \cos 5x) + (\sin 5x)(e^x)$

$\frac{dy}{dx} = 5e^x \cos 5x + e^x \sin 5x$

We can factor out $e^x$:

$\frac{dy}{dx} = e^x (5 \cos 5x + \sin 5x)$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(e^x (5 \cos 5x + \sin 5x))$

Again, we use the product rule. Let $u' = e^x$ and $v' = 5 \cos 5x + \sin 5x$.

We find the derivatives of $u'$ and $v'$ with respect to $x$:

$\frac{du'}{dx} = \frac{d}{dx}(e^x) = e^x$

$\frac{dv'}{dx} = \frac{d}{dx}(5 \cos 5x + \sin 5x)$

Using the sum rule:

$\frac{dv'}{dx} = \frac{d}{dx}(5 \cos 5x) + \frac{d}{dx}(\sin 5x)$

For $\frac{d}{dx}(5 \cos 5x)$, use the constant multiple rule and chain rule:

$\frac{d}{dx}(5 \cos 5x) = 5 \cdot \frac{d}{dx}(\cos 5x) = 5 \cdot (-\sin 5x) \cdot \frac{d}{dx}(5x) = 5 \cdot (-\sin 5x) \cdot 5 = -25 \sin 5x$

For $\frac{d}{dx}(\sin 5x)$, use the chain rule (as calculated before):

$\frac{d}{dx}(\sin 5x) = 5 \cos 5x$

Combining the terms for $\frac{dv'}{dx}$:

$\frac{dv'}{dx} = -25 \sin 5x + 5 \cos 5x$

Applying the product rule to $e^x (5 \cos 5x + \sin 5x)$:

$\frac{d^2y}{dx^2} = (e^x)(-25 \sin 5x + 5 \cos 5x) + (5 \cos 5x + \sin 5x)(e^x)$

$\frac{d^2y}{dx^2} = -25e^x \sin 5x + 5e^x \cos 5x + 5e^x \cos 5x + e^x \sin 5x$

Combine like terms (terms with $e^x \sin 5x$ and terms with $e^x \cos 5x$):

$\frac{d^2y}{dx^2} = (-25e^x \sin 5x + e^x \sin 5x) + (5e^x \cos 5x + 5e^x \cos 5x)$

$\frac{d^2y}{dx^2} = -24e^x \sin 5x + 10e^x \cos 5x$

We can factor out $e^x$:

$\frac{d^2y}{dx^2} = e^x (10 \cos 5x - 24 \sin 5x)$

Thus, the second order derivative of the function $e^x \sin 5x$ is $e^x (10 \cos 5x - 24 \sin 5x)$.

The final answer is $\boxed{e^x (10 \cos 5x - 24 \sin 5x)}$.

Let the given function be denoted by $y$.

$y = e^{6x} \cos 3x$

To find the second order derivative, we first find the first order derivative.

We use the product rule for differentiation: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Here, let $u=e^{6x}$ and $v=\cos 3x$.

We find the derivatives of $u$ and $v$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(e^{6x}) = e^{6x} \cdot \frac{d}{dx}(6x) = e^{6x} \cdot 6 = 6e^{6x}$

$\frac{dv}{dx} = \frac{d}{dx}(\cos 3x) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -\sin(3x) \cdot 3 = -3 \sin 3x$

Applying the product rule to $y = e^{6x} \cos 3x$:

$\frac{dy}{dx} = (e^{6x})(-3 \sin 3x) + (\cos 3x)(6e^{6x})$

$\frac{dy}{dx} = -3e^{6x} \sin 3x + 6e^{6x} \cos 3x$

$\frac{dy}{dx} = e^{6x} (6 \cos 3x - 3 \sin 3x)$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(e^{6x} (6 \cos 3x - 3 \sin 3x))$

Again, we use the product rule. Let $u' = e^{6x}$ and $v' = 6 \cos 3x - 3 \sin 3x$.

We find the derivatives of $u'$ and $v'$ with respect to $x$:

$\frac{du'}{dx} = \frac{d}{dx}(e^{6x}) = 6e^{6x}$

$\frac{dv'}{dx} = \frac{d}{dx}(6 \cos 3x - 3 \sin 3x)$

Using the sum/difference rule and chain rule:

$\frac{dv'}{dx} = 6 \frac{d}{dx}(\cos 3x) - 3 \frac{d}{dx}(\sin 3x)$

$\frac{dv'}{dx} = 6 (-3 \sin 3x) - 3 (3 \cos 3x)$

$\frac{dv'}{dx} = -18 \sin 3x - 9 \cos 3x$

Applying the product rule to $e^{6x} (6 \cos 3x - 3 \sin 3x)$:

$\frac{d^2y}{dx^2} = (e^{6x})(-18 \sin 3x - 9 \cos 3x) + (6 \cos 3x - 3 \sin 3x)(6e^{6x})$

$\frac{d^2y}{dx^2} = -18e^{6x} \sin 3x - 9e^{6x} \cos 3x + 36e^{6x} \cos 3x - 18e^{6x} \sin 3x$

Combine the terms:

$\frac{d^2y}{dx^2} = (-18e^{6x} \sin 3x - 18e^{6x} \sin 3x) + (-9e^{6x} \cos 3x + 36e^{6x} \cos 3x)$

$\frac{d^2y}{dx^2} = -36e^{6x} \sin 3x + 27e^{6x} \cos 3x$

We can factor out $9e^{6x}$:

$\frac{d^2y}{dx^2} = 9e^{6x} (3 \cos 3x - 4 \sin 3x)$

Thus, the second order derivative of the function $e^{6x} \cos 3x$ is $9e^{6x} (3 \cos 3x - 4 \sin 3x)$.

The final answer is $\boxed{9e^{6x} (3 \cos 3x - 4 \sin 3x)}$.

Let the given function be denoted by $y$.

$y = \tan^{-1} x$

To find the second order derivative, we first find the first order derivative.

Differentiating $y$ with respect to $x$, we get:

$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1} x)$

The standard derivative of $\tan^{-1} x$ is:

$\frac{dy}{dx} = \frac{1}{1+x^2}$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{1+x^2}\right)$

We can rewrite the expression as $(1+x^2)^{-1}$ and use the chain rule.

Let $u = 1+x^2$. Then $\frac{1}{1+x^2} = u^{-1}$.

Using the chain rule, $\frac{d}{dx}(u^{-1}) = \frac{d}{du}(u^{-1}) \cdot \frac{du}{dx}$.

$\frac{d}{du}(u^{-1}) = -1 \cdot u^{-1-1} = -u^{-2}$

$\frac{du}{dx} = \frac{d}{dx}(1+x^2) = 0 + 2x = 2x$

Substituting $u = 1+x^2$ back into the expression:

$\frac{d^2y}{dx^2} = -(1+x^2)^{-2} \cdot (2x)$

$\frac{d^2y}{dx^2} = -\frac{2x}{(1+x^2)^2}$

Alternatively, using the quotient rule, $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

Here, $f(x) = 1$ and $g(x) = 1+x^2$.

$f'(x) = \frac{d}{dx}(1) = 0$

$g'(x) = \frac{d}{dx}(1+x^2) = 2x$

Applying the quotient rule:

$\frac{d^2y}{dx^2} = \frac{(0)(1+x^2) - (1)(2x)}{(1+x^2)^2}$

$\frac{d^2y}{dx^2} = \frac{0 - 2x}{(1+x^2)^2}$

$\frac{d^2y}{dx^2} = -\frac{2x}{(1+x^2)^2}$

Thus, the second order derivative of the function $\tan^{-1} x$ is $-\frac{2x}{(1+x^2)^2}$.

The final answer is $\boxed{-\frac{2x}{(1+x^2)^2}}$.

Let the given function be denoted by $y$.

$y = \log (\log x)$

(Assuming $\log x$ refers to the natural logarithm, $\log_e x$)

$y = \log_e (\log_e x)$

To find the second order derivative, we first find the first order derivative.

We use the chain rule. Let the outer function be $\log_e u$ and the inner function be $u = \log_e x$.

The derivative of the outer function with respect to $u$ is $\frac{d}{du}(\log_e u) = \frac{1}{u}$.

The derivative of the inner function with respect to $x$ is $\frac{du}{dx} = \frac{d}{dx}(\log_e x) = \frac{1}{x}$.

Applying the chain rule, $\frac{dy}{dx} = \frac{d}{du}(\log_e u) \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \frac{1}{u} \cdot \frac{1}{x}$

Substitute $u = \log_e x$ back into the expression:

$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{1}{x}$

$\frac{dy}{dx} = \frac{1}{x \log_e x}$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x \log_e x}\right)$

We use the quotient rule, which states $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

Here, let $f(x) = 1$ and $g(x) = x \log_e x$.

We find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(1) = 0$

$g'(x) = \frac{d}{dx}(x \log_e x)$. We use the product rule for $g'(x)$: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Let $u = x$ and $v = \log_e x$. Then $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{1}{x}$.

$g'(x) = (x)\left(\frac{1}{x}\right) + (\log_e x)(1)$

$g'(x) = 1 + \log_e x$

Now, apply the quotient rule to find $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{(0)(x \log_e x) - (1)(1 + \log_e x)}{(x \log_e x)^2}$

$\frac{d^2y}{dx^2} = \frac{0 - (1 + \log_e x)}{(x \log_e x)^2}$

$\frac{d^2y}{dx^2} = -\frac{1 + \log_e x}{(x \log_e x)^2}$

Thus, assuming $\log x = \log_e x$, the second order derivative of the function $\log (\log x)$ is $-\frac{1 + \log_e x}{(x \log_e x)^2}$.

The final answer is $\boxed{-\frac{1 + \log_e x}{(x \log_e x)^2}}$.

Let the given function be denoted by $y$.

$y = \sin (\log x)$

(Assuming $\log x$ represents the natural logarithm, $\log_e x$)

$y = \sin (\log_e x)$

To find the second order derivative, we first find the first order derivative.

We use the chain rule. Let $u = \log_e x$. Then $y = \sin u$.

Differentiating $y$ with respect to $x$ using the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$

$\frac{du}{dx} = \frac{d}{dx}(\log_e x) = \frac{1}{x}$

Substituting these into the chain rule formula:

$\frac{dy}{dx} = (\cos u) \cdot \left(\frac{1}{x}\right)$

Substitute $u = \log_e x$ back into the expression:

$\frac{dy}{dx} = \cos(\log_e x) \cdot \frac{1}{x}$

$\frac{dy}{dx} = \frac{\cos(\log_e x)}{x}$

Now, we find the second order derivative by differentiating the first order derivative $\frac{dy}{dx} = \frac{\cos(\log_e x)}{x}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{\cos(\log_e x)}{x}\right)$

We use the quotient rule, which states $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

Here, let $f(x) = \cos(\log_e x)$ and $g(x) = x$.

First, find the derivative of $f(x) = \cos(\log_e x)$ using the chain rule:

$f'(x) = \frac{d}{dx}(\cos(\log_e x))$

Let $v = \log_e x$. Then $f(x) = \cos v$. $\frac{df}{dx} = \frac{df}{dv} \cdot \frac{dv}{dx}$.

$\frac{df}{dv} = \frac{d}{dv}(\cos v) = -\sin v$

$\frac{dv}{dx} = \frac{d}{dx}(\log_e x) = \frac{1}{x}$

$f'(x) = (-\sin v) \cdot \left(\frac{1}{x}\right) = -\sin(\log_e x) \cdot \frac{1}{x} = -\frac{\sin(\log_e x)}{x}$

Next, find the derivative of $g(x) = x$:

$g'(x) = \frac{d}{dx}(x) = 1$

Now, apply the quotient rule:

$\frac{d^2y}{dx^2} = \frac{\left(-\frac{\sin(\log_e x)}{x}\right)(x) - (\cos(\log_e x))(1)}{x^2}$

Simplify the numerator:

$\frac{d^2y}{dx^2} = \frac{-\sin(\log_e x) - \cos(\log_e x)}{x^2}$

We can factor out $-1$ from the numerator:

$\frac{d^2y}{dx^2} = -\frac{\sin(\log_e x) + \cos(\log_e x)}{x^2}$

Thus, assuming $\log x = \log_e x$, the second order derivative of the function $\sin (\log x)$ is $-\frac{\sin(\log_e x) + \cos(\log_e x)}{x^2}$.

The final answer is $\boxed{-\frac{\sin(\log_e x) + \cos(\log_e x)}{x^2}}$.

Given:

To Prove:

$\frac{d^2y}{dx^2} + y = 0$

Proof:

We are given the function:

$y = 5 \cos x - 3 \sin x$

First, we find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the difference rule and constant multiple rule.

$\frac{dy}{dx} = \frac{d}{dx}(5 \cos x - 3 \sin x)$

$\frac{dy}{dx} = 5 \frac{d}{dx}(\cos x) - 3 \frac{d}{dx}(\sin x)$

We know that $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\sin x) = \cos x$.

$\frac{dy}{dx} = 5 (-\sin x) - 3 (\cos x)$

$\frac{dy}{dx} = -5 \sin x - 3 \cos x$

Next, we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$. Use the difference rule and constant multiple rule again.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-5 \sin x - 3 \cos x)$

$\frac{d^2y}{dx^2} = -5 \frac{d}{dx}(\sin x) - 3 \frac{d}{dx}(\cos x)$

We know that $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$.

$\frac{d^2y}{dx^2} = -5 (\cos x) - 3 (-\sin x)$

$\frac{d^2y}{dx^2} = -5 \cos x + 3 \sin x$

Now, we substitute the expressions for $\frac{d^2y}{dx^2}$ and $y$ into the equation $\frac{d^2y}{dx^2} + y$:

$\frac{d^2y}{dx^2} + y = (-5 \cos x + 3 \sin x) + (5 \cos x - 3 \sin x)$

Combine the terms:

$\frac{d^2y}{dx^2} + y = (-5 \cos x + 5 \cos x) + (3 \sin x - 3 \sin x)$

$\frac{d^2y}{dx^2} + y = 0 + 0$

$\frac{d^2y}{dx^2} + y = 0$

Since we have shown that $\frac{d^2y}{dx^2} + y$ simplifies to 0, the equation is proven.

Hence, proven.

Given:

This implies that $x = \cos y$.

To Find:

$\frac{d^2y}{dx^2}$ in terms of $y$ alone.

Solution:

We are given $y = \cos^{-1} x$.

First, find the first derivative $\frac{dy}{dx}$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1} x)$

The standard derivative of $\cos^{-1} x$ is:

$\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$

Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{d}{dx}((1-x^2)^{-\frac{1}{2}})$

We use the chain rule. Let $u = 1-x^2$. Then $(1-x^2)^{-\frac{1}{2}} = u^{-\frac{1}{2}}$.

$\frac{d}{dx}(u^{-\frac{1}{2}}) = \frac{d}{du}(u^{-\frac{1}{2}}) \cdot \frac{du}{dx}$

$\frac{d}{du}(u^{-\frac{1}{2}}) = -\frac{1}{2} u^{-\frac{1}{2} - 1} = -\frac{1}{2} u^{-\frac{3}{2}}$

$\frac{du}{dx} = \frac{d}{dx}(1-x^2) = 0 - 2x = -2x$

Applying the chain rule:

$\frac{d}{dx}((1-x^2)^{-\frac{1}{2}}) = \left(-\frac{1}{2} (1-x^2)^{-\frac{3}{2}}\right) \cdot (-2x)$

$\frac{d}{dx}((1-x^2)^{-\frac{1}{2}}) = x (1-x^2)^{-\frac{3}{2}}$

So, the second derivative is:

$\frac{d^2y}{dx^2} = -[x (1-x^2)^{-\frac{3}{2}}]$

$\frac{d^2y}{dx^2} = -\frac{x}{(1-x^2)^{\frac{3}{2}}}$

We need to express $\frac{d^2y}{dx^2}$ in terms of $y$ alone. We know that $x = \cos y$.

Substitute $x = \cos y$ into the expression for $\frac{d^2y}{dx^2}$.

$\frac{d^2y}{dx^2} = -\frac{\cos y}{(1-(\cos y)^2)^{\frac{3}{2}}}$

Using the trigonometric identity $\sin^2 y + \cos^2 y = 1$, we have $1 - \cos^2 y = \sin^2 y$.

$\frac{d^2y}{dx^2} = -\frac{\cos y}{(\sin^2 y)^{\frac{3}{2}}}$

$(\sin^2 y)^{\frac{3}{2}} = (\sin^2 y)^{\frac{1}{2} \cdot 3} = ((\sin^2 y)^{\frac{1}{2}})^3 = (|\sin y|)^3$.

However, the domain of $y = \cos^{-1} x$ is typically $[0, \pi]$, where $\sin y \geq 0$. Thus, $|\sin y| = \sin y$.

So, $(\sin^2 y)^{\frac{3}{2}} = (\sin y)^3 = \sin^3 y$.

$\frac{d^2y}{dx^2} = -\frac{\cos y}{\sin^3 y}$

We can rewrite this expression using trigonometric identities: $\cot y = \frac{\cos y}{\sin y}$ and $\text{cosec} y = \frac{1}{\sin y}$.

$\frac{d^2y}{dx^2} = -\frac{\cos y}{\sin y \cdot \sin^2 y} = -\frac{\cos y}{\sin y} \cdot \left(\frac{1}{\sin y}\right)^2$

$\frac{d^2y}{dx^2} = - \cot y \cdot (\text{cosec} y)^2$

$\frac{d^2y}{dx^2} = -\cot y \ \text{cosec}^2 y$

Thus, the second order derivative of $y = \cos^{-1} x$ in terms of $y$ alone is $-\frac{\cos y}{\sin^3 y}$ or $-\cot y \ \text{cosec}^2 y$.

The final answer is $\boxed{-\cot y \ \text{cosec}^2 y}$.

Given:

(Assuming $\log x$ represents the natural logarithm, $\log_e x$)

$y = 3 \cos (\log_e x) + 4 \sin (\log_e x)$

To Show:

$x^2 y_2 + xy_1 + y = 0$, where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$.

Proof:

We are given the function:

$y = 3 \cos (\log_e x) + 4 \sin (\log_e x)$

First, find the first derivative $y_1 = \frac{dy}{dx}$ by differentiating $y$ with respect to $x$. We use the chain rule for $\cos(\log_e x)$ and $\sin(\log_e x)$, remembering that $\frac{d}{dx}(\log_e x) = \frac{1}{x}$.

$\frac{dy}{dx} = \frac{d}{dx}(3 \cos (\log_e x)) + \frac{d}{dx}(4 \sin (\log_e x))$

$\frac{dy}{dx} = 3 \frac{d}{dx}(\cos (\log_e x)) + 4 \frac{d}{dx}(\sin (\log_e x))$

$\frac{dy}{dx} = 3 (-\sin (\log_e x) \cdot \frac{1}{x}) + 4 (\cos (\log_e x) \cdot \frac{1}{x})$

$y_1 = -\frac{3 \sin (\log_e x)}{x} + \frac{4 \cos (\log_e x)}{x}$