Chapter 5 Continuity And Differentiability

Given:

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

We need to check the continuity at the point $x = 1$.

To Check:

Whether the function $f(x)$ is continuous at $x = 1$.

Solution:

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

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

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

3. The limit must equal the function value, i.e., $\lim\limits_{x \to c} f(x) = f(c)$.

Here, $c=1$. Let's check these conditions.

Step 1: Value of the function at $x = 1$

First, we find the value of the function at $x=1$ by direct substitution:

$f(1) = 2(1) + 3 = 2 + 3 = 5$

The function is defined at $x=1$ and its value is 5.

Step 2: Limit of the function as $x \to 1$

Next, we find the limit of the function as $x$ approaches 1.

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

Since $f(x)$ is a polynomial function, its limit can be evaluated by direct substitution:

$\lim\limits_{x \to 1} (2x + 3) = 2(1) + 3 = 5$

The limit exists and is equal to 5.

Step 3: Compare the limit and the function value

We compare the results from Step 1 and Step 2.

$\lim\limits_{x \to 1} f(x) = 5$ and $f(1) = 5$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, all three conditions for continuity are met.

Conclusion:

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

Given:

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

We need to examine its continuity at the point $x = 0$.

To Examine:

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

Solution:

We need to check if $\lim\limits_{x \to 0} f(x) = f(0)$.

Step 1: Value of the function at $x = 0$

Substitute $x=0$ into the function:

$f(0) = (0)^2 = 0$

The function is defined at $x=0$.

Step 2: Limit of the function as $x \to 0$

We find the limit of $f(x)$ as $x$ approaches 0.

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

Since $f(x)$ is a polynomial function, we can use direct substitution:

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

The limit exists and is equal to 0.

Step 3: Compare the limit and the function value

We compare the results from the previous steps:

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

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the condition for continuity is satisfied.

Conclusion:

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

Given:

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

We need to discuss its continuity at the point $x = 0$.

To Discuss:

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

Solution:

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

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

To check for continuity at $x = 0$, where the definition changes, we must check if the left-hand limit (LHL), right-hand limit (RHL), and the function value $f(0)$ are all equal.

Step 1: Value of the function at $x = 0$

Using the definition for $x \geq 0$, we have:

$f(0) = 0$

Step 2: Left-Hand Limit (LHL)

As $x$ approaches 0 from the left, $x < 0$. So we use $f(x) = -x$.

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

Step 3: Right-Hand Limit (RHL)

As $x$ approaches 0 from the right, $x > 0$. So we use $f(x) = x$.

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

Step 4: Compare the values

We have LHL = 0, RHL = 0, and $f(0) = 0$.

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

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

Conclusion:

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

Given:

The function $f$ is defined as:

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

To Show:

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

Proof:

A function is continuous at a point $c$ if $\lim\limits_{x \to c} f(x) = f(c)$. We will check this condition for $c = 0$.

Step 1: Value of the function at $x = 0$

From the definition of the function, when $x = 0$, the value of the function is given directly:

$f(0) = 1$

Step 2: Limit of the function as $x \to 0$

To find the limit as $x$ approaches 0, we consider values of $x$ close to 0 but not equal to 0 (i.e., $x \neq 0$). For these values, the function is defined as $f(x) = x^3 + 3$.

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

By direct substitution:

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

Step 3: Compare the limit and the function value

We compare the results:

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

$f(0) = 1$

Since $\lim\limits_{x \to 0} f(x) \neq f(0)$ (because $3 \neq 1$), the condition for continuity is not met.

Conclusion:

Because the limit of the function at $x=0$ does not equal the value of the function at $x=0$, the function $f(x)$ is not continuous at $x = 0$.

Given:

The constant function $f(x) = k$, where $k$ is a real constant. The domain is all real numbers, $\mathbb{R}$.

To Check:

The points of continuity for the function $f(x) = k$.

Solution:

To check for continuity, we pick an arbitrary point $c$ from the domain $\mathbb{R}$ and verify if $\lim\limits_{x \to c} f(x) = f(c)$.

Step 1: Value of the function at $x = c$

By definition of the function, for any input $x$, the output is $k$. Therefore:

$f(c) = k$

Step 2: Limit of the function as $x \to c$

We evaluate the limit of $f(x)$ as $x$ approaches $c$.

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

The limit of a constant is the constant itself, so:

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

Step 3: Compare the limit and the function value

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

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

Since $c$ was an arbitrary real number, this result holds for every point in the domain.

Conclusion:

The constant function $f(x) = k$ is continuous at every real number.

Given:

The identity function $f(x) = x$, defined on the set of all real numbers, $\mathbb{R}$.

To Prove:

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

Proof:

To prove continuity for all real numbers, we select an arbitrary real number $c$ and show that the function is continuous at $c$. The condition for continuity is $\lim\limits_{x \to c} f(x) = f(c)$.

Step 1: Value of the function at $x = c$

For the identity function, the output is the same as the input. So:

$f(c) = c$

Step 2: Limit of the function as $x \to c$

We evaluate the limit of $f(x)$ as $x$ approaches $c$.

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

By the fundamental properties of limits:

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

Step 3: Compare the limit and the function value

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

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

Since $c$ was an arbitrary real number, this property holds for all real numbers.

Conclusion:

The identity function $f(x) = x$ is continuous at every real number.

Given:

The function $f(x) = |x|$, defined for all real numbers.

To Determine:

Whether $f(x) = |x|$ is a continuous function. This requires checking for continuity at all points in its domain, $\mathbb{R}$.

Solution:

We analyze the continuity by considering three cases for an arbitrary point $c \in \mathbb{R}$.

The function is defined as $f(x) = \begin{cases} x & , & x \geq 0 \\ -x & , & x < 0 \end{cases}$

Case 1: $c > 0$

Let $c$ be any positive real number. For values of $x$ near $c$, $x$ will also be positive, so $f(x) = x$.

The function value is $f(c) = c$.

The limit is $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} x = c$.

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

Case 2: $c < 0$

Let $c$ be any negative real number. For values of $x$ near $c$, $x$ will also be negative, so $f(x) = -x$.

The function value is $f(c) = -c$. (Note: since $c$ is negative, $-c$ is positive).

The limit is $\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (-x) = -c$.

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

Case 3: $c = 0$

This is the point where the function's definition changes. We must check the left and right-hand limits.

Function value: $f(0) = 0$.

Left-Hand Limit (LHL): For $x \to 0^-$, $x<0$, so $f(x) = -x$.

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

Right-Hand Limit (RHL): For $x \to 0^+$, $x>0$, so $f(x) = x$.

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

Since LHL = RHL = $f(0)$, the function is continuous at $c = 0$.

As the function is continuous for $c > 0$, $c < 0$, and $c = 0$, it is continuous for all real numbers.

Conclusion:

Yes, the function $f(x) = |x|$ is a continuous function.

Given:

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

To Discuss:

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

Solution:

The given function $f(x) = x^3 + x^2 - 1$ is a polynomial function. We can prove its continuity in two ways.

Method 1: Using the property of polynomial functions

A fundamental theorem in calculus states that all polynomial functions are continuous for all real numbers. Since $f(x)$ is a polynomial, it is continuous everywhere on its domain, which is $\mathbb{R}$.

Method 2: Direct proof using limits

Let $c$ be an arbitrary real number. We need to show that $\lim\limits_{x \to c} f(x) = f(c)$.

First, the value of the function at $x = c$ is:

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

Next, we evaluate the limit as $x$ approaches $c$:

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

Using the properties of limits (sum, difference, and power rules):

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

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

Comparing the two results, we see that $\lim\limits_{x \to c} f(x) = f(c)$.

Since $c$ was an arbitrary real number, the function is continuous everywhere.

Conclusion:

The function $f(x) = x^3 + x^2 - 1$ is continuous for all real numbers.

Given:

The function $f(x) = \frac{1}{x}$.

The domain of the function is specified as all real numbers except 0, i.e., $x \in \mathbb{R} \setminus \{0\}$.

To Discuss:

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

Solution:

To discuss the continuity, we must check every point within the function's domain.

Let $c$ be an arbitrary real number in the domain of $f$, which means $c \neq 0$.

We need to verify if $\lim\limits_{x \to c} f(x) = f(c)$.

Step 1: Value of the function at $x = c$

Since $c$ is in the domain, $f(c)$ is defined:

$f(c) = \frac{1}{c}$

Step 2: Limit of the function as $x \to c$

We evaluate the limit of $f(x)$ as $x$ approaches $c$.

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

Since $c \neq 0$, we can use the quotient rule for limits (or direct substitution for rational functions where the denominator is non-zero):

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

Step 3: Compare the limit and the function value

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

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

As $c$ was any arbitrary point in the domain, the function is continuous at every point of its domain.

Note: The function is not defined at $x=0$, so the question of continuity at $x=0$ does not arise.

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:

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

To Discuss:

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

Solution:

The function is defined by two different polynomial expressions. Polynomials are continuous everywhere. Therefore, the only point where a discontinuity might occur is at $x=1$, where the definition of the function changes.

We analyze the continuity in three parts:

Case 1: $c < 1$

For any $c < 1$, $f(x) = x+2$ in the neighborhood of $c$. Since this is a polynomial, the function is continuous for all $x < 1$.

Case 2: $c > 1$

For any $c > 1$, $f(x) = x-2$ in the neighborhood of $c$. Since this is a polynomial, the function is continuous for all $x > 1$.

Case 3: $c = 1$

We must check the continuity at $x=1$ by comparing the function value, the left-hand limit (LHL), and the right-hand limit (RHL).

Function value at $x=1$:

The definition for $x \le 1$ is $f(x) = x+2$. So, $f(1) = 1+2 = 3$.

Left-Hand Limit (LHL):

As $x \to 1^-$, $x<1$, so we use $f(x) = x+2$.

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

Right-Hand Limit (RHL):

As $x \to 1^+$, $x>1$, so we use $f(x) = x-2$.

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

Since the Left-Hand Limit (3) is not equal to the Right-Hand Limit (-1), the overall limit $\lim\limits_{x \to 1} f(x)$ does not exist. Therefore, the function is not continuous at $x=1$.

Conclusion:

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

Given:

The function $f(x)$ is defined as:

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

To Find:

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

Solution:

The function is defined by polynomials for $x < 1$ and $x > 1$, so it is continuous on the intervals $(-\infty, 1)$ and $(1, \infty)$. The only potential point of discontinuity is at $x=1$.

We check for continuity at $x=1$ by finding the function value, LHL, and RHL.

Function value at $x=1$:

From the definition, $f(1) = 0$.

Left-Hand Limit (LHL):

As $x \to 1^-$, $x < 1$, so we use $f(x) = x+2$.

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

Right-Hand Limit (RHL):

As $x \to 1^+$, $x>1$, so we use $f(x) = x-2$.

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

Since the LHL ($3$) is not equal to the RHL ($-1$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist. A function cannot be continuous at a point where the limit does not exist.

Therefore, the function is discontinuous at $x=1$.

Conclusion:

The only point of discontinuity for the function $f(x)$ is $x = 1$.

Given:

The function $f(x)$ is defined as:

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

To Discuss:

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

Solution:

The domain of this function is all real numbers except $x=0$, as the function is not defined at $x=0$. The domain is $\mathbb{R} \setminus \{0\}$.

We discuss continuity at points within its domain.

Case 1: $c < 0$

For any $c < 0$, the function is defined as $f(x) = x+2$. This is a polynomial and is continuous for all $x < 0$.

Case 2: $c > 0$

For any $c > 0$, the function is defined as $f(x) = -x+2$. This is also a polynomial and is continuous for all $x > 0$.

At the point $x=0$:

The function $f(x)$ is not defined at $x=0$. A necessary condition for continuity at a point is that the function must be defined at that point. Since $f(0)$ is not defined, the function is not continuous at $x=0$.

However, the question asks to discuss the continuity of the function, which usually means on its domain of definition.

Conclusion:

The function $f(x)$ is continuous at every point in its domain, which is $(-\infty, 0) \cup (0, \infty)$. There is a discontinuity at $x=0$ because the function is not defined there.

Given:

The function $f(x)$ is defined as:

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

To Discuss:

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

Solution:

The only point where a discontinuity might occur is at $x=0$, where the definition changes. For all other points, the function is defined by a polynomial ($x$ or $x^2$) and is therefore continuous.

We check for continuity at $x=0$.

Function value at $x=0$:

For $x \ge 0$, $f(x) = x$. Thus, $f(0) = 0$.

Left-Hand Limit (LHL):

As $x \to 0^-$, $x<0$, so we use $f(x) = x^2$.

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

Right-Hand Limit (RHL):

As $x \to 0^+$, $x>0$, so we use $f(x) = x$.

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

Since LHL = RHL = $f(0) = 0$, the function is continuous at $x=0$.

As the function is continuous for $x<0$, $x>0$, and at $x=0$, it is continuous for all real numbers.

Conclusion:

The function $f(x)$ is a continuous function for all real numbers.

Given:

A general polynomial function $P(x)$ of degree $n$:

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

where $a_0, a_1, \dots, a_n$ are real coefficients. The domain of any polynomial is $\mathbb{R}$.

To Prove:

Every polynomial function is continuous for all real numbers.

Proof:

Let $c$ be an arbitrary real number. To prove continuity at $c$, we must show that $\lim\limits_{x \to c} P(x) = P(c)$.

Step 1: Value of the function at $x=c$

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

Step 2: Limit of the function as $x \to c$

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

Using the properties of limits (sum rule, constant multiple rule, power rule):

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

We know that $\lim\limits_{x \to c} x^k = c^k$ for any positive integer $k$, and $\lim\limits_{x \to c} a_0 = a_0$. Substituting these results:

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

Step 3: Compare the limit and the function value

From Step 1 and Step 2, we see that:

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

Since $c$ was an arbitrary real number, this holds true for every point in the domain of the polynomial.

Conclusion:

Every polynomial function is continuous on the set of all real numbers ($\mathbb{R}$).

Given:

The greatest integer function $f(x) = [x]$. The domain is $\mathbb{R}$.

To Find:

All points where the function $f(x) = [x]$ is discontinuous.

Solution:

We analyze the continuity by considering two cases for an arbitrary point $c \in \mathbb{R}$.

Case 1: $c$ is an integer

Let $c=n$, where $n$ is an integer. We check the condition for continuity at $x=n$.

Function value: $f(n) = [n] = n$.

Left-Hand Limit (LHL): As $x \to n^-$, $x$ is slightly less than $n$ (e.g., $n-0.001$). For such $x$, the greatest integer less than or equal to $x$ is $n-1$.

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

Right-Hand Limit (RHL): As $x \to n^+$, $x$ is slightly greater than $n$ (e.g., $n+0.001$). For such $x$, the greatest integer less than or equal to $x$ is $n$.

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

Since LHL ($n-1$) $\neq$ RHL ($n$), the limit $\lim\limits_{x \to n} f(x)$ does not exist. Therefore, the function is discontinuous at every integer.

Case 2: $c$ is not an integer

Let $c$ be a real number that is not an integer. Then there exists an integer $n$ such that $n < c < n+1$.

Function value: $f(c) = [c] = n$.

Limit: For any $x$ in a small neighborhood around $c$ (e.g., for $x$ in $(n, n+1)$), the value of $[x]$ is always $n$.

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} [x] = n$.

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

Conclusion:

The greatest integer function $f(x) = [x]$ is discontinuous at every integer point.

Given:

A rational function is a function of the form $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomial functions and $Q(x) \neq 0$.

The domain of $f(x)$ is all real numbers $x$ such that $Q(x) \neq 0$.

To Prove:

Every rational function is continuous on its domain.

Proof:

We use the properties of continuous functions.

1. From Example 14, we know that every polynomial function is continuous for all real numbers. Therefore, $P(x)$ and $Q(x)$ are both continuous functions on $\mathbb{R}$.

2. A theorem on the algebra of continuous functions states that if two functions, say $g(x)$ and $h(x)$, are continuous at a point $c$, then their quotient $\frac{g(x)}{h(x)}$ is also continuous at $c$, provided that $h(c) \neq 0$.

Let $c$ be an arbitrary point in the domain of the rational function $f(x) = \frac{P(x)}{Q(x)}$.

By the definition of the domain, $Q(c) \neq 0$.

Since $P(x)$ and $Q(x)$ are polynomials, they are continuous at $c$.

Applying the quotient rule theorem for continuity, since $P(x)$ and $Q(x)$ are continuous at $c$ and $Q(c) \neq 0$, the function $f(x) = \frac{P(x)}{Q(x)}$ must be continuous at $c$.

Since $c$ was an arbitrary point in the domain of $f(x)$, the function is continuous at every point of its domain.

Conclusion:

Every rational function is continuous at every point in its domain.

Given:

The sine function, $f(x) = \sin(x)$. Its domain is all real numbers, $\mathbb{R}$.

To Discuss:

The continuity of the sine function.

Solution:

Let $c$ be an arbitrary real number. We need to show that $\lim\limits_{x \to c} \sin(x) = \sin(c)$.

Let $x = c + h$. As $x \to c$, it implies that $h \to 0$. We can rewrite the limit in terms of $h$.

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

Using the trigonometric sum 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 for limits:

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

Since $\sin c$ and $\cos c$ are constants with respect to $h$, we can factor them out:

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

We use the two fundamental trigonometric limits: $\lim\limits_{h \to 0} \cos h = 1$ and $\lim\limits_{h \to 0} \sin h = 0$.

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

So, we have shown that $\lim\limits_{x \to c} \sin(x) = \sin(c)$.

Since this holds for any arbitrary real number $c$, the sine function is continuous everywhere.

Conclusion:

The sine function, $f(x) = \sin(x)$, is continuous for all real numbers.

Given:

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

To Prove:

The function $f(x) = \tan x$ is a continuous function on its domain.

Proof:

The tangent function can be expressed as a ratio of the sine and cosine functions:

$f(x) = \tan x = \frac{\sin x}{\cos x}$

The domain of $\tan x$ is the set of all real numbers $x$ for which the denominator is not zero, i.e., $\cos x \neq 0$. This occurs when $x \neq \frac{\pi}{2} + n\pi$ for any integer $n$.

We know from the properties of continuous functions:

1. The function $g(x) = \sin x$ is continuous for all real numbers.

2. The function $h(x) = \cos x$ is continuous for all real numbers (proof similar to sine).

3. The quotient of two continuous functions $\frac{g(x)}{h(x)}$ is continuous at every point $c$ where both are continuous and $h(c) \neq 0$.

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

Since $\sin x$ and $\cos x$ are continuous at $c$ and $\cos c \neq 0$, their quotient $f(x) = \frac{\sin x}{\cos x} = \tan x$ is continuous at $c$.

As $c$ was an arbitrary point in the domain, the tangent function is continuous at every point of its domain.

Conclusion:

The function $f(x) = \tan x$ is a continuous function on its domain.

Given:

The function $f(x) = \sin(x^2)$. Its domain is all real numbers, $\mathbb{R}$.

To Show:

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

Proof:

We can prove this by recognizing that $f(x)$ is a composition of two functions.

Let's define two functions:

$g(x) = x^2$

$h(u) = \sin u$

Then the function $f(x)$ can be written as the composition of $h$ and $g$:

$f(x) = h(g(x)) = (h \circ g)(x)$

Now we use the theorem on the continuity of composite functions, which states: If $g$ is continuous at a point $c$ and $h$ is continuous at $g(c)$, then the composite function $(h \circ g)$ is continuous at $c$.

Let's check the continuity of $g(x)$ and $h(u)$:

1. Continuity of $g(x) = x^2$: This is a polynomial function, so it is continuous for all real numbers.

2. Continuity of $h(u) = \sin u$: This is the sine function, which is continuous for all real numbers.

Now, let $c$ be any arbitrary real number.

The function $g(x)$ is continuous at $c$. The function $h(u)$ is continuous at all real numbers, so it is also continuous at the point $g(c) = c^2$.

According to the theorem, since both conditions are met, the composite function $f(x) = (h \circ g)(x) = \sin(x^2)$ is continuous at $c$.

Since $c$ was arbitrary, the function is continuous for all real numbers.

Conclusion:

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

Given:

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

To Show:

The function $f(x)$ is continuous for all real numbers.

Proof (Method 1: Using Composition of Functions):

We can show that $f(x)$ is continuous by building it from simpler functions that are known to be continuous.

1. The functions $g(x) = 1$ (constant), $h(x) = x$ (identity), and $k(x) = |x|$ (absolute value) are all continuous for all real numbers.

2. The sum and difference of continuous functions are continuous. Therefore, the function $m(x) = 1 - x + |x|$ (i.e., $g(x) - h(x) + k(x)$) is continuous for all real numbers.

3. The composition of continuous functions is continuous. The function $f(x)$ is the composition of the absolute value function $k(x)$ and the function $m(x)$ we just defined. That is, $f(x) = |m(x)| = k(m(x))$.

Since $m(x)$ is continuous everywhere and $k(x)=|x|$ is continuous everywhere, their composition $f(x)$ is continuous for all real numbers.

Proof (Method 2: Using Piecewise Definition):

We can rewrite the function by considering the definition of $|x|$.

Case 1: $x \geq 0$

If $x \geq 0$, then $|x| = x$.

$f(x) = |1 - x + x| = |1| = 1$

Case 2: $x < 0$

If $x < 0$, then $|x| = -x$.

$f(x) = |1 - x + (-x)| = |1 - 2x|$

So, the piecewise definition of $f(x)$ is:

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

The function is continuous for $x > 0$ (constant function) and for $x < 0$ (composition of absolute value and polynomial). We only need to check for continuity at $x = 0$.

Function value: $f(0) = 1$.

LHL: $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} |1-2x| = |1 - 2(0)| = 1$.

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

Since LHL = RHL = $f(0) = 1$, the function is continuous at $x=0$.

Conclusion:

Both methods show that the function $f(x) = |1 - x + |x||$ is continuous for all real numbers.

Given:

The function is $f(x) = 5x - 3$.

To Prove:

The function $f(x)$ is continuous at $x = 0$, $x = -3$, and $x = 5$.

Proof:

A function $f(x)$ is continuous at a point $x=c$ if $\lim\limits_{x \to c} f(x) = f(c)$.

The given function $f(x) = 5x - 3$ is a polynomial function. We know that polynomial functions are continuous for all real numbers. However, we will prove it for the specified points.

Case 1: Continuity at $x = 0$

Value of the function at $x=0$:

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

Limit of the function as $x \to 0$:

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

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

Case 2: Continuity at $x = -3$

Value of the function at $x=-3$:

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

Limit of the function as $x \to -3$:

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

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

Case 3: Continuity at $x = 5$

Value of the function at $x=5$:

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

Limit of the function as $x \to 5$:

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

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

Given:

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

The point to check is $x = 3$.

To Examine:

The continuity of $f(x)$ at $x = 3$.

Solution:

For a function to be continuous at a point $x = c$, we must have $\lim\limits_{x \to c} f(x) = f(c)$.

Here, $c=3$.

Step 1: Find the value of the function at $x = 3$.

$f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$

Step 2: Find the limit of the function as $x \to 3$.

Since $f(x)$ is a polynomial, the limit can be found by direct substitution.

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

Step 3: Compare the limit and the function value.

We see that $\lim\limits_{x \to 3} f(x) = 17$ and $f(3) = 17$.

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

Conclusion:

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

(a) f(x) = x – 5

The function $f(x) = x-5$ is a polynomial function. Since all polynomial functions are continuous for all real numbers, $f(x)$ is a continuous function on its domain $\mathbb{R}$.

(b) f(x) = $\frac{1}{x-5}$ , x ≠ 5

The function $f(x) = \frac{1}{x-5}$ is a rational function. A rational function is continuous at every point in its domain. The domain of this function is all real numbers except where the denominator is zero, i.e., $x-5 \neq 0 \implies x \neq 5$.

Thus, $f(x)$ is continuous on its domain, which is $\mathbb{R} \setminus \{5\}$.

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

The domain is given as all real numbers except $x=-5$. For any $x$ in the domain, we can simplify the function:

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

So, for all $x \neq -5$, the function is equivalent to the polynomial $g(x) = x-5$. Since polynomials are continuous everywhere, $f(x)$ is continuous at every point in its domain.

Therefore, $f(x)$ is continuous on its domain, which is $\mathbb{R} \setminus \{-5\}$.

(d) f(x) = | x – 5 |

This function is a composition of two continuous functions:

1. The polynomial function $g(x) = x-5$, which is continuous for all $x \in \mathbb{R}$.

2. The absolute value function $h(u) = |u|$, which is continuous for all $u \in \mathbb{R}$.

The given function is $f(x) = h(g(x))$. The composition of two continuous functions is always continuous. Therefore, $f(x) = |x-5|$ is a continuous function for all real numbers.

Given:

The function $f(x) = x^n$, where $n$ is a positive integer.

To Prove:

The function $f(x)$ is continuous at the point $x = n$.

Proof:

To prove continuity at $x=n$, we need to show that $\lim\limits_{x \to n} f(x) = f(n)$.

Step 1: Find the value of the function at $x = n$.

$f(n) = n^n$

Step 2: Find the limit of the function as $x \to n$.

The function $f(x) = x^n$ is a polynomial function for any positive integer $n$. For any polynomial function, the limit at a point can be evaluated by direct substitution.

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

Step 3: Compare the limit and the function value.

We found that $\lim\limits_{x \to n} f(x) = n^n$ and $f(n) = n^n$.

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

Conclusion:

The function $f(x) = x^n$ is continuous at x = n for any positive integer $n$.

Given:

The function is defined as $f(x) = \begin{cases} x,& if \;x ≤ 1\\5,& if\; x > 1 \end{cases}$.

Solution:

Continuity at x = 0:

For values of $x$ near 0, the rule $f(x) = x$ applies since $0 \le 1$.

The value of the function is $f(0) = 0$.

The limit is $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} x = 0$.

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

Continuity at x = 1:

This is the point where the function's definition changes, so we must check the left-hand and right-hand limits.

The value of the function is $f(1) = 1$ (using the rule for $x \le 1$).

The Left-Hand Limit (LHL): $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} x = 1$.

The Right-Hand Limit (RHL): $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 5 = 5$.

Since LHL $\neq$ RHL ($1 \neq 5$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist. Therefore, the function is not continuous at x = 1.

Continuity at x = 2:

For values of $x$ near 2, the rule $f(x) = 5$ applies since $2 > 1$.

The value of the function is $f(2) = 5$.

The limit is $\lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} 5 = 5$.

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

Given:

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

To Find:

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

Solution:

The function is defined by polynomials in the intervals $(-\infty, 2)$ and $(2, \infty)$, so it is continuous in these intervals. The only potential point of discontinuity is at $x=2$, where the definition changes.

We check for continuity at $x=2$.

Left-Hand Limit (LHL):

As $x \to 2^-$, we use $f(x) = 2x+3$.

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

Right-Hand Limit (RHL):

As $x \to 2^+$, we use $f(x) = 2x-3$.

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

Since the Left-Hand Limit (7) is not equal to the Right-Hand Limit (1), the limit $\lim\limits_{x \to 2} f(x)$ does not exist.

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

Conclusion:

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

Given:

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

To Find:

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

Solution:

The potential points of discontinuity are at $x=-3$ and $x=3$. For $x \le -3$, we can write $|x| = -x$, so $f(x) = -x+3$. The function is defined by polynomials on the open intervals $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$, so it is continuous there.

Continuity at $x = -3$:

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

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

Function Value: $f(-3) = |-3|+3 = 3+3 = 6$.

Since LHL = RHL = $f(-3)$, the function is continuous at $x=-3$.

Continuity at $x = 3$:

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

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

Since LHL $\neq$ RHL ($-6 \neq 20$), the limit at $x=3$ does not exist, and the function is discontinuous at $x=3$.

Conclusion:

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

Given:

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

To Find:

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

Solution:

Let's first simplify the function definition. For $x > 0$, $|x|=x$, so $f(x) = \frac{x}{x} = 1$. For $x < 0$, $|x|=-x$, so $f(x) = \frac{-x}{x} = -1$.

The function can be rewritten as:

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

The function is constant (and thus continuous) for all $x < 0$ and all $x > 0$. The only potential point of discontinuity is at $x=0$.

We check for continuity at $x=0$.

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

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

Since LHL $\neq$ RHL ($-1 \neq 1$), the limit at $x=0$ does not exist, and the function is discontinuous at $x=0$.

Conclusion:

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

Given:

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

To Find:

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

Solution:

Let's simplify the function definition for $x < 0$. If $x < 0$, then $|x| = -x$.

So, for $x < 0$, $f(x) = \frac{x}{|x|} = \frac{x}{-x} = -1$.

The function can be rewritten as:

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

This shows that $f(x) = -1$ for all real numbers $x$. This is a constant function.

Constant functions are continuous everywhere. There is no break or jump at $x=0$ because the value from the left, from the right, and at the point itself is -1.

Conclusion:

The function is continuous for all real numbers. There are no points of discontinuity.

Given:

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

To Find:

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

Solution:

The function is defined by polynomials on the intervals $(-\infty, 1)$ and $(1, \infty)$, so it is continuous there. The only potential point of discontinuity is at $x=1$.

We check for continuity at $x=1$.

LHL: As $x \to 1^-$, we use $f(x) = x^2+1$.

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

RHL: As $x \to 1^+$, we use $f(x) = x+1$.

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

Function Value: $f(1) = 1+1 = 2$.

Since LHL = RHL = $f(1) = 2$, the function is continuous at $x=1$.

Conclusion:

The function is continuous for all real numbers. There are no points of discontinuity.

Given:

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

To Find:

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

Solution:

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

We check for continuity at $x=2$.

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

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

Function Value: $f(2) = (2)^3-3 = 5$.

Since LHL = RHL = $f(2) = 5$, the function is continuous at $x=2$.

Conclusion:

The function is continuous for all real numbers. There are no points of discontinuity.

Given:

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

To Find:

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

Solution:

The only potential point of discontinuity is at $x=1$.

We check for continuity at $x=1$.

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

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

Since LHL $\neq$ RHL ($0 \neq 1$), the limit at $x=1$ does not exist.

Conclusion:

The function is discontinuous at $x=1$. The only point of discontinuity is $x=1$.

Given:

$f(x) = \begin{cases} x+5,& if\; x ≤ 1\\x−5,& if\; x > 1 \end{cases}$

To Determine:

Whether $f(x)$ is a continuous function.

Solution:

A function is continuous if it is continuous at every point in its domain. The only potential point of discontinuity for this function is at $x=1$.

We check for continuity at $x=1$.

LHL: $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+5) = 1+5 = 6$.

RHL: $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x-5) = 1-5 = -4$.

Since LHL $\neq$ RHL ($6 \neq -4$), the limit at $x=1$ does not exist, and the function is discontinuous at this point.

Conclusion:

Since the function is not continuous at $x=1$, it is not a continuous function.

Given:

$f(x) = \begin{cases} 3,& if\; 0 ≤ x ≤ 1\\4,& if\; 1 < x < 3\\5,& if\; 3 ≤ x ≤ 10 \end{cases}$

The domain of the function is $[0, 10]$.

To Discuss:

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

Solution:

The function is a step function. It is constant and therefore continuous on the open intervals $(0, 1)$, $(1, 3)$, and $(3, 10)$. We only need to check the points where the definition changes: $x=1$ and $x=3$.

Continuity at $x = 1$:

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

RHL: $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 4 = 4$.

Since LHL $\neq$ RHL, the function is discontinuous at $x=1$.

Continuity at $x = 3$:

LHL: $\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} 4 = 4$.

RHL: $\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} 5 = 5$.

Since LHL $\neq$ RHL, the function is discontinuous at $x=3$.

Conclusion:

The points of discontinuity of the function $f(x)$ on its domain $[0, 10]$ are at $x=1$ and $x=3$.

Given:

$f(x) = \begin{cases} 2x,& if\; x < 0\\0,& if\; 0 ≤ x ≤ 1\\4x,& if\; x > 1 \end{cases}$

To Find:

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

Solution:

The potential points of discontinuity are at $x=0$ and $x=1$.

Continuity at $x = 0$:

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

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

Function Value: $f(0) = 0$.

Since LHL = RHL = $f(0)$, the function is continuous at $x=0$.

Continuity at $x = 1$:

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

RHL: $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (4x) = 4(1) = 4$.

Since LHL $\neq$ RHL ($0 \neq 4$), the function is discontinuous at $x=1$.

Conclusion:

The only point of discontinuity is $x=1$.

Given:

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

To Find:

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

Solution:

The potential points of discontinuity are at $x=-1$ and $x=1$.

Continuity at $x = -1$:

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

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

Function Value: $f(-1) = -2$.

Since LHL = RHL = $f(-1)$, the function is continuous at $x=-1$.

Continuity at $x = 1$:

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

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

Function Value: $f(1) = 2(1) = 2$.

Since LHL = RHL = $f(1)$, the function is continuous at $x=1$.

Conclusion:

The function is continuous at all real numbers. There are no points of discontinuity.

Given:

The function $f(x) = \begin{cases} ax+1,& if\; x ≤ 3\\bx+3,& if\; x > 3 \end{cases}$ is continuous at $x = 3$.

To Find:

The relationship between $a$ and $b$.

Solution:

For the function to be continuous at $x=3$, the left-hand limit, right-hand limit, and the function value at $x=3$ must all be equal.

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

Left-Hand Limit (LHL):

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

Right-Hand Limit (RHL):

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

Since the function is continuous, LHL = RHL.

$3a+1 = 3b+3$

$3a - 3b = 3 - 1$

$3a - 3b = 2$

Dividing by 3 gives:

$a - b = \frac{2}{3}$

This can also be written as $a = b + \frac{2}{3}$.

Conclusion:

The relationship between $a$ and $b$ is $a = b + \frac{2}{3}$ or $3a - 3b = 2$.

Given:

$f(x) = \begin{cases} λ(x^2−2x),& if\; x ≤ 0\\4x+1,& if\; x > 0 \end{cases}$

Solution:

Part 1: Continuity at $x = 0$

For the function to be continuous at $x=0$, we must have $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x)$.

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

RHL: $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (4x+1) = 4(0)+1 = 1$.

For continuity, we would need LHL = RHL, which means $0 = 1$. This is a contradiction.

The left-hand and right-hand limits are different regardless of the value of λ. Therefore, the function cannot be made continuous at $x=0$.

Part 2: Continuity at $x = 1$

For values of $x$ near 1, the rule $f(x) = 4x+1$ applies since $1 > 0$.

The value of the function is $f(1) = 4(1)+1 = 5$.

The limit is $\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (4x+1) = 4(1)+1 = 5$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the function is continuous at $x=1$. This is true for any value of λ, as the definition of the function around $x=1$ does not involve λ.

Conclusion:

There is no value of λ for which the function is continuous at $x=0$.

The function is continuous at $x=1$ for all values of λ.

Given:

The function $g(x) = x - [x]$, where $[x]$ is the greatest integer function.

To Show:

The function $g(x)$ is discontinuous at all integer points.

Proof:

Let $n$ be an arbitrary integer. To show discontinuity at $x=n$, we will show that the left-hand limit and right-hand limit are not equal.

Left-Hand Limit (LHL) at $x=n$:

As $x \to n^-$, $x$ is slightly less than $n$. For such values, $[x] = n-1$.

$\lim\limits_{x \to n^-} g(x) = \lim\limits_{x \to n^-} (x - [x]) = n - (n-1) = 1$.

Right-Hand Limit (RHL) at $x=n$:

As $x \to n^+$, $x$ is slightly greater than $n$. For such values, $[x] = n$.

$\lim\limits_{x \to n^+} g(x) = \lim\limits_{x \to n^+} (x - [x]) = n - n = 0$.

Since the LHL (1) is not equal to the RHL (0), the limit $\lim\limits_{x \to n} g(x)$ does not exist.

Because the limit does not exist at $x=n$, the function is discontinuous at $x=n$.

Since $n$ was an arbitrary integer, this holds for all integers.

Conclusion:

The function $g(x) = x - [x]$ is discontinuous at all integral points.

Given:

The function $f(x) = x^2 - \sin x + 5$.

To Determine:

Whether $f(x)$ is continuous at $x = \pi$.

Solution:

The function $f(x)$ is a combination of three functions:

1. $g(x) = x^2$ (a polynomial, continuous everywhere)

2. $h(x) = \sin x$ (the sine function, continuous everywhere)

3. $k(x) = 5$ (a constant function, continuous everywhere)

The function $f(x)$ is formed by the sum and difference of these three functions: $f(x) = g(x) - h(x) + k(x)$.

Since the sum and difference of continuous functions are always continuous, $f(x)$ is continuous for all real numbers.

Because the function is continuous for all real numbers, it must also be continuous at the specific point $x = \pi$.

Alternatively, we can check the condition directly:

Value of the function at $x=\pi$:

$f(\pi) = \pi^2 - \sin(\pi) + 5 = \pi^2 - 0 + 5 = \pi^2 + 5$.

Limit of the function as $x \to \pi$:

$\lim\limits_{x \to \pi} f(x) = \lim\limits_{x \to \pi} (x^2 - \sin x + 5) = \pi^2 - \sin(\pi) + 5 = \pi^2 + 5$.

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

Conclusion:

Yes, the function $f(x) = x^2 - \sin x + 5$ is continuous at x = π.

Solution:

We know that the functions $g(x) = \sin x$ and $h(x) = \cos x$ are both continuous for all real numbers, $x \in \mathbb{R}$.

We use the algebra of continuous functions, which states that if two functions are continuous on a domain, then their sum, difference, and product are also continuous on that domain.

(a) f(x) = sin x + cos x

This function is the sum of two continuous functions, $\sin x$ and $\cos x$. Therefore, $f(x)$ is continuous for all real numbers.

(b) f(x) = sin x – cos x

This function is the difference of two continuous functions, $\sin x$ and $\cos x$. Therefore, $f(x)$ is continuous for all real numbers.

(c) f(x) = sin x . cos x

This function is the product of two continuous functions, $\sin x$ and $\cos x$. Therefore, $f(x)$ is continuous for all real numbers.

We discuss the continuity of each trigonometric function based on the properties of continuous functions and the known continuity of the sine and cosine functions.

We know that the sine function $g(x) = \sin x$ is continuous for all real numbers ($\mathbb{R}$).

We also know that the cosine function $h(x) = \cos x$ is continuous for all real numbers ($\mathbb{R}$).

A key property of continuous functions states that if two functions are continuous at a point, then their sum, difference, and product are also continuous at that point. Furthermore, their quotient is continuous at any point where both functions are continuous and the denominator is non-zero.

Continuity of the Cosine Function:

The cosine function is $f(x) = \cos x$.

Its domain is $\mathbb{R}$.

To show it is continuous on its domain, consider an arbitrary real number $c$. We need to show $\lim\limits_{x \to c} \cos x = \cos c$.

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

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

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

Using the standard limits $\lim\limits_{h \to 0} \cos h = 1$ and $\lim\limits_{h \to 0} \sin h = 0$:

$= \cos c \cdot 1 - \sin c \cdot 0 = \cos c$

Since $\lim\limits_{x \to c} \cos x = \cos c$ for any $c \in \mathbb{R}$, the cosine function is continuous at every real number.

Conclusion: The cosine function $f(x) = \cos x$ is continuous on its domain, which is all real numbers ($\mathbb{R}$).

Continuity of the Cosecant Function:

The cosecant function is $f(x) = \text{cosec } x = \frac{1}{\sin x}$.

Its domain is the set of all real numbers $x$ for which $\sin x \neq 0$. This excludes $x = n\pi$, where $n$ is any integer.

The numerator is the constant function $g(x) = 1$, which is continuous on $\mathbb{R}$.

The denominator is $h(x) = \sin x$, which is continuous on $\mathbb{R}$.

The cosecant function is the quotient of two continuous functions. It is continuous at all points where the denominator is non-zero.

The denominator $\sin x$ is non-zero exactly on the domain of $\text{cosec } x$.

Therefore, the cosecant function is continuous at every point in its domain.

Conclusion: The cosecant function $f(x) = \text{cosec } x$ is continuous on its domain, which is $\mathbb{R} \setminus \{n\pi \mid n \in \mathbb{Z}\}$.

Continuity of the Secant Function:

The secant function is $f(x) = \sec x = \frac{1}{\cos x}$.

Its domain is the set of all real numbers $x$ for which $\cos x \neq 0$. This excludes $x = \frac{\pi}{2} + n\pi$, where $n$ is any integer.

The numerator is the constant function $g(x) = 1$, which is continuous on $\mathbb{R}$.

The denominator is $h(x) = \cos x$, which is continuous on $\mathbb{R}$.

The secant function is the quotient of two continuous functions. It is continuous at all points where the denominator is non-zero.

The denominator $\cos x$ is non-zero exactly on the domain of $\sec x$.

Therefore, the secant function is continuous at every point in its domain.

Conclusion: The secant function $f(x) = \sec x$ is continuous on its domain, which is $\mathbb{R} \setminus \{\frac{\pi}{2} + n\pi \mid n \in \mathbb{Z}\}$.

Continuity of the Cotangent Function:

The cotangent function is $f(x) = \cot x = \frac{\cos x}{\sin x}$.

Its domain is the set of all real numbers $x$ for which $\sin x \neq 0$. This excludes $x = n\pi$, where $n$ is any integer.

The numerator is $g(x) = \cos x$, which is continuous on $\mathbb{R}$.

The denominator is $h(x) = \sin x$, which is continuous on $\mathbb{R}$.

The cotangent function is the quotient of two continuous functions. It is continuous at all points where the denominator is non-zero.

The denominator $\sin x$ is non-zero exactly on the domain of $\cot x$.

Therefore, the cotangent function is continuous at every point in its domain.

Conclusion: The cotangent function $f(x) = \cot x$ is continuous on its domain, which is $\mathbb{R} \setminus \{n\pi \mid n \in \mathbb{Z}\}$.

Given:

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

To Find:

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

Solution:

The only potential point of discontinuity is at $x=0$.

We check for continuity at $x=0$.

LHL: $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{\sin x}{x}$. This is a standard limit which equals 1.

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

Function Value: $f(0) = 0+1 = 1$.

Since LHL = RHL = $f(0) = 1$, the function is continuous at $x=0$.

For $x < 0$, $f(x) = \frac{\sin x}{x}$ is a quotient of continuous functions with a non-zero denominator, so it's continuous. For $x > 0$, $f(x)=x+1$ is a polynomial, so it's continuous.

Conclusion:

The function is continuous for all real numbers. There are no points of discontinuity.

Given:

$f(x) = \begin{cases} x^2\sin \frac{1}{x},& if\; x \neq 0\\0,& if\; x = 0 \end{cases}$

To Determine:

Whether $f(x)$ is a continuous function.

Solution:

For $x \neq 0$, the function is a product and composition of continuous functions ($x^2, \sin x, 1/x$), so it is continuous. The only point to check is $x=0$.

We need to check if $\lim\limits_{x \to 0} f(x) = f(0)$.

We know $f(0) = 0$.

To find the limit, we use the Squeeze Theorem. We know that the range of the sine function is $[-1, 1]$.

$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$

Multiplying by $x^2$ (which is always non-negative), the inequalities remain:

$-x^2 \leq x^2\sin\left(\frac{1}{x}\right) \leq x^2$

Now, we take the limit as $x \to 0$ for the outer functions:

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

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

By the Squeeze Theorem, the limit of the function in the middle must also be 0.

$\lim\limits_{x \to 0} x^2\sin\left(\frac{1}{x}\right) = 0$

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

Conclusion:

Yes, the function is continuous for all real numbers, so it is a continuous function.

Given:

$f(x) = \begin{cases} \sin x − \cos x,& if\; x \neq 0\\−1,& if\; x = 0 \end{cases}$

To Examine:

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

Solution:

For $x \neq 0$, $f(x)$ is a difference of two continuous functions ($\sin x$ and $\cos x$), so it is continuous. We only need to check the continuity at $x=0$.

We check if $\lim\limits_{x \to 0} f(x) = f(0)$.

Function Value: $f(0) = -1$.

Limit: $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (\sin x - \cos x) = \sin(0) - \cos(0) = 0 - 1 = -1$.

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

Conclusion:

The function is continuous for all real numbers. It is a continuous function.

Given:

The function $f(x) = \begin{cases} \frac{k\cos x}{\pi-2x},& if\; x \neq \frac{\pi}{2} \\3,& if\; x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$.

To Find:

The value of $k$.

Solution:

For the function to be continuous at $x=\frac{\pi}{2}$, we must have $\lim\limits_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$.

We are given $f(\frac{\pi}{2}) = 3$.

Now, we evaluate the limit. As $x \to \frac{\pi}{2}$, we get the indeterminate form $\frac{0}{0}$.

Let $x = \frac{\pi}{2} - h$. As $x \to \frac{\pi}{2}$, $h \to 0$.

$\lim\limits_{h \to 0} \frac{k\cos(\frac{\pi}{2}-h)}{\pi - 2(\frac{\pi}{2}-h)} = \lim\limits_{h \to 0} \frac{k\sin h}{\pi - \pi + 2h} = \lim\limits_{h \to 0} \frac{k\sin h}{2h}$

$= \frac{k}{2} \lim\limits_{h \to 0} \frac{\sin h}{h} = \frac{k}{2} \cdot 1 = \frac{k}{2}$

For continuity, we set the limit equal to the function value:

$\frac{k}{2} = 3$

$k = 6$

Conclusion:

The value of $k$ is 6.

Given:

The function $f(x) = \begin{cases} kx^2,& if\; x ≤ 2\\3,& if\; x > 2 \end{cases}$ is continuous at $x = 2$.

To Find:

The value of $k$.

Solution:

For continuity at $x=2$, we must have $\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^+} f(x)$.

LHL: $\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (kx^2) = k(2)^2 = 4k$.

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

Equating the limits:

$4k = 3$

$k = \frac{3}{4}$

Conclusion:

The value of $k$ is $\frac{3}{4}$.

Given:

The function $f(x) = \begin{cases} kx+1,& if\; x ≤ \pi\\\cos x,& if\; x > \pi \end{cases}$ is continuous at $x = \pi$.

To Find:

The value of $k$.

Solution:

For continuity at $x=\pi$, we must have $\lim\limits_{x \to \pi^-} f(x) = \lim\limits_{x \to \pi^+} f(x)$.

LHL: $\lim\limits_{x \to \pi^-} f(x) = \lim\limits_{x \to \pi^-} (kx+1) = k\pi+1$.

RHL: $\lim\limits_{x \to \pi^+} f(x) = \lim\limits_{x \to \pi^+} (\cos x) = \cos(\pi) = -1$.

Equating the limits:

$k\pi+1 = -1$

$k\pi = -2$

$k = -\frac{2}{\pi}$

Conclusion:

The value of $k$ is $-\frac{2}{\pi}$.

Given:

The function $f(x) = \begin{cases} kx+1,& if\; x ≤ 5\\3x-5,& if\; x > 5 \end{cases}$ is continuous at $x = 5$.

To Find:

The value of $k$.

Solution:

For continuity at $x=5$, we must have $\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x)$.

LHL: $\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (kx+1) = 5k+1$.

RHL: $\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (3x-5) = 3(5)-5 = 15-5 = 10$.

Equating the limits:

$5k+1 = 10$

$5k = 9$

$k = \frac{9}{5}$

Conclusion:

The value of $k$ is $\frac{9}{5}$.

Given:

The function $f(x) = \begin{cases} 5,& if\; x ≤ 2\\ax+b,& if\; 2 < x < 10\\21,& if\; x ≥ 10 \end{cases}$ is a continuous function.

To Find:

The values of $a$ and $b$.

Solution:

Since the function is continuous, it must be continuous at the points where its definition changes, i.e., at $x=2$ and $x=10$.

Continuity at $x=2$:

LHL must equal RHL.

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

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (ax+b) = 2a+b$.

Therefore, we get our first equation:

Continuity at $x=10$:

LHL must equal RHL.

$\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^-} (ax+b) = 10a+b$.

$\lim\limits_{x \to 10^+} f(x) = \lim\limits_{x \to 10^+} (21) = 21$.

Therefore, we get our second equation:

Now we solve the system of linear equations. Subtracting equation (i) from equation (ii):

$(10a+b) - (2a+b) = 21 - 5$

$8a = 16$

$a = 2$

Substituting $a=2$ into equation (i):

$2(2)+b = 5$

$4+b = 5$

$b = 1$

Conclusion:

The required values are $a=2$ and $b=1$.

Given:

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

To Show:

The function $f(x)$ is a continuous function.

Proof:

We can write $f(x)$ as a composition of two functions, $(h \circ g)(x)$, where:

$g(x) = x^2$ (the inner function)

$h(u) = \cos u$ (the outer function)

We know that:

1. The polynomial function $g(x)=x^2$ is continuous for all real numbers.

2. The cosine function $h(u)=\cos u$ is continuous for all real numbers.

A theorem on continuity states that the composition of two continuous functions is also continuous.

Since $f(x)$ is the composition of two functions that are continuous everywhere, $f(x)$ is also continuous for all real numbers.

Conclusion:

The function $f(x) = \cos(x^2)$ is a continuous function.

Given:

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

To Show:

The function $f(x)$ is a continuous function.

Proof:

We can write $f(x)$ as a composition of two functions, $(h \circ g)(x)$, where:

$g(x) = \cos x$ (the inner function)

$h(u) = |u|$ (the outer function)

We know that:

1. The cosine function $g(x)=\cos x$ is continuous for all real numbers.

2. The absolute value function $h(u)=|u|$ is continuous for all real numbers.

The composition of two continuous functions is a continuous function.

Since $f(x)$ is the composition of two continuous functions, it is also continuous for all real numbers.

Conclusion:

The function $f(x) = |\cos x|$ is a continuous function.

Given:

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

To Examine:

Whether the function $f(x)$ is a continuous function.

Solution:

We can write $f(x)$ as a composition of two functions, $(h \circ g)(x)$, where:

$g(x) = |x|$ (the inner function)

$h(u) = \sin u$ (the outer function)

We know that:

1. The absolute value function $g(x)=|x|$ is continuous for all real numbers.

2. The sine function $h(u)=\sin u$ is continuous for all real numbers.

The composition of two continuous functions is a continuous function.

Since $f(x)$ is the composition of two continuous functions, it is also continuous for all real numbers.

Conclusion:

The function $f(x) = \sin|x|$ is a continuous function.

Given:

The function $f(x) = |x| - |x+1|$.

To Find:

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

Solution:

The function $f(x)$ is the difference of two functions:

1. $g(x) = |x|$

2. $h(x) = |x+1|$

The function $g(x)=|x|$ is the absolute value function, which is known to be continuous for all real numbers.

The function $h(x)=|x+1|$ is a composition of the absolute value function and the polynomial function $x+1$. Since both are continuous for all real numbers, their composition $h(x)$ is also continuous for all real numbers.

The difference of two continuous functions is always a continuous function.

Since $f(x)$ is the difference of $g(x)$ and $h(x)$, which are both continuous everywhere, $f(x)$ must also be continuous for all real numbers.

Conclusion:

The function is continuous for all real numbers. There are no points of discontinuity.

Given:

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

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(x^2)$ is a composite function. We can find its derivative using the Chain Rule.

The Chain Rule states that if $f(x) = h(g(x))$, then $f'(x) = h'(g(x)) \cdot g'(x)$.

Let the outer function be $h(u) = \sin u$ and the inner function be $g(x) = x^2$.

Then, $f(x) = h(g(x))$.

First, find the derivative of the outer function $h(u)$ with respect to $u$:

$h'(u) = \frac{d}{du}(\sin u) = \cos u$

Next, find the derivative of the inner function $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(x^2)$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$g'(x) = 2x^{2-1} = 2x$

Now, apply the Chain Rule: $f'(x) = h'(g(x)) \cdot g'(x)$.

Substitute $g(x) = x^2$ into $h'(u) = \cos u$ to get $h'(g(x)) = \cos(x^2)$.

Substitute $g'(x) = 2x$ into the Chain Rule formula.

$f'(x) = \cos(x^2) \cdot (2x)$

Rearranging the terms, we get:

$f'(x) = 2x \cos(x^2)$

Conclusion:

The derivative of the function $f(x) = \sin(x^2)$ is $f'(x) = 2x \cos(x^2)$.

Given:

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

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \tan(2x+3)$ is a composite function. We can find its derivative using the Chain Rule.

The Chain Rule states that if $f(x) = h(g(x))$, then $f'(x) = h'(g(x)) \cdot g'(x)$.

Let the outer function be $h(u) = \tan u$ and the inner function be $g(x) = 2x+3$.

Then, $f(x) = h(g(x))$.

First, find the derivative of the outer function $h(u)$ with respect to $u$:

$h'(u) = \frac{d}{du}(\tan u) = \sec^2 u$

Next, find the derivative of the inner function $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(2x+3)$

Using the sum rule and constant multiple rule:

$g'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(3) = 2 \frac{d}{dx}(x) + 0 = 2(1) + 0 = 2$

$g'(x) = 2$

Now, apply the Chain Rule: $f'(x) = h'(g(x)) \cdot g'(x)$.

Substitute $g(x) = 2x+3$ into $h'(u) = \sec^2 u$ to get $h'(g(x)) = \sec^2(2x+3)$.

Substitute $g'(x) = 2$ into the Chain Rule formula.

$f'(x) = \sec^2(2x+3) \cdot (2)$

Rearranging the terms, we get:

$f'(x) = 2 \sec^2(2x+3)$

Conclusion:

The derivative of $\tan(2x+3)$ is $2 \sec^2(2x+3)$.

Given:

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

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(\cos(x^2))$ is a composite function involving multiple layers. We will use the Chain Rule repeatedly to find its derivative.

Let's break down the function into layers:

Outer function: $\sin(\text{expression})$

Middle function: $\cos(\text{inner expression})$

Inner function: $x^2$

Let $u = \cos(x^2)$. Then $f(x) = \sin u$.

Using the Chain Rule, $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

First, find $\frac{df}{du}$:

$\frac{df}{du} = \frac{d}{du}(\sin u) = \cos u$

Now, we need to find $\frac{du}{dx}$, where $u = \cos(x^2)$. This is another composite function.

Let $v = x^2$. Then $u = \cos v$.

Using the Chain Rule again, $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$.

Find $\frac{du}{dv}$:

$\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v$

Find $\frac{dv}{dx}$:

$\frac{dv}{dx} = \frac{d}{dx}(x^2) = 2x$

Now, combine these to find $\frac{du}{dx}$:

$\frac{du}{dx} = (-\sin v) \cdot (2x)$

Substitute back $v = x^2$:

$\frac{du}{dx} = -\sin(x^2) \cdot (2x) = -2x \sin(x^2)$

Finally, combine $\frac{df}{du}$ and $\frac{du}{dx}$ to find $\frac{df}{dx}$:

$\frac{df}{dx} = \cos u \cdot (-2x \sin(x^2))$

Substitute back $u = \cos(x^2)$:

$\frac{df}{dx} = \cos(\cos(x^2)) \cdot (-2x \sin(x^2))$

Rearranging the terms, we get the derivative:

$f'(x) = -2x \sin(x^2) \cos(\cos(x^2))$

Conclusion:

The derivative of $\sin(\cos(x^2))$ with respect to $x$ is $-2x \sin(x^2) \cos(\cos(x^2))$.

Given:

The function $f(x) = \sin(x^2+5)$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(x^2+5)$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \sin u$ and the inner function be $g(x) = x^2+5$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\sin u) = \cos u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^2+5)$

Using the sum rule and power rule:

$= \frac{d}{dx}(x^2) + \frac{d}{dx}(5) = 2x + 0 = 2x$

Step 3: Apply the Chain Rule. Substitute $u = g(x) = x^2+5$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

$f'(x) = \cos(u) \cdot (2x)$

Substitute back $u = x^2+5$:

$f'(x) = \cos(x^2+5) \cdot (2x)$

Rearranging the terms:

$f'(x) = 2x \cos(x^2+5)$

Conclusion:

The derivative of $\sin(x^2+5)$ is $2x \cos(x^2+5)$.

Given:

The function $f(x) = \cos(\sin x)$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \cos(\sin x)$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \cos u$ and the inner function be $g(x) = \sin x$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\cos u) = -\sin u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

$\frac{d}{dx}(g(x)) = \frac{d}{dx}(\sin x) = \cos x$

Step 3: Apply the Chain Rule. Substitute $u = g(x) = \sin x$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

$f'(x) = (-\sin u) \cdot (\cos x)$

Substitute back $u = \sin x$:

$f'(x) = (-\sin (\sin x)) \cdot (\cos x)$

Rearranging the terms:

$f'(x) = -\cos x \sin(\sin x)$

Conclusion:

The derivative of $\cos(\sin x)$ is $-\cos x \sin(\sin x)$.

Given:

The function $f(x) = \sin(ax+b)$, where $a$ and $b$ are constants.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(ax+b)$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \sin u$ and the inner function be $g(x) = ax+b$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\sin u) = \cos u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

$\frac{d}{dx}(g(x)) = \frac{d}{dx}(ax+b)$

Using the sum rule, constant multiple rule, and power rule:

$= \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a \frac{d}{dx}(x) + 0 = a(1) + 0 = a$

Step 3: Apply the Chain Rule. Substitute $u = g(x) = ax+b$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

$f'(x) = (\cos u) \cdot (a)$

Substitute back $u = ax+b$:

$f'(x) = \cos(ax+b) \cdot a$

Rearranging the terms:

$f'(x) = a \cos(ax+b)$

Conclusion:

The derivative of $\sin(ax+b)$ is $a \cos(ax+b)$.

Given:

The function $f(x) = \sec(\tan(\sqrt{x}))$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sec(\tan(\sqrt{x}))$ is a composite function with multiple layers. We will use the Chain Rule repeatedly.

Let's break down the function into layers:

Outer function: $\sec(\text{expression})$

Middle function: $\tan(\text{inner expression})$

Inner function: $\sqrt{x} = x^{1/2}$

Let $u = \tan(\sqrt{x})$. Then $f(x) = \sec u$.

Using the Chain Rule, $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

Step 1: Find the derivative of the outermost function with respect to $u$:

$\frac{df}{du} = \frac{d}{du}(\sec u) = \sec u \tan u$

Step 2: Find $\frac{du}{dx}$, where $u = \tan(\sqrt{x})$. This is another composite function.

Let $v = \sqrt{x} = x^{1/2}$. Then $u = \tan v$.

Using the Chain Rule again, $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$.

Find $\frac{du}{dv}$:

$\frac{du}{dv} = \frac{d}{dv}(\tan v) = \sec^2 v$

Find $\frac{dv}{dx}$:

$\frac{dv}{dx} = \frac{d}{dx}(x^{1/2})$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$= \frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$

Now, combine these to find $\frac{du}{dx}$:

$\frac{du}{dx} = (\sec^2 v) \cdot (\frac{1}{2\sqrt{x}})$

Substitute back $v = \sqrt{x}$:

$\frac{du}{dx} = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}$

Step 3: Apply the outermost Chain Rule: $f'(x) = \frac{df}{du} \cdot \frac{du}{dx}$.

Substitute $u = \tan(\sqrt{x})$ into $\frac{df}{du} = \sec u \tan u$ to get $\sec(\tan(\sqrt{x})) $$ \tan(\tan(\sqrt{x}))$.

Substitute $\frac{du}{dx} = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}$ into the formula.

$f'(x) = (\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))) \cdot \left(\frac{\sec^2(\sqrt{x})}{2\sqrt{x}}\right)$

Rearranging the terms:

$f'(x) = \frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$

Conclusion:

The derivative of $\sec(\tan(\sqrt{x}))$ is $\frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$.

Solution:

Let the given function be $y$.

$y = \frac{\sin (ax+b)}{\cos (cx+d)}$

This function is a quotient of two functions. We will use the Quotient Rule to find its derivative. The quotient rule is given by:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Here, let:

$u = \sin(ax+b)$

$v = \cos(cx+d)$

Now, we need to find the derivatives of $u$ and $v$ with respect to $x$. We will use the Chain Rule for this.

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin(ax+b))$

$\frac{du}{dx} = \cos(ax+b) \cdot \frac{d}{dx}(ax+b)$

$\frac{du}{dx} = \cos(ax+b) \cdot a = a\cos(ax+b)$

Differentiating $v$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(\cos(cx+d))$

$\frac{dv}{dx} = -\sin(cx+d) \cdot \frac{d}{dx}(cx+d)$

$\frac{dv}{dx} = -\sin(cx+d) \cdot c = -c\sin(cx+d)$

Now, substituting these derivatives into the Quotient Rule formula:

$\frac{dy}{dx} = \frac{\cos(cx+d) \cdot [a\cos(ax+b)] - \sin(ax+b) \cdot [-c\sin(cx+d)]}{[\cos(cx+d)]^2}$

Simplifying the numerator:

$\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

This is the derivative of the given function.

Alternate Representation:

The derivative can also be written by splitting the fraction:

$\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)} + \frac{c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

$\frac{dy}{dx} = a\cos(ax+b) \cdot \frac{1}{\cos(cx+d)} + c\sin(ax+b) \cdot \frac{\sin(cx+d)}{\cos(cx+d)} \cdot \frac{1}{\cos(cx+d)}$

Using the identities $\sec(x) = \frac{1}{\cos(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$:

$\frac{dy}{dx} = a\cos(ax+b)\sec(cx+d) + c\sin(ax+b)\tan(cx+d)\sec(cx+d)$

Given:

Let the function be $y = \cos(x^3) \cdot \sin^2(x^5)$.

This can be written as $y = \cos(x^3) \cdot [\sin(x^5)]^2$.

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

The function is a product of two functions, so we will use the Product Rule for differentiation, which states:

$\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$

Here, let:

$u = \cos(x^3)$

$v = [\sin(x^5)]^2$

First, we find the derivative of $u$ with respect to $x$ using the Chain Rule.

$\frac{du}{dx} = \frac{d}{dx}(\cos(x^3)) = -\sin(x^3) \cdot \frac{d}{dx}(x^3)$

$\frac{du}{dx} = -\sin(x^3) \cdot (3x^2) = -3x^2 \sin(x^3)$

Next, we find the derivative of $v$ with respect to $x$ using the Chain Rule multiple times.

$\frac{dv}{dx} = \frac{d}{dx}([\sin(x^5)]^2)$

Applying the power rule first:

$\frac{dv}{dx} = 2[\sin(x^5)]^1 \cdot \frac{d}{dx}(\sin(x^5))$

Now, differentiating $\sin(x^5)$:

$\frac{dv}{dx} = 2\sin(x^5) \cdot [\cos(x^5) \cdot \frac{d}{dx}(x^5)]$

Finally, differentiating $x^5$:

$\frac{dv}{dx} = 2\sin(x^5) \cdot \cos(x^5) \cdot (5x^4)$

$\frac{dv}{dx} = 10x^4 \sin(x^5) \cos(x^5)$

Now, we substitute the expressions for $u, v, \frac{du}{dx}, \frac{dv}{dx}$ into the Product Rule formula:

$\frac{dy}{dx} = \cos(x^3) \cdot [10x^4 \sin(x^5) \cos(x^5)] + [\sin(x^5)]^2 \cdot [-3x^2 \sin(x^3)]$

Rearranging the terms for clarity:

$\frac{dy}{dx} = 10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin^2(x^5) \sin(x^3)$

Alternate Form:

We can use the double angle identity $2\sin A \cos A = \sin(2A)$ to simplify the first term.

$10x^4 \cos(x^3) \sin(x^5) \cos(x^5) = 5x^4 \cos(x^3) \cdot [2\sin(x^5)\cos(x^5)] $$ = 5x^4 \cos(x^3) \sin(2x^5)$

So, the derivative can also be written as:

$\frac{dy}{dx} = 5x^4 \cos(x^3) \sin(2x^5) - 3x^2 \sin^2(x^5) \sin(x^3)$

Given:

Let the function be $y = 2\sqrt{\cot(x^2)}$.

This can be written as $y = 2[\cot(x^2)]^{1/2}$.

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use the Chain Rule multiple times to differentiate this function.

The Chain Rule states that if $y = f(g(h(x)))$, then $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$.

Differentiating with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} \left( 2[\cot(x^2)]^{1/2} \right)$

Using the constant multiple rule:

$\frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left( [\cot(x^2)]^{1/2} \right)$

First, apply the power rule (outermost layer):

$\frac{dy}{dx} = 2 \cdot \frac{1}{2}[\cot(x^2)]^{\frac{1}{2}-1} \cdot \frac{d}{dx}(\cot(x^2))$

$\frac{dy}{dx} = [\cot(x^2)]^{-1/2} \cdot \frac{d}{dx}(\cot(x^2))$

Next, differentiate the middle layer, $\cot(x^2)$:

$\frac{d}{dx}(\cot(x^2)) = -\text{cosec}^2(x^2) \cdot \frac{d}{dx}(x^2)$

Finally, differentiate the innermost layer, $x^2$:

$\frac{d}{dx}(x^2) = 2x$

Substitute this back:

$\frac{d}{dx}(\cot(x^2)) = -\text{cosec}^2(x^2) \cdot (2x) = -2x \text{cosec}^2(x^2)$

Now, substitute this result into the main derivative expression:

$\frac{dy}{dx} = [\cot(x^2)]^{-1/2} \cdot (-2x \text{cosec}^2(x^2))$

Rewrite the term with the negative exponent as a square root in the denominator:

$\frac{dy}{dx} = \frac{1}{\sqrt{\cot(x^2)}} \cdot (-2x \text{cosec}^2(x^2))$

$\frac{dy}{dx} = \frac{-2x \text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$

Given:

Let the function be $y = \cos(\sqrt{x})$.

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

This is a composite function, so we will use the Chain Rule for differentiation.

The Chain Rule states: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.

Here, the outer function is $f(u) = \cos u$ and the inner function is $g(x) = \sqrt{x}$.

Differentiating with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\cos(\sqrt{x}))$

First, differentiate the outer function, $\cos(\cdot)$, keeping the inner function, $\sqrt{x}$, unchanged:

$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$

Next, differentiate the inner function, $\sqrt{x} = x^{1/2}$:

$\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Now, substitute this result back into the expression:

$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$

Combining into a single fraction gives the final answer:

$\frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}$

Given:

The function $f(x) = |x - 1|$, where $x \in \mathbb{R}$.

To Prove:

The function $f(x)$ is not differentiable at $x = 1$.

Proof:

A function $f(x)$ is differentiable at a point $x=c$ if its derivative, defined by the following limit, exists:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

For this limit to exist, the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) at $x=c$ must exist and be equal.

LHD at $x=c$: $\lim\limits_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$

RHD at $x=c$: $\lim\limits_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$

We will check the differentiability at $x=1$. Here, $c=1$.

First, we find the values of $f(1)$ and $f(1+h)$:

$f(1) = |1 - 1| = |0| = 0$

$f(1+h) = |(1+h) - 1| = |h|$

Now, let's find the LHD and RHD at $x=1$.

Left-Hand Derivative (LHD) at $x=1$:

As $h \to 0^-$, $h$ is a small negative number. Therefore, $|h| = -h$.

LHD = $\lim\limits_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^-} \frac{|h| - 0}{h} = \lim\limits_{h \to 0^-} \frac{-h}{h}$

LHD = $\lim\limits_{h \to 0^-} (-1) = -1$

Right-Hand Derivative (RHD) at $x=1$:

As $h \to 0^+$, $h$ is a small positive number. Therefore, $|h| = h$.

RHD = $\lim\limits_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^+} \frac{|h| - 0}{h} = \lim\limits_{h \to 0^+} \frac{h}{h}$

RHD = $\lim\limits_{h \to 0^+} (1) = 1$

Since the Left-Hand Derivative ($-1$) is not equal to the Right-Hand Derivative ($1$), the derivative $f'(1)$ does not exist.

Conclusion:

As LHD $\neq$ RHD at $x=1$, the function $f(x) = |x - 1|$ is not differentiable at x = 1.

Given:

The greatest integer function $f(x) = [x]$, defined on the interval $0 < x < 3$.

To Prove:

The function $f(x)$ is not differentiable at $x = 1$ and $x = 2$.

Proof:

A function is not differentiable at a point if it is not continuous at that point. Let's first check for continuity.

Case 1: At $x = 1$

Left-Hand Limit (LHL):

As $x \to 1^-$, $x$ is slightly less than 1 (e.g., 0.99). So, $[x] = 0$.

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

Right-Hand Limit (RHL):

As $x \to 1^+$, $x$ is slightly greater than 1 (e.g., 1.01). So, $[x] = 1$.

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

Since LHL $\neq$ RHL, the function is not continuous at $x=1$. A function that is not continuous at a point cannot be differentiable at that point.

Case 2: At $x = 2$

Left-Hand Limit (LHL):

As $x \to 2^-$, $x$ is slightly less than 2 (e.g., 1.99). So, $[x] = 1$.

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

Right-Hand Limit (RHL):

As $x \to 2^+$, $x$ is slightly greater than 2 (e.g., 2.01). So, $[x] = 2$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} [x] = 2$.

Since LHL $\neq$ RHL, the function is not continuous at $x=2$. Therefore, it is also not differentiable at $x=2$.

Alternate Solution (Using the definition of derivative):

We check if the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) are equal.

At $x = 1$:

LHD = $\lim\limits_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^-} \frac{[1+h] - [1]}{h} = \lim\limits_{h \to 0^-} \frac{0 - 1}{h} = \lim\limits_{h \to 0^-} \frac{-1}{h}$, which approaches $+\infty$.

RHD = $\lim\limits_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^+} \frac{[1+h] - [1]}{h} = \lim\limits_{h \to 0^+} \frac{1 - 1}{h} = \lim\limits_{h \to 0^+} \frac{0}{h} = 0$.

Since LHD $\neq$ RHD, the function is not differentiable at $x=1$.

At $x = 2$:

LHD = $\lim\limits_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = \lim\limits_{h \to 0^-} \frac{[2+h] - [2]}{h} = \lim\limits_{h \to 0^-} \frac{1 - 2}{h} = \lim\limits_{h \to 0^-} \frac{-1}{h}$, which approaches $+\infty$.

RHD = $\lim\limits_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = \lim\limits_{h \to 0^+} \frac{[2+h] - [2]}{h} = \lim\limits_{h \to 0^+} \frac{2 - 2}{h} = \lim\limits_{h \to 0^+} \frac{0}{h} = 0$.

Since LHD $\neq$ RHD, the function is not differentiable at $x=2$.

Conclusion:

The function $f(x)=[x]$ is discontinuous at all integers. Since differentiability implies continuity, the function is not differentiable at the integer points within its domain. Therefore, $f(x)$ is not differentiable at x = 1 and x = 2.

Given:

The equation $x - y = \pi$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We can find $\frac{dy}{dx}$ using implicit differentiation. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (and using the Chain Rule where necessary).

Differentiate both sides of the equation $x - y = \pi$ with respect to $x$:

$\frac{d}{dx}(x - y) = \frac{d}{dx}(\pi)$

Using the difference rule on the left side and the constant rule on the right side:

$\frac{d}{dx}(x) - \frac{d}{dx}(y) = 0$

The derivative of $x$ with respect to $x$ is 1:

$1 - \frac{d}{dx}(y) = 0$

The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$:

$1 - \frac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

$1 = \frac{dy}{dx}$

$\frac{dy}{dx} = 1$

Alternatively, we could solve the given equation explicitly for $y$ first:

$x - y = \pi$

$-y = \pi - x$

$y = x - \pi$

Now, differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x - \pi)$

Using the difference rule and constant rule:

$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\pi) = 1 - 0 = 1$

Both methods yield the same result.

Conclusion:

If $x - y = \pi$, then $\frac{dy}{dx} = 1$.

Given:

The equation $y + \sin y = \cos x$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (and using the Chain Rule where necessary).

Differentiate both sides of the equation $y + \sin y = \cos x$ with respect to $x$:

$\frac{d}{dx}(y + \sin y) = \frac{d}{dx}(\cos x)$

Using the sum rule on the left side:

$\frac{d}{dx}(y) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(\cos x)$

The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

For the term $\frac{d}{dx}(\sin y)$, we use the Chain Rule. Treat $\sin y$ as a function of $y$, where $y$ is a function of $x$. The derivative of $\sin u$ with respect to $u$ is $\cos u$. So, $\frac{d}{dx}(\sin y) = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$.

The derivative of $\cos x$ with respect to $x$ is $-\sin x$.

Substituting these derivatives back into the equation:

$\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = -\sin x$

Now, we need to solve for $\frac{dy}{dx}$. Factor out $\frac{dy}{dx}$ from the terms on the left side:

$\frac{dy}{dx} (1 + \cos y) = -\sin x$

Divide both sides by $(1 + \cos y)$, assuming $1 + \cos y \neq 0$ (which means $\cos y \neq -1$, or $y \neq \pi + 2n\pi$ for integer $n$):

$\frac{dy}{dx} = \frac{-\sin x}{1 + \cos y}$

Conclusion:

If $y + \sin y = \cos x$, then $\frac{dy}{dx} = \frac{-\sin x}{1 + \cos y}$, provided $1 + \cos y \neq 0$.

Given:

The function $f(x) = \sin^{-1} x$ (also written as $\arcsin x$).

The domain of $f(x)$ is $[-1, 1]$ and the standard range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We assume the derivative exists, which is for $x \in (-1, 1)$.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx}(\sin^{-1} x)$.

Solution:

Let $y = \sin^{-1} x$.

By the definition of the inverse sine function, this means:

$x = \sin y$

where $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $x \in [-1, 1]$. We are assuming the derivative exists, so we consider $x \in (-1, 1)$, which implies $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

We differentiate both sides of the equation $x = \sin y$ with respect to $x$. We use implicit differentiation on the right side, treating $y$ as a function of $x$ and applying the Chain Rule.

$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$

$1 = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx}$

We know that $\frac{d}{dy}(\sin y) = \cos y$.

$1 = \cos y \cdot \frac{dy}{dx}$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\cos y}$

We need to express $\cos y$ in terms of $x$. We use the identity $\sin^2 y + \cos^2 y = 1$.

$\cos^2 y = 1 - \sin^2 y$

Since $x = \sin y$, we have $\sin^2 y = x^2$.

$\cos^2 y = 1 - x^2$

Taking the square root of both sides:

$\cos y = \pm \sqrt{1 - x^2}$

Since the range of $y = \sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and for $y$ in the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos y$ is positive, we take the positive square root.

$\cos y = \sqrt{1 - x^2}$ for $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$ (which corresponds to $x \in (-1, 1)$)

Substituting this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

This derivative exists for all $x$ in the interval $(-1, 1)$. At the endpoints $x = \pm 1$, the derivative is undefined, corresponding to a vertical tangent line on the graph of $y = \sin^{-1} x$.

Conclusion:

The derivative of the function $f(x) = \sin^{-1} x$, where it exists (for $x \in (-1, 1)$), is $f'(x) = \frac{1}{\sqrt{1 - x^2}}$.

Given:

The function $f(x) = \tan^{-1} x$ (also written as $\arctan x$).

The domain of $f(x)$ is the set of all real numbers ($\mathbb{R}$), and the standard range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We assume the derivative exists.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx}(\tan^{-1} x)$.

Solution:

Let $y = \tan^{-1} x$.

By the definition of the inverse tangent function, this means:

$x = \tan y$

where $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Since $x$ can be any real number, the derivative exists for all $x \in \mathbb{R}$.

We differentiate both sides of the equation $x = \tan y$ with respect to $x$. We use implicit differentiation on the right side, treating $y$ as a function of $x$ and applying the Chain Rule.

$\frac{d}{dx}(x) = \frac{d}{dx}(\tan y)$

$1 = \frac{d}{dy}(\tan y) \cdot \frac{dy}{dx}$

We know that $\frac{d}{dy}(\tan y) = \sec^2 y$.

$1 = \sec^2 y \cdot \frac{dy}{dx}$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\sec^2 y}$

We need to express $\sec^2 y$ in terms of $x$. We use the trigonometric identity relating tangent and secant: $\sec^2 y = 1 + \tan^2 y$.

Since $x = \tan y$, we have $\tan^2 y = x^2$.

$\sec^2 y = 1 + x^2$

Substituting this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{1 + x^2}$

This derivative exists for all real numbers $x$.

Conclusion:

The derivative of the function $f(x) = \tan^{-1} x$ is $f'(x) = \frac{1}{1 + x^2}$.

Given:

The equation $2x + 3y = \sin x$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (and using the Chain Rule where necessary).

Differentiate both sides of the equation $2x + 3y = \sin x$ with respect to $x$:

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin x)$

Using the sum rule on the left side:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$

Using the constant multiple rule:

$2 \frac{d}{dx}(x) + 3 \frac{d}{dx}(y) = \frac{d}{dx}(\sin x)$

We know that $\frac{d}{dx}(x) = 1$. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$. The derivative of $\sin x$ with respect to $x$ is $\cos x$.

$2(1) + 3 \frac{dy}{dx} = \cos x$

$2 + 3 \frac{dy}{dx} = \cos x$

Now, we need to solve for $\frac{dy}{dx}$. Subtract 2 from both sides:

$3 \frac{dy}{dx} = \cos x - 2$

Divide both sides by 3:

$\frac{dy}{dx} = \frac{\cos x - 2}{3}$

Conclusion:

If $2x + 3y = \sin x$, then $\frac{dy}{dx} = \frac{\cos x - 2}{3}$.

Given:

The equation $2x + 3y = \sin y$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (and using the Chain Rule where necessary).

Differentiate both sides of the equation $2x + 3y = \sin y$ with respect to $x$:

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin y)$

Using the sum rule on the left side and the Chain Rule on the right side:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx}$

Using the constant multiple rule on the left side:

$2 \frac{d}{dx}(x) + 3 \frac{d}{dx}(y) = \cos y \cdot \frac{dy}{dx}$

We know that $\frac{d}{dx}(x) = 1$. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

$2(1) + 3 \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

$2 + 3 \frac{dy}{dx} = \cos y \frac{dy}{dx}$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side.

$2 = \cos y \frac{dy}{dx} - 3 \frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the terms on the right side:

$2 = \frac{dy}{dx} (\cos y - 3)$

Divide both sides by $(\cos y - 3)$, assuming $\cos y - 3 \neq 0$ (which is always true since $-1 \leq \cos y \leq 1$, so $\cos y - 3$ is between $-4$ and $-2$).

$\frac{dy}{dx} = \frac{2}{\cos y - 3}$

Conclusion:

If $2x + 3y = \sin y$, then $\frac{dy}{dx} = \frac{2}{\cos y - 3}$.

Given:

The equation $ax + by^2 = \cos y$, where $a$ and $b$ are constants.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $ax + by^2 = \cos y$ with respect to $x$:

$\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)$

Using the sum rule on the left side:

$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$

Using the constant multiple rule:

$a \frac{d}{dx}(x) + b \frac{d}{dx}(y^2) = \frac{d}{dx}(\cos y)$

We know that $\frac{d}{dx}(x) = 1$.

For the term $\frac{d}{dx}(y^2)$, we use the Chain Rule. Treat $y^2$ as a function of $y$, where $y$ is a function of $x$. The derivative of $u^2$ with respect to $u$ is $2u$. So, $\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}$. Thus, $b \frac{d}{dx}(y^2) = b(2y \frac{dy}{dx}) = 2by \frac{dy}{dx}$.

For the term $\frac{d}{dx}(\cos y)$, we use the Chain Rule. Treat $\cos y$ as a function of $y$, where $y$ is a function of $x$. The derivative of $\cos u$ with respect to $u$ is $-\sin u$. So, $\frac{d}{dx}(\cos y) = \frac{d}{dy}(\cos y) \cdot \frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$a(1) + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

$a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. Move the $2by \frac{dy}{dx}$ term to the right side:

$a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the terms on the right side:

$a = \frac{dy}{dx} (-\sin y - 2by)$

Divide both sides by $(-\sin y - 2by)$, assuming this expression is not zero:

$\frac{dy}{dx} = \frac{a}{-\sin y - 2by}$

We can factor out $-1$ from the denominator:

$\frac{dy}{dx} = \frac{a}{-(2by + \sin y)}$

$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Conclusion:

If $ax + by^2 = \cos y$, then $\frac{dy}{dx} = -\frac{a}{2by + \sin y}$, provided $2by + \sin y \neq 0$.

Given:

The equation $xy + y^2 = \tan x + y$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $xy + y^2 = \tan x + y$ with respect to $x$:

$\frac{d}{dx}(xy + y^2) = \frac{d}{dx}(\tan x + y)$

Using the sum rule on both sides:

$\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y)$

For the term $\frac{d}{dx}(xy)$, we use the Product Rule, treating $x$ as a function of $x$ and $y$ as a function of $x$. $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x$ and $v=y$. $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{dy}{dx}$.

$\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}$

For the term $\frac{d}{dx}(y^2)$, we use the Chain Rule. $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.

For the term $\frac{d}{dx}(\tan x)$, the derivative is $\sec^2 x$.

For the term $\frac{d}{dx}(y)$, the derivative is $\frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$(y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$

$y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. Let's move $\frac{dy}{dx}$ terms to the left side and the $y$ term to 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:

$\frac{dy}{dx} (x + 2y - 1) = \sec^2 x - y$

Divide both sides by $(x + 2y - 1)$, assuming this expression is not zero:

$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$

Conclusion:

If $xy + y^2 = \tan x + y$, then $\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$, provided $x + 2y - 1 \neq 0$.

Given:

The equation $x^2 + xy + y^2 = 100$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $x^2 + xy + y^2 = 100$ with respect to $x$:

$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)$

Using the sum rule on the left side and the constant rule on the right side:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$

We know that $\frac{d}{dx}(x^2) = 2x$.

For the term $\frac{d}{dx}(xy)$, we use the Product Rule, treating $x$ as a function of $x$ and $y$ as a function of $x$. $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x$ and $v=y$. $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{dy}{dx}$.

$\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}$

For the term $\frac{d}{dx}(y^2)$, we use the Chain Rule. Treat $y^2$ as a function of $y$, where $y$ is a function of $x$. The derivative of $u^2$ with respect to $u$ is $2u$. So, $\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$2x + (y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0$

$2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. Move the terms without $\frac{dy}{dx}$ to the right side:

$x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - y$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$\frac{dy}{dx} (x + 2y) = -2x - y$

Divide both sides by $(x + 2y)$, assuming this expression is not zero:

$\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$

We can factor out $-1$ from the numerator:

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

Conclusion:

If $x^2 + xy + y^2 = 100$, then $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$, provided $x + 2y \neq 0$.

Given:

The equation $x^3 + x^2y + xy^2 + y^3 = 81$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We use implicit differentiation. Differentiate both sides of the equation with respect to $x$.

$\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81)$

Differentiate each term, using the product rule for $x^2y$ and $xy^2$:

$3x^2 + \left(2xy + x^2\frac{dy}{dx}\right) + \left(y^2 + x(2y\frac{dy}{dx})\right) + 3y^2\frac{dy}{dx} = 0$

$3x^2 + 2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$

Group the terms with $\frac{dy}{dx}$:

$x^2\frac{dy}{dx} + 2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -3x^2 - 2xy - y^2$

Factor out $\frac{dy}{dx}$:

$(x^2 + 2xy + 3y^2)\frac{dy}{dx} = -(3x^2 + 2xy + y^2)$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

Given:

The equation $\sin^2 y + \cos(xy) = \kappa$, where $\kappa$ is a constant.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We use implicit differentiation. Differentiate both sides of the equation with respect to $x$.

$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)$

Differentiate each term using the chain rule:

$2\sin y \cdot \cos y \cdot \frac{dy}{dx} - \sin(xy) \cdot \frac{d}{dx}(xy) = 0$

Using the double angle identity and the product rule for $xy$:

$\sin(2y)\frac{dy}{dx} - \sin(xy)\left(1 \cdot y + x \cdot \frac{dy}{dx}\right) = 0$

$\sin(2y)\frac{dy}{dx} - y\sin(xy) - x\sin(xy)\frac{dy}{dx} = 0$

Group the terms with $\frac{dy}{dx}$:

$\sin(2y)\frac{dy}{dx} - x\sin(xy)\frac{dy}{dx} = y\sin(xy)$

Factor out $\frac{dy}{dx}$:

$(\sin(2y) - x\sin(xy))\frac{dy}{dx} = y\sin(xy)$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{y\sin(xy)}{\sin(2y) - x\sin(xy)}$

Given:

The equation $\sin^2 x + \cos^2 y = 1$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We use implicit differentiation. Differentiate both sides of the equation with respect to $x$.

$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1)$

Differentiate each term using the chain rule:

$2\sin x \cdot \cos x + 2\cos y \cdot (-\sin y) \cdot \frac{dy}{dx} = 0$

Using double angle identities:

$\sin(2x) - \sin(2y)\frac{dy}{dx} = 0$

Solve for $\frac{dy}{dx}$:

$\sin(2x) = \sin(2y)\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}$

Given:

The function $y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

The expression $\frac{2x}{1+x^2}$ is reminiscent of the double angle identity for sine, $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.

Let's use the substitution $x = \tan\theta$. This implies $\theta = \tan^{-1}x$.

Substituting into the function:

$y = \sin^{-1} \left( \frac{2\tan\theta}{1+\tan^2\theta} \right) = \sin^{-1}(\sin(2\theta))$

For the principal value range, if $-1 < x < 1$, then $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, which means $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. In this range, $\sin^{-1}(\sin(2\theta)) = 2\theta$.

So, the function simplifies to $y = 2\theta$.

Substituting back $\theta = \tan^{-1}x$:

$y = 2\tan^{-1}x$

Now, we differentiate this simplified function with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}x) = 2 \cdot \frac{d}{dx}(\tan^{-1}x)$

$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$

This result is valid for $-1 < x < 1$.

Given:

The function $y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ for $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

The expression inside the function is the triple angle identity for tangent: $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Let's use the substitution $x = \tan\theta$. This implies $\theta = \tan^{-1}x$.

The given domain for $x$ is $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

This means $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$. Multiplying by 3, we get $-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$.

Substituting into the function:

$y = \tan^{-1}(\tan(3\theta))$

Since $3\theta$ is within the principal value range of $\tan^{-1}$, we can simplify this to:

$y = 3\theta$

Substituting back $\theta = \tan^{-1}x$:

$y = 3\tan^{-1}x$

Now, we differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(3\tan^{-1}x) = 3 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = \frac{3}{1+x^2}$

Given:

The function $y = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$ for $0 < x < 1$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

The expression inside the function is the double angle identity for cosine in terms of tangent: $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.

Let's use the substitution $x = \tan\theta$. This implies $\theta = \tan^{-1}x$.

The given domain for $x$ is $0 < x < 1$.

This means $0 < \theta < \frac{\pi}{4}$. Multiplying by 2, we get $0 < 2\theta < \frac{\pi}{2}$.

Substituting into the function:

$y = \cos^{-1}(\cos(2\theta))$

Since $2\theta$ is within the principal value range of $\cos^{-1}$ ($[0, \pi]$), we can simplify this to:

$y = 2\theta$

Substituting back $\theta = \tan^{-1}x$:

$y = 2\tan^{-1}x$

Now, we differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}x) = 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = \frac{2}{1+x^2}$

Given:

The function $y = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$ for $0 < x < 1$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Let's use the substitution $x = \tan\theta$. This implies $\theta = \tan^{-1}x$.

The inner expression becomes $\frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta)$.

$y = \sin^{-1}(\cos(2\theta))$

Using the identity $\cos(A) = \sin(\frac{\pi}{2} - A)$, we get:

$y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right)$

The domain for $x$ is $0 < x < 1$, so $0 < \theta < \frac{\pi}{4}$. This means $0 < 2\theta < \frac{\pi}{2}$, and further $0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2}$.

Since $\frac{\pi}{2} - 2\theta$ is in the principal value range of $\sin^{-1}$, we can simplify:

$y = \frac{\pi}{2} - 2\theta$

Substituting back $\theta = \tan^{-1}x$:

$y = \frac{\pi}{2} - 2\tan^{-1}x$

Now, we differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\tan^{-1}x\right) = 0 - 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = -\frac{2}{1+x^2}$

Given:

The function $y = \cos^{-1} \left( \frac{2x}{1 + x^2} \right)$ for $-1 < x < 1$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

Let's use the substitution $x = \tan\theta$. This implies $\theta = \tan^{-1}x$.

The inner expression becomes $\frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta)$.

$y = \cos^{-1}(\sin(2\theta))$

Using the identity $\sin(A) = \cos(\frac{\pi}{2} - A)$, we get:

$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right)$

The domain for $x$ is $-1 < x < 1$, so $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. This means $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$, and further $0 < \frac{\pi}{2} - 2\theta < \pi$.

Since $\frac{\pi}{2} - 2\theta$ is in the principal value range of $\cos^{-1}$, we can simplify:

$y = \frac{\pi}{2} - 2\theta$

Substituting back $\theta = \tan^{-1}x$:

$y = \frac{\pi}{2} - 2\tan^{-1}x$

Now, we differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\tan^{-1}x\right) = 0 - 2 \cdot \frac{1}{1+x^2}$

$\frac{dy}{dx} = -\frac{2}{1+x^2}$

Given:

The function $y = \sin^{-1} \left( 2x \sqrt{1 - x^2} \right)$ for $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

The form of the inner expression suggests the substitution $x = \sin\theta$. This implies $\theta = \sin^{-1}x$.

Then $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = |\cos\theta|$.

The domain for $x$ is $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, which means $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. In this interval, $\cos\theta$ is positive, so $|\cos\theta| = \cos\theta$.

Substituting into the function:

$y = \sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin(2\theta))$

From the range of $\theta$, we get $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. This is within the principal value range of $\sin^{-1}$.

So, we can simplify to $y = 2\theta$.

Substituting back $\theta = \sin^{-1}x$:

$y = 2\sin^{-1}x$

Now, we differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1}x) = 2 \cdot \frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$

Given:

The function $y = \sec^{-1}\left( \frac{1}{2x^2 - 1} \right)$ for $0 < x < \frac{1}{\sqrt{2}}$.

To Find:

The derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

The form of the inner expression suggests the substitution $x = \cos\theta$. This implies $\theta = \cos^{-1}x$.

The inner expression becomes $\frac{1}{2\cos^2\theta - 1} = \frac{1}{\cos(2\theta)} = \sec(2\theta)$.

$y = \sec^{-1}(\sec(2\theta))$

The domain for $x$ is $0 < x < \frac{1}{\sqrt{2}}$. Since $\cos\theta$ is decreasing on $[0, \pi]$, this means $\cos^{-1}(\frac{1}{\sqrt{2}}) < \theta < \cos^{-1}(0)$, which is $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.

Multiplying by 2, we get $\frac{\pi}{2} < 2\theta < \pi$.

This range for $2\theta$ is within the principal value range of $\sec^{-1}$, which is $[0, \pi] - \{\frac{\pi}{2}\}$.

So, we can simplify to $y = 2\theta$.

Substituting back $\theta = \cos^{-1}x$:

$y = 2\cos^{-1}x$

Now, we differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2\cos^{-1}x) = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$

$\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}$

To Determine:

Whether the equation $x = e^{\log x}$ is true for all real values of $x$.

Solution:

The notation $\log x$ without a specified base typically refers to the natural logarithm, $\log_e x$ or $\log x$.

The function $f(x) = \log x$ is the inverse of the exponential function $g(x) = e^x$. The domain of the logarithmic function is the set of all positive real numbers. This means that $\log x$ is defined only when $x > 0$.

Let's analyze the given equation: $x = e^{\log x}$.

The left side of the equation, $x$, can be any real number.

However, the right side of the equation, $e^{\log x}$, involves the term $\log x$. Since $\log x$ is only defined for $x > 0$, the entire expression $e^{\log x}$ is also only defined for $x > 0$.

For any value $x \leq 0$, the right side of the equation is undefined in the real number system. For example, if we try to check for $x=-2$, the expression $e^{\log(-2)}$ is not defined because $\log(-2)$ is not a real number.

The property that $e^{\log x} = x$ is a fundamental property of logarithms and exponentials, but it is only valid for values of $x$ for which both sides of the equation are defined. This condition is met only when $x > 0$.

Conclusion:

No, it is not true that $x = e^{\log x}$ for all real $x$. The statement is only true for all positive real numbers ($x > 0$).

(i) Differentiate $e^{-x}$ w.r.t. x

Let $y = e^{-x}$. We use the chain rule.

Let $u = -x$. Then $y = e^u$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = e^u$ and $\frac{du}{dx} = -1$.

$\frac{dy}{dx} = e^u \cdot (-1) = e^{-x} \cdot (-1) = -e^{-x}$

Answer: The derivative of $e^{-x}$ is $-e^{-x}$.

(ii) Differentiate $\sin(\log x)$, x > 0 w.r.t. x

Let $y = \sin(\log x)$. We use the chain rule.

Let $u = \log x$. Then $y = \sin u$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = \cos u$ and $\frac{du}{dx} = \frac{1}{x}$.

$\frac{dy}{dx} = \cos u \cdot \frac{1}{x} = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x}$

Answer: The derivative of $\sin(\log x)$ is $\frac{\cos(\log x)}{x}$.

(iii) Differentiate $\cos^{-1}(e^x)$ w.r.t. x

Let $y = \cos^{-1}(e^x)$. We use the chain rule.

Let $u = e^x$. Then $y = \cos^{-1} u$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

We know that $\frac{d}{du}(\cos^{-1} u) = -\frac{1}{\sqrt{1-u^2}}$ and $\frac{d}{dx}(e^x) = e^x$.

$\frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot e^x$

Substituting back $u = e^x$:

$\frac{dy}{dx} = -\frac{1}{\sqrt{1-(e^x)^2}} \cdot e^x = -\frac{e^x}{\sqrt{1-e^{2x}}}$

Answer: The derivative of $\cos^{-1}(e^x)$ is $-\frac{e^x}{\sqrt{1-e^{2x}}}$.

(iv) Differentiate $e^{\cos x}$ w.r.t. x

Let $y = e^{\cos x}$. We use the chain rule.

Let $u = \cos x$. Then $y = e^u$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = e^u$ and $\frac{du}{dx} = -\sin x$.

$\frac{dy}{dx} = e^u \cdot (-\sin x)$

Substituting back $u = \cos x$:

$\frac{dy}{dx} = e^{\cos x} \cdot (-\sin x) = -\sin x \ e^{\cos x}$

Answer: The derivative of $e^{\cos x}$ is $-\sin x \ e^{\cos x}$.

Given:

The function to differentiate is $y = \frac{e^x}{\sin x}$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \frac{e^x}{\sin x}$ with respect to $x$. This function is in the form of a quotient $\frac{u}{v}$, where $u = e^x$ and $v = \sin x$.

We will use the quotient rule for differentiation, which states that if $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{v(x) \frac{du}{dx} - u(x) \frac{dv}{dx}}{(v(x))^2}$.

Let $u = e^x$. The derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x$

Let $v = \sin x$. The derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the quotient rule:

$\frac{dy}{dx} = \frac{(\sin x) \cdot \frac{d}{dx}(e^x) - (e^x) \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2}$

Substitute the derivatives we found:

$\frac{dy}{dx} = \frac{(\sin x) \cdot (e^x) - (e^x) \cdot (\cos x)}{\sin^2 x}$

Factor out $e^x$ from the numerator:

$\frac{dy}{dx} = \frac{e^x (\sin x - \cos x)}{\sin^2 x}$

Final Answer:

The derivative of $\frac{e^x}{\sin x}$ with respect to $x$ is $\frac{e^x (\sin x - \cos x)}{\sin^2 x}$.

Given:

The function to differentiate is $y = e^{\sin^{-1} x}$.

Note: The domain of $\sin^{-1} x$ is $[-1, 1]$. The function $e^u$ is defined for all real $u$. Thus, the domain of the given function is $[-1, 1]$. For the derivative to exist, we consider $x \in (-1, 1)$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = e^{\sin^{-1} x}$ with respect to $x$. We will use the chain rule.

Let $u = \sin^{-1} x$. Then the function becomes $y = e^u$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, differentiate $y = e^u$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$

Next, differentiate $u = \sin^{-1} x$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$

Now, multiply these results to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = e^u \cdot \frac{1}{\sqrt{1-x^2}}$

Substitute back $u = \sin^{-1} x$:

$\frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$

So, the derivative is:

$\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}$

Final Answer:

The derivative of $e^{\sin^{-1} x}$ with respect to $x$ is $\frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}$.

Given:

The function to differentiate is $y = e^{x^3}$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = e^{x^3}$ with respect to $x$. We will use the chain rule.

The function is in the form $e^u$, where $u$ is a function of $x$. Let $u = x^3$.

Then the function is $y = e^u$.

According to the chain rule, the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

First, we find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(e^u)$

The derivative of $e^u$ with respect to $u$ is $e^u$ itself.

$\frac{dy}{du} = e^u$

Next, we find the derivative of $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^3)$

Using the power rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{du}{dx} = 3x^{3-1} = 3x^2$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to get $\frac{dy}{dx}$:

$\frac{dy}{dx} = e^u \cdot 3x^2$

Substitute back $u = x^3$ into the expression:

$\frac{dy}{dx} = e^{x^3} \cdot 3x^2$

We can write this as:

$\frac{dy}{dx} = 3x^2 e^{x^3}$

Final Answer:

The derivative of $e^{x^3}$ with respect to $x$ is $3x^2 e^{x^3}$.

Given:

The function to differentiate is $y = \sin(\tan^{-1} e^{-x})$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \sin(\tan^{-1} e^{-x})$ with respect to $x$. We will use the chain rule.

Let's apply the chain rule step by step.

The outermost function is $\sin(u)$, where $u = \tan^{-1} e^{-x}$.

The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$.

So, $\frac{dy}{du} = \cos(\tan^{-1} e^{-x})$.

Now we need to differentiate $u = \tan^{-1} e^{-x}$ with respect to $x$. This is of the form $\tan^{-1}(v)$, where $v = e^{-x}$.

The derivative of $\tan^{-1}(v)$ with respect to $v$ is $\frac{1}{1+v^2}$.

So, $\frac{du}{dv} = \frac{1}{1+(e^{-x})^2} = \frac{1}{1+e^{-2x}}$.

Next, we need to differentiate $v = e^{-x}$ with respect to $x$. This is of the form $e^w$, where $w = -x$.

The derivative of $e^w$ with respect to $w$ is $e^w$.

So, $\frac{dv}{dw} = e^{-x}$.

Finally, we differentiate $w = -x$ with respect to $x$.

$\frac{dw}{dx} = \frac{d}{dx}(-x) = -1$.

Now, combining these using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$

$\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1+e^{-2x}} \cdot e^{-x} \cdot (-1)$

Multiplying these terms gives:

$\frac{dy}{dx} = -\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}$

Final Answer:

The derivative of $\sin(\tan^{-1} e^{-x})$ with respect to $x$ is $-\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}$.

Given:

The function to differentiate is $y = \log(\cos e^x)$.

Note: For the logarithm to be defined, its argument must be positive. Thus, $\cos e^x > 0$. Also, $e^x$ is defined for all real $x$. The domain of $\cos(u)$ is all real numbers. So, the domain of the function is the set of $x$ such that $\cos(e^x) > 0$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \log(\cos e^x)$ with respect to $x$. We will use the chain rule.

The function is a composition of functions. Let's break it down:

Let $u = \cos e^x$. Then $y = \log u$.

Let $v = e^x$. Then $u = \cos v$.

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.

First, differentiate $y = \log u$ with respect to $u$. Assuming $\log u$ is the natural logarithm (base $e$):

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}$

Substitute $u = \cos e^x$ back:

$\frac{dy}{du} = \frac{1}{\cos e^x}$

Next, differentiate $u = \cos v$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v$

Substitute $v = e^x$ back:

$\frac{du}{dv} = -\sin e^x$

Finally, differentiate $v = e^x$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(e^x) = e^x$

Now, multiply the results together:

$\frac{dy}{dx} = \frac{1}{\cos e^x} \cdot (-\sin e^x) \cdot e^x$

Simplify the expression:

$\frac{dy}{dx} = -\frac{\sin e^x}{\cos e^x} \cdot e^x$

Recall that $\frac{\sin \theta}{\cos \theta} = \tan \theta$. So, $\frac{\sin e^x}{\cos e^x} = \tan e^x$.

$\frac{dy}{dx} = -\tan(e^x) \cdot e^x$

Rearranging the terms:

$\frac{dy}{dx} = -e^x \tan(e^x)$

Final Answer:

The derivative of $\log(\cos e^x)$ with respect to $x$ is $-e^x \tan(e^x)$.

Given:

The function to differentiate is $y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of the function $y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$ with respect to $x$.

We can differentiate each term of the sum separately:

$\frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(e^{x^3}) + \frac{d}{dx}(e^{x^4}) + \frac{d}{dx}(e^{x^5})$

We use the chain rule for differentiation, which states that $\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)$.

For the first term, $\frac{d}{dx}(e^x)$: Here $f(x) = x$, so $f'(x) = \frac{d}{dx}(x) = 1$.

$\frac{d}{dx}(e^x) = e^x \cdot 1 = e^x$

For the second term, $\frac{d}{dx}(e^{x^2})$: Here $f(x) = x^2$, so $f'(x) = \frac{d}{dx}(x^2) = 2x$.

$\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot 2x = 2xe^{x^2}$

For the third term, $\frac{d}{dx}(e^{x^3})$: Here $f(x) = x^3$, so $f'(x) = \frac{d}{dx}(x^3) = 3x^2$.

$\frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot 3x^2 = 3x^2e^{x^3}$

For the fourth term, $\frac{d}{dx}(e^{x^4})$: Here $f(x) = x^4$, so $f'(x) = \frac{d}{dx}(x^4) = 4x^3$.

$\frac{d}{dx}(e^{x^4}) = e^{x^4} \cdot 4x^3 = 4x^3e^{x^4}$

For the fifth term, $\frac{d}{dx}(e^{x^5})$: Here $f(x) = x^5$, so $f'(x) = \frac{d}{dx}(x^5) = 5x^4$.

$\frac{d}{dx}(e^{x^5}) = e^{x^5} \cdot 5x^4 = 5x^4e^{x^5}$

Now, sum the derivatives of all terms:

$\frac{dy}{dx} = e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}$

Final Answer:

The derivative of $e^x + e^{x^2} + … + e^{x^5}$ with respect to $x$ is $e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}$.

Given:

Let the function be $y = \sqrt{e^{\sqrt{x}}}$, where $x > 0$.

This can be written using exponents as $y = \left(e^{x^{1/2}}\right)^{1/2}$.

Using the law of exponents $(a^m)^n = a^{mn}$, we can simplify the function:

$y = e^{x^{1/2} \cdot \frac{1}{2}} = e^{\frac{1}{2}x^{1/2}}$

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to differentiate $y = e^{\frac{1}{2}\sqrt{x}}$ with respect to $x$. We will use the Chain Rule.

Let $u = \frac{1}{2}\sqrt{x}$. Then the function is $y = e^u$.

The Chain Rule states: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Step 1: Differentiate $y$ with respect to $u$.

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$

Step 2: Differentiate $u$ with respect to $x$.

$u = \frac{1}{2}x^{1/2}$

$\frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{2}x^{1/2}\right) = \frac{1}{2} \cdot \frac{d}{dx}(x^{1/2})$

Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{du}{dx} = \frac{1}{2} \cdot \left(\frac{1}{2}x^{\frac{1}{2}-1}\right) = \frac{1}{4}x^{-1/2} = \frac{1}{4\sqrt{x}}$

Step 3: Apply the Chain Rule.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^u \cdot \frac{1}{4\sqrt{x}}$

Step 4: Substitute back $u = \frac{1}{2}\sqrt{x}$.

$\frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{1}{4\sqrt{x}}$

We can rewrite $e^{\frac{1}{2}\sqrt{x}}$ as its original form $\sqrt{e^{\sqrt{x}}}$.

$\frac{dy}{dx} = \frac{e^{\frac{\sqrt{x}}{2}}}{4\sqrt{x}}$ or $\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$

Alternate Solution (Without simplifying first):

Let $y = \sqrt{e^{\sqrt{x}}}$. This involves three nested functions.

Outermost: $\sqrt{u}$

Middle: $e^v$

Innermost: $\sqrt{x}$

$\frac{dy}{dx} = \frac{d}{dx} \left( \sqrt{e^{\sqrt{x}}} \right)$

$= \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot \frac{d}{dx} \left( e^{\sqrt{x}} \right)$

$= \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x})$

$= \frac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$

$= \frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}}$

Since $\frac{a}{\sqrt{a}} = \sqrt{a}$, we can simplify $\frac{e^{\sqrt{x}}}{\sqrt{e^{\sqrt{x}}}} = \sqrt{e^{\sqrt{x}}}$.

$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$

This matches the previous result.

Given:

The function to differentiate is $y = \log(\log x)$.

The domain for $x$ is $x > 1$. This ensures that $\log x > 0$, so $\log(\log x)$ is defined.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \log(\log x)$ with respect to $x$. We will use the chain rule.

Let the outer function be $f(u) = \log u$, and the inner function be $u = \log x$. Then $y = f(u(x))$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, we find the derivative of the outer function $y = \log u$ with respect to $u$. Assuming $\log u$ is the natural logarithm (base $e$):

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}$

Substitute back $u = \log x$ into this expression:

$\frac{dy}{du} = \frac{1}{\log x}$

Next, we find the derivative of the inner function $u = \log x$ with respect to $x$. Assuming $\log x$ is the natural logarithm (base $e$):

$\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \left(\frac{1}{\log x}\right) \cdot \left(\frac{1}{x}\right)$

Simplify the expression:

$\frac{dy}{dx} = \frac{1}{x \log x}$

Final Answer:

The derivative of $\log(\log x)$ with respect to $x$ is $\frac{1}{x \log x}$.

Given:

The function to differentiate is $y = \frac{\cos x}{\log x}$.

The domain for $x$ is $x > 0$. Note that for the function to be defined and the derivative to exist, we also require $\log x \neq 0$, which means $x \neq 1$. Thus, the domain for differentiation is $x > 0, x \neq 1$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \frac{\cos x}{\log x}$ with respect to $x$. This function is in the form of a quotient $\frac{u}{v}$, where $u = \cos x$ and $v = \log x$. We assume $\log x$ refers to the natural logarithm (base $e$).

We will use the quotient rule for differentiation, which states that if $y = \frac{u(x)}{v(x)}$, then:

$\frac{dy}{dx} = \frac{v(x) \frac{du}{dx} - u(x) \frac{dv}{dx}}{(v(x))^2}$

Let $u = \cos x$. The derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x$

Let $v = \log x$. The derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, apply the quotient rule:

$\frac{dy}{dx} = \frac{(\log x) \cdot \frac{d}{dx}(\cos x) - (\cos x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2}$

Substitute the derivatives we found:

$\frac{dy}{dx} = \frac{(\log x) \cdot (-\sin x) - (\cos x) \cdot \left(\frac{1}{x}\right)}{(\log x)^2}$

Simplify the numerator:

$\frac{dy}{dx} = \frac{-\sin x \log x - \frac{\cos x}{x}}{(\log x)^2}$

To express the numerator as a single fraction, find a common denominator:

$\frac{dy}{dx} = \frac{\frac{-x \sin x \log x - \cos x}{x}}{(\log x)^2}$

Multiply the numerator by the reciprocal of the denominator $(\log x)^2 = \frac{(\log x)^2}{1}$:

$\frac{dy}{dx} = \frac{-x \sin x \log x - \cos x}{x} \cdot \frac{1}{(\log x)^2}$

$\frac{dy}{dx} = \frac{-(x \sin x \log x + \cos x)}{x(\log x)^2}$

Final Answer:

The derivative of $\frac{\cos x}{\log x}$ with respect to $x$ is $\frac{-(x \sin x \log x + \cos x)}{x(\log x)^2}$.

Given:

The function to differentiate is $y = \cos(\log x + e^x)$.

The domain for $x$ is $x > 0$. This ensures that $\log x$ is defined.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \cos(\log x + e^x)$ with respect to $x$. We will use the chain rule.

Let the outer function be $\cos(u)$, where $u$ is the argument of the cosine function.

Let $u = \log x + e^x$.

Then the function is $y = \cos u$.

According to the chain rule, the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

First, we find the derivative of $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u$

Substitute back $u = \log x + e^x$:

$\frac{dy}{du} = -\sin(\log x + e^x)$

Next, we find the derivative of $u$ with respect to $x$. We differentiate each term in the sum $\log x + e^x$ separately:

$\frac{du}{dx} = \frac{d}{dx}(\log x + e^x)$

$\frac{du}{dx} = \frac{d}{dx}(\log x) + \frac{d}{dx}(e^x)$

Assuming $\log x$ is the natural logarithm (base $e$):

$\frac{d}{dx}(\log x) = \frac{1}{x}$

The derivative of $e^x$ is:

$\frac{d}{dx}(e^x) = e^x$

So, the derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = \frac{1}{x} + e^x$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = (-\sin(\log x + e^x)) \cdot \left(\frac{1}{x} + e^x\right)$

We can write the term $\left(\frac{1}{x} + e^x\right)$ with a common denominator:

$\frac{1}{x} + e^x = \frac{1}{x} + \frac{xe^x}{x} = \frac{1 + xe^x}{x}$

Substitute this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\sin(\log x + e^x) \cdot \left(\frac{1 + xe^x}{x}\right)$

Rearranging the terms gives:

$\frac{dy}{dx} = -\frac{1 + xe^x}{x} \sin(\log x + e^x)$

Final Answer:

The derivative of $\cos(\log x + e^x)$ with respect to $x$ is $-\frac{1 + xe^x}{x} \sin(\log x + e^x)$.

Given:

The function to differentiate is $y = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$. This differentiation can be simplified by using logarithmic differentiation.

First, take the natural logarithm of both sides of the equation:

$\log y = \log \left( \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \right)$

Rewrite the square root as a power of $\frac{1}{2}$:

$\log y = \log \left( \left(\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}\right)^{1/2} \right)$

Using the logarithm property $\log(a^b) = b \log a$:

$\log y = \frac{1}{2} \log \left( \frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5} \right)$

Using the logarithm property $\log(\frac{a}{b}) = \log a - \log b$:

$\log y = \frac{1}{2} \left[ \log((x − 3) (x^2 + 4)) - \log(3x^2 + 4x + 5) \right]$

Using the logarithm property $\log(ab) = \log a + \log b$:

$\log y = \frac{1}{2} \left[ \log(x − 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5) \right]$

Now, 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 each term within the brackets:

$\frac{d}{dx}(\log y) = \frac{d}{dx} \left( \frac{1}{2} \left[ \log(x − 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5) \right] \right)$

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log(x − 3)) + \frac{d}{dx}(\log(x^2 + 4)) - \frac{d}{dx}(\log(3x^2 + 4x + 5)) \right]$

Differentiate each term using the chain rule $\frac{d}{dx}(\log f(x)) = \frac{f'(x)}{f(x)}$:

$\frac{d}{dx}(\log(x − 3)) = \frac{\frac{d}{dx}(x-3)}{x-3} = \frac{1}{x-3}$

$\frac{d}{dx}(\log(x^2 + 4)) = \frac{\frac{d}{dx}(x^2+4)}{x^2+4} = \frac{2x}{x^2+4}$

$\frac{d}{dx}(\log(3x^2 + 4x + 5)) = \frac{\frac{d}{dx}(3x^2+4x+5)}{3x^2+4x+5} = \frac{6x+4}{3x^2+4x+5}$

Substitute these derivatives back into the equation:

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

Finally, 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 $y$ back into the equation:

$\frac{dy}{dx} = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \cdot \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

Final Answer:

The derivative of $\sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \left( \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right)$

Given:

The function to differentiate is $y = a^x$, where $a$ is a positive constant.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = a^x$ with respect to $x$. We can use logarithmic differentiation to solve this.

Take the natural logarithm of both sides of the equation:

$\log y = \log (a^x)$

Using the logarithm property $\log(b^c) = c \log b$:

$\log y = x \log a$

Here, $\log a$ is a constant since $a$ is a constant.

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule:

$\frac{d}{dx}(\log y) = \frac{d}{dx}(x \log a)$

For the left side, let $u = y$. Then $\log y = \log u$. $\frac{d}{dx}(\log u) = \frac{d}{du}(\log u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$.

For 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, the derivative of the right side is:

$\frac{d}{dx}(x \log a) = \log a \cdot 1 = \log a$

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 the original expression for $y$, which is $a^x$:

$\frac{dy}{dx} = a^x \log a$

Note that $\log a$ represents the natural logarithm of $a$, which is $\log_e a$ or $\log a$. The formula can also be written as $\frac{d}{dx}(a^x) = a^x \log a$.

Final Answer:

The derivative of $a^x$ with respect to $x$ is $a^x \log a$.

Given:

The function to differentiate is $y = x^{\sin x}$.

The domain for $x$ is $x > 0$.

To Find:

The derivative of the given function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We need to find the derivative of $y = x^{\sin x}$ with respect to $x$. This is a function of the form $f(x)^{g(x)}$, which can be differentiated using logarithmic differentiation.

Take the natural logarithm of both sides of the equation:

$\log y = \log (x^{\sin x})$

Using the logarithm property $\log(a^b) = b \log a$:

$\log y = \sin x \log x$

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule:

$\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$

On the right side, we have a product of two functions, $\sin x$ and $\log x$. We use the product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Let $u = \sin x$ and $v = \log x$.

The derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

The derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, apply the product rule to the right side of the equation $\log y = \sin x \log x$:

$\frac{d}{dx}(\sin x \log x) = (\sin x) \cdot \frac{d}{dx}(\log x) + (\log x) \cdot \frac{d}{dx}(\sin x)$

$\frac{d}{dx}(\sin x \log x) = (\sin x) \cdot \left(\frac{1}{x}\right) + (\log x) \cdot (\cos x)$

$\frac{d}{dx}(\sin x \log x) = \frac{\sin x}{x} + \cos x \log x$

Equating the derivatives of both sides:

$\frac{1}{y} \frac{dy}{dx} = \frac{\sin x}{x} + \cos x \log x$

Now, 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 the original expression for $y$, which is $x^{\sin x}$:

$\frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$

Final Answer:

The derivative of $x^{\sin x}$ with respect to $x$ is $x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$.

Given:

The equation is $y^x + x^y + x^x = a^b$, where $a$ and $b$ are constants.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given the implicit relation $y^x + x^y + x^x = a^b$. To find $\frac{dy}{dx}$, we will differentiate both sides of the equation with respect to $x$.

Let $u = y^x$, $v = x^y$, and $w = x^x$. The equation becomes $u + v + w = a^b$.

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(u + v + w) = \frac{d}{dx}(a^b)$

$\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$

(Since $a^b$ is a constant, its derivative is 0).

Now we need to find the derivative of each term separately using logarithmic differentiation.

For $u = y^x$:

Take the natural logarithm of both sides:

$\log u = \log(y^x)$

$\log u = x \log y$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \log y)$

$\frac{1}{u} \frac{du}{dx} = (1) \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx}$

$\frac{du}{dx} = u \left( \log y + \frac{x}{y} \frac{dy}{dx} \right)$

Substitute back $u = y^x$:

$\frac{du}{dx} = y^x \log y + y^x \frac{x}{y} \frac{dy}{dx}$

$\frac{du}{dx} = y^x \log y + x y^{x-1} \frac{dy}{dx}$

For $v = x^y$:

Take the natural logarithm of both sides:

$\log v = \log(x^y)$

$\log v = y \log x$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(y \log x)$

$\frac{1}{v} \frac{dv}{dx} = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x}$

$\frac{dv}{dx} = v \left( \log x \frac{dy}{dx} + \frac{y}{x} \right)$

Substitute back $v = x^y$:

$\frac{dv}{dx} = x^y \log x \frac{dy}{dx} + x^y \frac{y}{x}$

$\frac{dv}{dx} = x^y \log x \frac{dy}{dx} + y x^{y-1}$

For $w = x^x$:

Take the natural logarithm of both sides:

$\log w = \log(x^x)$

$\log w = x \log x$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right:

$\frac{1}{w} \frac{dw}{dx} = \frac{d}{dx}(x \log x)$

$\frac{1}{w} \frac{dw}{dx} = (1) \cdot \log x + x \cdot \frac{1}{x}$

$\frac{1}{w} \frac{dw}{dx} = \log x + 1$

$\frac{dw}{dx} = w (\log x + 1)$

Substitute back $w = x^x$:

$\frac{dw}{dx} = x^x (\log x + 1)$

Now substitute $\frac{du}{dx}$, $\frac{dv}{dx}$, and $\frac{dw}{dx}$ back into the 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 without $\frac{dy}{dx}$:

$x y^{x-1} \frac{dy}{dx} + x^y \log x \frac{dy}{dx} = -y^x \log y - y x^{y-1} - x^x (\log x + 1)$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$\frac{dy}{dx} (x y^{x-1} + x^y \log x) = -(y^x \log y + y x^{y-1} + x^x (\log x + 1))$

Solve for $\frac{dy}{dx}$ by dividing by the coefficient of $\frac{dy}{dx}$:

$\frac{dy}{dx} = - \frac{y^x \log y + y x^{y-1} + x^x (\log x + 1)}{x y^{x-1} + x^y \log x}$

Final Answer:

The derivative $\frac{dy}{dx}$ is $- \frac{y^x \log y + y x^{y-1} + x^x (\log x + 1)}{x y^{x-1} + x^y \log x}$.

Given:

Let the given function be denoted by $y$.

$y = \cos x \cdot \cos 2x \cdot \cos 3x$

Solution:

To differentiate the product of multiple functions, it is convenient to use logarithmic differentiation.

Taking the natural logarithm on both sides of the equation, we get:

$\log y = \log (\cos x \cdot \cos 2x \cdot \cos 3x)$

Using the property of logarithms $\log(abc) = \log a + \log b + \log c$, we expand the right side:

$\log y = \log(\cos x) + \log(\cos 2x) + \log(\cos 3x)$

Now, we differentiate both sides of this equation with respect to $x$:

$\frac{d}{dx}(\log y) = \frac{d}{dx}(\log(\cos x)) + \frac{d}{dx}(\log(\cos 2x)) + \frac{d}{dx}(\log(\cos 3x))$

Applying the chain rule, $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$, to each term:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} \frac{d}{dx}(\cos x) + \frac{1}{\cos 2x} \frac{d}{dx}(\cos 2x) + \frac{1}{\cos 3x} \frac{d}{dx}(\cos 3x)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} (-\sin x) + \frac{1}{\cos 2x} (-\sin 2x \cdot 2) + \frac{1}{\cos 3x} (-\sin 3x \cdot 3)$

Using the identity $\frac{\sin \theta}{\cos \theta} = \tan \theta$, we simplify the expression:

$\frac{1}{y} \frac{dy}{dx} = -\tan x - 2\tan 2x - 3\tan 3x$

To find $\frac{dy}{dx}$, we multiply both sides by $y$:

$\frac{dy}{dx} = y (-\tan x - 2\tan 2x - 3\tan 3x)$

Finally, we substitute the original expression for $y$ back into the equation and factor out the negative sign:

$\frac{dy}{dx} = -(\cos x \cdot \cos 2x \cdot \cos 3x) (\tan x + 2\tan 2x + 3\tan 3x)$

Given:

Let the given function be denoted by $y$.

$y = \sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}} = \left(\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}\right)^{1/2}$

Solution:

Due to the complex structure involving products, quotients, and a root, logarithmic differentiation is the most efficient method.

Take the natural logarithm of both sides:

$\log y = \log \left[\left(\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}\right)^{1/2}\right]$

Using logarithm properties to expand the expression:

$\log y = \frac{1}{2} \log \left(\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}\right)$

$\log y = \frac{1}{2} [\log((x − 1) (x − 2)) - \log((x − 3) (x − 4) (x − 5))]$

$\log y = \frac{1}{2} [\log(x − 1) + \log(x − 2) - (\log(x − 3) + \log(x − 4) $$ + \log(x − 5))]$

$\log y = \frac{1}{2} [\log(x − 1) + \log(x − 2) - \log(x − 3) - \log(x − 4) $$ - \log(x − 5)]$

Now, differentiate both sides with respect to $x$:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right]$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = \frac{1}{2} y \left[ \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right]$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}} \left( \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right)$

Given:

Let the function be $y = (\log x)^{\cos x}$. We assume $\log x$ is the natural logarithm, $\log x$.

Solution:

The function is of the form $f(x)^{g(x)}$, which requires logarithmic differentiation.

Take the natural logarithm on both sides:

$\log y = \log((\log x)^{\cos x})$

$\log y = \cos x \cdot \log(\log x)$

Differentiate both sides with respect to $x$. We use the product rule on the right side.

$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \log(\log x) + \cos x \cdot \frac{d}{dx}(\log(\log x))$

The derivative of $\cos x$ is $-\sin x$.

Using the chain rule, the derivative of $\log(\log x)$ is $\frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.

Substituting these into the equation:

$\frac{1}{y} \frac{dy}{dx} = (-\sin x) \cdot \log(\log x) + \cos x \cdot \left(\frac{1}{x \log x}\right)$

$\frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{x \log x} - \sin x \log(\log x)$

Solve for $\frac{dy}{dx}$ by multiplying by $y$:

$\frac{dy}{dx} = y \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)$

Substitute the original expression for $y$ back:

$\frac{dy}{dx} = (\log x)^{\cos x} \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)$

Given:

Let $y = x^x - 2^{\sin x}$.

We cannot use logarithmic differentiation directly on the entire expression. We must differentiate each term separately.

Let $u = x^x$ and $v = 2^{\sin x}$. Then $y = u - v$, and $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.

Solution:

Step 1: Differentiate $u = x^x$

We use logarithmic differentiation for this term.

$\log u = \log(x^x) = x \log x$

Differentiating with respect to $x$ using the product rule:

$\frac{1}{u}\frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$

$\frac{du}{dx} = u(\log x + 1) = x^x(\log x + 1)$

Step 2: Differentiate $v = 2^{\sin x}$

We use the formula for the derivative of $a^{f(x)}$, which is $a^{f(x)} \log a \cdot f'(x)$.

Here, $a=2$ and $f(x) = \sin x$. The derivative $f'(x)$ is $\cos x$.

$\frac{dv}{dx} = 2^{\sin x} \cdot \log 2 \cdot \cos x$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$

$\frac{dy}{dx} = x^x(\log x + 1) - (2^{\sin x} \log 2 \cos x)$

Given:

Let the given function be denoted by $y$.

$y = (x + 3)^2 (x + 4)^3 (x + 5)^4$

Solution:

Since the function is a product of terms raised to powers, logarithmic differentiation is a suitable method.

Taking the natural logarithm on both sides:

$\log y = \log[(x + 3)^2 (x + 4)^3 (x + 5)^4]$

Using the logarithm property $\log(abc) = \log a + \log b + \log c$:

$\log y = \log(x + 3)^2 + \log(x + 4)^3 + \log(x + 5)^4$

Using the logarithm property $\log(a^n) = n \log a$:

$\log y = 2 \log(x + 3) + 3 \log(x + 4) + 4 \log(x + 5)$

Now, differentiate both sides with respect to $x$. Remember that $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(x+c) = 1$ for any constant $c$.

$\frac{d}{dx}(\log y) = \frac{d}{dx}[2 \log(x + 3) + 3 \log(x + 4) + 4 \log(x + 5)]$

$\frac{1}{y} \frac{dy}{dx} = 2 \frac{d}{dx}(\log(x + 3)) + 3 \frac{d}{dx}(\log(x + 4)) + 4 \frac{d}{dx}(\log(x + 5))$

$\frac{1}{y} \frac{dy}{dx} = 2 \left(\frac{1}{x + 3} \cdot \frac{d}{dx}(x+3)\right) + 3 \left(\frac{1}{x + 4} \cdot \frac{d}{dx}(x+4)\right) $$ + 4 \left(\frac{1}{x + 5} \cdot \frac{d}{dx}(x+5)\right)$

$\frac{1}{y} \frac{dy}{dx} = 2 \left(\frac{1}{x + 3} \cdot 1\right) + 3 \left(\frac{1}{x + 4} \cdot 1\right) + 4 \left(\frac{1}{x + 5} \cdot 1\right)$

$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right)$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right)$

Final Answer:

The derivative of the given function 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:

Let $y = \left( x + \frac{1}{x} \right)^x + x^{\left(1 + \frac{1}{x} \right)}$.

Let $u = \left( x + \frac{1}{x} \right)^x$ and $v = x^{\left(1 + \frac{1}{x} \right)}$. Then $y = u + v$, so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

Solution:

Step 1: Differentiate $u = \left( x + \frac{1}{x} \right)^x$

Using logarithmic differentiation:

$\log u = x \log \left( x + \frac{1}{x} \right)$

Differentiating with respect to $x$ using the product rule:

$\frac{1}{u}\frac{du}{dx} = 1 \cdot \log\left(x+\frac{1}{x}\right) + x \cdot \frac{1}{x+\frac{1}{x}} \cdot \left(1-\frac{1}{x^2}\right)$

$\frac{1}{u}\frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + x \cdot \frac{x}{x^2+1} \cdot \frac{x^2-1}{x^2}$

$\frac{1}{u}\frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1}$

$\frac{du}{dx} = u \left[ \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \right] = \left(x+\frac{1}{x}\right)^x \left[ \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \right]$

Step 2: Differentiate $v = x^{\left(1 + \frac{1}{x} \right)}$

Using logarithmic differentiation:

$\log v = \left(1 + \frac{1}{x}\right) \log x$

Differentiating with respect to $x$ using the product rule:

$\frac{1}{v}\frac{dv}{dx} = \left(-\frac{1}{x^2}\right) \log x + \left(1+\frac{1}{x}\right) \cdot \frac{1}{x}$

$\frac{1}{v}\frac{dv}{dx} = -\frac{\log x}{x^2} + \frac{1}{x} + \frac{1}{x^2} = \frac{x+1-\log x}{x^2}$

$\frac{dv}{dx} = v \left[ \frac{x+1-\log x}{x^2} \right] = x^{\left(1 + \frac{1}{x} \right)} \left[ \frac{x+1-\log x}{x^2} \right]$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

$\frac{dy}{dx} = \left(x+\frac{1}{x}\right)^x \left[ \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \right] + x^{\left(1 + \frac{1}{x} \right)} \left[ \frac{x+1-\log x}{x^2} \right]$

Given:

Let the given function be denoted by $y$.

$y = (\log x)^x + x^{\log x}$

We interpret $\log x$ as the natural logarithm, $\log x$.

Solution:

We can write the function as $y = u + v$, where $u = (\log x)^x$ and $v = x^{\log x}$.

To find the derivative $\frac{dy}{dx}$, we find the derivatives of $u$ and $v$ separately and then add them:

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = (\log x)^x$

This is of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log u = \log((\log x)^x)$

Using the property $\log(a^b) = b \log a$:

$\log u = x \log(\log x)$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\log u) = \frac{d}{dx}(x \log(\log x))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) \log(\log x) + x \left(\frac{d}{dx} \log(\log x)\right)$

We know $\frac{d}{dx} x = 1$. For the second term, use the chain rule. Let $w = \log x$. Then $\frac{d}{dx}(\log w) = \frac{1}{w} \frac{dw}{dx} = \frac{1}{\log x} \frac{d}{dx}(\log x)$.

Since $\frac{d}{dx}(\log x) = \frac{1}{x}$, we have:

$\frac{d}{dx}(\log(\log x)) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$

Substitute these back:

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\log x) + x \cdot \left(\frac{1}{x \log x}\right)$

$\frac{1}{u} \frac{du}{dx} = \log(\log x) + \frac{1}{\log x}$

Solve for $\frac{du}{dx}$:

$\frac{du}{dx} = u \left( \log(\log x) + \frac{1}{\log x} \right)$

Substitute $u = (\log x)^x$ back:

$\frac{du}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right)$

Step 2: Differentiate $v = x^{\log x}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log v = \log(x^{\log x})$

Using the property $\log(a^b) = b \log a$:

$\log v = (\log x) (\log x) = (\log x)^2$

Differentiate both sides with respect to $x$ using the chain rule on the right side:

$\frac{d}{dx}(\log v) = \frac{d}{dx}((\log x)^2)$

$\frac{1}{v} \frac{dv}{dx} = 2 (\log x) \cdot \frac{d}{dx}(\log x)$

$\frac{1}{v} \frac{dv}{dx} = 2 (\log x) \cdot \frac{1}{x} = \frac{2 \log x}{x}$

Solve for $\frac{dv}{dx}$:

$\frac{dv}{dx} = v \left( \frac{2 \log x}{x} \right)$

Substitute $v = x^{\log x}$ back:

$\frac{dv}{dx} = x^{\log x} \left( \frac{2 \log x}{x} \right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

$\frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) + x^{\log x} \left( \frac{2 \log x}{x} \right)$

Final Answer:

The derivative of the given function 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:

Let the given function be denoted by $y$.

$y = (\sin x)^x + \sin^{-1} \sqrt{x}$

We can express this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = (\sin x)^x$ and $v = \sin^{-1} \sqrt{x}$.

Solution:

To find the derivative $\frac{dy}{dx}$, we find the derivatives of $u$ and $v$ with respect to $x$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = (\sin x)^x$

This function is in the form of $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log u = \log((\sin x)^x)$

Using the logarithm property $\log(a^b) = b \log a$:

$\log u = x \log(\sin x)$

Differentiate both sides with respect to $x$. Use the product rule on the right side, $\frac{d}{dx}(ab) = a'b + ab'$, where $a=x$ and $b=\log(\sin x)$.

$\frac{d}{dx}(\log u) = \frac{d}{dx}(x \log(\sin x))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) \log(\sin x) + x \left(\frac{d}{dx} \log(\sin x)\right)$

We know $\frac{d}{dx} x = 1$. For the second term, use the chain rule: $\frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$.

$\frac{d}{dx}(\sin x) = \cos x$

So, $\frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

Substitute these back into the derivative equation for $\log u$:

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\sin x) + x \cdot (\cot x)$

$\frac{1}{u} \frac{du}{dx} = \log(\sin x) + x \cot x$

Solve for $\frac{du}{dx}$ by multiplying both sides by $u$:

$\frac{du}{dx} = u (\log(\sin x) + x \cot x)$

Substitute the expression for $u$ back:

$\frac{du}{dx} = (\sin x)^x (\log(\sin x) + x \cot x)$

Step 2: Differentiate $v = \sin^{-1} \sqrt{x}$

This requires the chain rule. The derivative of $\sin^{-1} w$ with respect to $w$ is $\frac{1}{\sqrt{1 - w^2}}$. Here $w = \sqrt{x} = x^{1/2}$.

The derivative of $\sin^{-1} \sqrt{x}$ with respect to $x$ is $\frac{d}{dw}(\sin^{-1} w) \cdot \frac{dw}{dx}$, where $w = \sqrt{x}$.

$\frac{dw}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$

The derivative of $\sin^{-1} w$ with $w = \sqrt{x}$ is $\frac{1}{\sqrt{1 - (\sqrt{x})^2}} = \frac{1}{\sqrt{1 - x}}$.

Combine these using the chain rule:

$\frac{dv}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}$

$\frac{dv}{dx} = \frac{1}{2\sqrt{x(1 - x)}}$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = (\sin x)^x (\log(\sin x) + x \cot x) + \frac{1}{2\sqrt{x(1 - x)}}$

Final Answer:

The derivative of the given function 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:

Let the given function be denoted by $y$.

$y = x^{\sin x} + (\sin x)^{\cos x}$

We can write this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$.

Solution:

To find the derivative of $y$ with respect to $x$, we need to find the derivatives of $u$ and $v$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = x^{\sin x}$

This is of the form $f(x)^{g(x)}$, so we use logarithmic differentiation.

Taking natural logarithm on both sides:

$\log u = \log(x^{\sin x})$

Using the property $\log(a^b) = b \log a$:

$\log u = \sin x \log x$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\log u) = \frac{d}{dx}(\sin x \log x)$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} \sin x\right) \log x + \sin x \left(\frac{d}{dx} \log x\right)$

We know $\frac{d}{dx} \sin x = \cos x$ and $\frac{d}{dx} \log x = \frac{1}{x}$.

Substitute these back:

$\frac{1}{u} \frac{du}{dx} = (\cos x) \log x + \sin x \left(\frac{1}{x}\right)$

$\frac{1}{u} \frac{du}{dx} = \cos x \log x + \frac{\sin x}{x}$

Solve for $\frac{du}{dx}$:

$\frac{du}{dx} = u \left(\cos x \log x + \frac{\sin x}{x}\right)$

Substitute $u = x^{\sin x}$ back:

$\frac{du}{dx} = x^{\sin x} \left(\cos x \log x + \frac{\sin x}{x}\right)$

Step 2: Differentiate $v = (\sin x)^{\cos x}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Taking natural logarithm on both sides:

$\log v = \log((\sin x)^{\cos x})$

Using the property $\log(a^b) = b \log a$:

$\log v = \cos x \log(\sin x)$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\log v) = \frac{d}{dx}(\cos x \log(\sin x))$

$\frac{1}{v} \frac{dv}{dx} = \left(\frac{d}{dx} \cos x\right) \log(\sin x) + \cos x \left(\frac{d}{dx} \log(\sin x)\right)$

We know $\frac{d}{dx} \cos x = -\sin x$. For the second term, 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 back:

$\frac{1}{v} \frac{dv}{dx} = (-\sin x) \log(\sin x) + \cos x (\cot x)$

$\frac{1}{v} \frac{dv}{dx} = -\sin x \log(\sin x) + \cos x \frac{\cos x}{\sin x}$

$\frac{1}{v} \frac{dv}{dx} = -\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}$

Solve for $\frac{dv}{dx}$:

$\frac{dv}{dx} = v \left(-\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Substitute $v = (\sin x)^{\cos x}$ back:

$\frac{dv}{dx} = (\sin x)^{\cos x} \left(-\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = x^{\sin x} \left(\cos x \log x + \frac{\sin x}{x}\right) + (\sin x)^{\cos x} \left(-\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = x^{\sin x} \left(\cos x \log x + \frac{\sin x}{x}\right) + (\sin x)^{\cos x} \left(-\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Given:

Let the given function be denoted by $y$.

$y = x^{x \cos x} + \frac{x^2 + 1}{x^2 − 1}$

We can write this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = x^{x \cos x}$ and $v = \frac{x^2 + 1}{x^2 − 1}$.

Solution:

To find the derivative of $y$ with respect to $x$, we need to find the derivatives of $u$ and $v$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = x^{x \cos x}$

This function is of the form $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log u = \log(x^{x \cos x})$

Using the logarithm property $\log(a^b) = b \log a$:

$\log u = (x \cos x) \log x$

Differentiate both sides with respect to $x$. Use the product rule on the right side, $\frac{d}{dx}(AB) = A'B + AB'$, where $A = x \cos x$ and $B = \log x$.

$\frac{d}{dx}(\log u) = \frac{d}{dx}((x \cos x) \log x)$

First, find the derivative of $A = x \cos x$ using the product rule:

$\frac{dA}{dx} = \frac{d}{dx}(x \cos x) = (1)(\cos x) + x(-\sin x) = \cos x - x \sin x$

The derivative of $B = \log x$ is:

$\frac{dB}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, apply the product rule to $\log u$:

$\frac{1}{u} \frac{du}{dx} = (\cos x - x \sin x) (\log x) + (x \cos x) \left(\frac{1}{x}\right)$

$\frac{1}{u} \frac{du}{dx} = (\cos x - x \sin x) \log x + \cos x$

Solve for $\frac{du}{dx}$ by multiplying both sides by $u$:

$\frac{du}{dx} = u \left((\cos x - x \sin x) \log x + \cos x\right)$

Substitute the expression for $u$ back:

$\frac{du}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \log x + \cos x\right)$

Step 2: Differentiate $v = \frac{x^2 + 1}{x^2 − 1}$

This function is a quotient. Use the quotient rule: $\frac{d}{dx}\left(\frac{p}{q}\right) = \frac{p'q - pq'}{q^2}$.

Here, $p = x^2 + 1$ and $q = x^2 - 1$.

Find the derivatives of $p$ and $q$:

$p' = \frac{d}{dx}(x^2 + 1) = 2x$

$q' = \frac{d}{dx}(x^2 - 1) = 2x$

Apply the quotient rule:

$\frac{dv}{dx} = \frac{(2x)(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$

Simplify the numerator:

Numerator = $2x(x^2 - 1) - 2x(x^2 + 1) = (2x^3 - 2x) - (2x^3 + 2x) $$ = 2x^3 - 2x - 2x^3 - 2x = -4x$

So, the derivative of $v$ is:

$\frac{dv}{dx} = \frac{-4x}{(x^2 - 1)^2}$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \log x + \cos x\right) + \frac{-4x}{(x^2 - 1)^2}$

$\frac{dy}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \log x + \cos x\right) - \frac{4x}{(x^2 - 1)^2}$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \log x + \cos x\right) - \frac{4x}{(x^2 - 1)^2}$

Given:

Let the given function be denoted by $y$.

$y = (x \cos x)^x + (x \sin x)^{\frac{1}{x}}$

We can express this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = (x \cos x)^x$ and $v = (x \sin x)^{\frac{1}{x}}$.

Solution:

To find the derivative of $y$ with respect to $x$, we need to find the derivatives of $u$ and $v$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = (x \cos x)^x$

This function is in the form of $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log u = \log((x \cos x)^x)$

Using the logarithm property $\log(a^b) = b \log a$:

$\log u = x \log(x \cos x)$

Using the logarithm property $\log(ab) = \log a + \log b$:

$\log u = x (\log x + \log(\cos x))$

$\log u = x \log x + x \log(\cos x)$

Differentiate both sides with respect to $x$. Use the product rule on both terms on the right side.

$\frac{d}{dx}(\log u) = \frac{d}{dx}(x \log x) + \frac{d}{dx}(x \log(\cos x))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) (\log x) + x \left(\frac{d}{dx} \log x\right) + \left(\frac{d}{dx} x\right) (\log(\cos x)) $$ + x \left(\frac{d}{dx} \log(\cos x)\right)$

$\frac{1}{u} \frac{du}{dx} = (1)(\log x) + x\left(\frac{1}{x}\right) + (1)(\log(\cos x)) + x\left(\frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)\right)$

$\frac{1}{u} \frac{du}{dx} = \log x + 1 + \log(\cos x) + x\left(\frac{1}{\cos x} \cdot (-\sin x)\right)$

$\frac{1}{u} \frac{du}{dx} = \log x + 1 + \log(\cos x) - x \frac{\sin x}{\cos x}$

$\frac{1}{u} \frac{du}{dx} = \log x + 1 + \log(\cos x) - x \tan x$

Solve for $\frac{du}{dx}$ by multiplying both sides by $u$:

$\frac{du}{dx} = u (\log x + 1 + \log(\cos x) - x \tan x)$

Substitute the expression for $u$ back:

$\frac{du}{dx} = (x \cos x)^x (\log x + 1 + \log(\cos x) - x \tan x)$

Step 2: Differentiate $v = (x \sin x)^{\frac{1}{x}}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Take the natural logarithm of both sides:

$\log v = \log((x \sin x)^{\frac{1}{x}})$

Using the property $\log(a^b) = b \log a$:

$\log v = \frac{1}{x} \log(x \sin x)$

Using the property $\log(ab) = \log a + \log b$:

$\log v = \frac{1}{x} (\log x + \log(\sin x))$

$\log v = \frac{\log x}{x} + \frac{\log(\sin x)}{x}$

Differentiate both sides with respect to $x$. Use the quotient rule on both terms on the right side.

$\frac{d}{dx}(\log v) = \frac{d}{dx}\left(\frac{\log x}{x}\right) + \frac{d}{dx}\left(\frac{\log(\sin x)}{x}\right)$

For the first term, $\frac{d}{dx}\left(\frac{\log x}{x}\right) = \frac{(\frac{1}{x})(x) - (\log x)(1)}{x^2} = \frac{1 - \log x}{x^2}$.

For the second term, $\frac{d}{dx}\left(\frac{\log(\sin x)}{x}\right) = \frac{(\frac{d}{dx}\log(\sin x))(x) - (\log(\sin x))(\frac{d}{dx}x)}{x^2}$.

Using the chain rule, $\frac{d}{dx}\log(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

So the second term derivative is: $\frac{(\cot x)(x) - (\log(\sin x))(1)}{x^2} = \frac{x \cot x - \log(\sin x)}{x^2}$.

Combine the derivatives for $\frac{1}{v} \frac{dv}{dx}$:

$\frac{1}{v} \frac{dv}{dx} = \frac{1 - \log x}{x^2} + \frac{x \cot x - \log(\sin x)}{x^2}$

$\frac{1}{v} \frac{dv}{dx} = \frac{1 - \log x + x \cot x - \log(\sin x)}{x^2}$

Solve for $\frac{dv}{dx}$ by multiplying both sides by $v$:

$\frac{dv}{dx} = v \left(\frac{1 - \log x + x \cot x - \log(\sin x)}{x^2}\right)$

Substitute the expression for $v$ back:

$\frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \left(\frac{1 - \log x + x \cot x - \log(\sin x)}{x^2}\right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = (x \cos x)^x (\log x + 1 + \log(\cos x) - x \tan x) $$ + (x \sin x)^{\frac{1}{x}} \left(\frac{1 - \log x + x \cot x - \log(\sin x)}{x^2}\right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = (x \cos x)^x (\log x + 1 + \log(\cos x) - x \tan x) $$ + (x \sin x)^{\frac{1}{x}} \left(\frac{1 - \log x + x \cot x - \log(\sin x)}{x^2}\right)$

Given:

The implicit equation is $x^y + y^x = 1$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Let the given equation be

Differentiating both sides of equation (1) with respect to $x$, we get:

$\frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = \frac{d}{dx}(1)$

$\frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = 0$

Let's evaluate each derivative separately.

For the term $x^y$, let $u = x^y$. Taking natural logarithm on both sides:

$\log u = \log(x^y)$

$\log u = y \log x$

Differentiating with respect to $x$ using the chain rule and product rule:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(y \log x)$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{dy}{dx}\right) \log x + y \left(\frac{1}{x}\right)$

$\frac{du}{dx} = u \left(\frac{dy}{dx} \log x + \frac{y}{x}\right)$

Substitute $u = x^y$ back:

$\frac{du}{dx} = x^y \left(\frac{dy}{dx} \log x + \frac{y}{x}\right)$

$\frac{du}{dx} = x^y \log x \frac{dy}{dx} + x^y \frac{y}{x}$

$\frac{du}{dx} = x^y \log x \frac{dy}{dx} + y x^{y-1}$

For the term $y^x$, let $v = y^x$. Taking natural logarithm on both sides:

$\log v = \log(y^x)$

$\log v = x \log y$

Differentiating with respect to $x$ using the chain rule and product rule:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(x \log y)$

$\frac{1}{v} \frac{dv}{dx} = (1) \log y + x \left(\frac{1}{y} \frac{dy}{dx}\right)$

$\frac{dv}{dx} = v \left(\log y + \frac{x}{y} \frac{dy}{dx}\right)$

Substitute $v = y^x$ back:

$\frac{dv}{dx} = y^x \left(\log y + \frac{x}{y} \frac{dy}{dx}\right)$

$\frac{dv}{dx} = y^x \log y + y^x \frac{x}{y} \frac{dy}{dx}$

$\frac{dv}{dx} = y^x \log y + x y^{x-1} \frac{dy}{dx}$

Now substitute these derivatives back into the differentiated equation:

$(x^y \log x \frac{dy}{dx} + y x^{y-1}) + (y^x \log y + x y^{x-1} \frac{dy}{dx}) = 0$

Group terms containing $\frac{dy}{dx}$ on one side and the remaining terms on the other side:

$x^y \log x \frac{dy}{dx} + x y^{x-1} \frac{dy}{dx} = - y x^{y-1} - y^x \log y$

Factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx} (x^y \log x + x y^{x-1}) = - (y x^{y-1} + y^x \log y)$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

$\frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$

Given:

The implicit equation is $y^x = x^y$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Since we have variables in the exponents, we take the natural logarithm of both sides of equation (1):

$\log(y^x) = \log(x^y)$

Using the logarithm property $\log(a^b) = b \log a$:

Now, differentiate both sides of equation (2) with respect to $x$. We will use the product rule on both sides, remembering that $y$ is a function of $x$, so we apply the chain rule when differentiating terms involving $y$.

$\frac{d}{dx}(x \log y) = \frac{d}{dx}(y \log x)$

Apply the product rule $\frac{d}{dx}(uv) = u'v + uv'$:

$\left(\frac{d}{dx} x\right) (\log y) + x \left(\frac{d}{dx} \log y\right) = \left(\frac{d}{dx} y\right) (\log x) + y \left(\frac{d}{dx} \log x\right)$

Calculate the individual derivatives:

$\frac{d}{dx} x = 1$

$\frac{d}{dx} \log y = \frac{1}{y} \frac{dy}{dx}$ (using chain rule)

$\frac{d}{dx} y = \frac{dy}{dx}$

$\frac{d}{dx} \log x = \frac{1}{x}$

Substitute these derivatives back into the differentiated equation:

$(1) (\log y) + x \left(\frac{1}{y} \frac{dy}{dx}\right) = \left(\frac{dy}{dx}\right) (\log x) + y \left(\frac{1}{x}\right)$

$\log y + \frac{x}{y} \frac{dy}{dx} = \log x \frac{dy}{dx} + \frac{y}{x}$

Now, collect all terms containing $\frac{dy}{dx}$ on one side and all other terms on the other side.

$\frac{x}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} - \log y$

Factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx} \left(\frac{x}{y} - \log x\right) = \frac{y}{x} - \log y$

Combine the terms within the parentheses and on the right side by finding a common denominator:

$\frac{dy}{dx} \left(\frac{x - y \log x}{y}\right) = \frac{y - x \log y}{x}$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $\frac{y}{x - y \log x}$:

$\frac{dy}{dx} = \frac{y - x \log y}{x} \cdot \frac{y}{x - y \log x}$

$\frac{dy}{dx} = \frac{y (y - x \log y)}{x (x - y \log x)}$

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

$\frac{dy}{dx} = \frac{y (y - x \log y)}{x (x - y \log x)}$

Given:

The implicit equation is $(\cos x)^y = (\cos y)^x$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Since we have variables in the exponents, we take the natural logarithm of both sides of equation (1):

$\log((\cos x)^y) = \log((\cos y)^x)$

Using the logarithm property $\log(a^b) = b \log a$:

Now, differentiate both sides of equation (2) with respect to $x$. We will use the product rule on both sides, remembering that $y$ is a function of $x$, so we apply the chain rule when differentiating terms involving $y$.

$\frac{d}{dx}(y \log(\cos x)) = \frac{d}{dx}(x \log(\cos y))$

Apply the product rule $\frac{d}{dx}(uv) = u'v + uv'$:

$\left(\frac{d}{dx} y\right) (\log(\cos x)) + y \left(\frac{d}{dx} \log(\cos x)\right) = \left(\frac{d}{dx} x\right) (\log(\cos y)) $$ + x \left(\frac{d}{dx} \log(\cos y)\right)$

Calculate the individual derivatives:

$\frac{d}{dx} y = \frac{dy}{dx}$

$\frac{d}{dx} \log(\cos x)$: Use the chain rule. Let $u = \cos x$, then $\frac{d}{du}(\log u) = \frac{1}{u}$ and $\frac{du}{dx} = -\sin x$. So, $\frac{d}{dx}\log(\cos x) = \frac{1}{\cos x}(-\sin x) = -\tan x$.

$\frac{d}{dx} x = 1$

$\frac{d}{dx} \log(\cos y)$: Use the chain rule. Let $v = \cos y$, then $\frac{d}{dv}(\log v) = \frac{1}{v}$ and $\frac{dv}{dx} = \frac{d}{dx}(\cos y) = -\sin y \frac{dy}{dx}$ (since $y$ is a function of $x$). So, $\frac{d}{dx}\log(\cos y) = \frac{1}{\cos y}(-\sin y \frac{dy}{dx}) = -\tan y \frac{dy}{dx}$.

Substitute these derivatives back into the differentiated equation:

$\left(\frac{dy}{dx}\right) (\log(\cos x)) + y (-\tan x) = (1) (\log(\cos y)) + x \left(-\tan y \frac{dy}{dx}\right)$

$\log(\cos x) \frac{dy}{dx} - y \tan x = \log(\cos y) - x \tan y \frac{dy}{dx}$

Collect all terms containing $\frac{dy}{dx}$ on one side and all other terms on the other side.

$\log(\cos x) \frac{dy}{dx} + x \tan y \frac{dy}{dx} = \log(\cos y) + y \tan x$

Factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx} (\log(\cos x) + x \tan y) = \log(\cos y) + y \tan x$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y}$

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

$\frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y}$

Given:

The implicit equation is $xy = e^{(x – y)}$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Differentiate both sides of equation (1) with respect to $x$:

$\frac{d}{dx}(xy) = \frac{d}{dx}(e^{x-y})$

For the left side, use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u=x$ and $v=y$:

$\frac{d}{dx}(xy) = \left(\frac{d}{dx} x\right) y + x \left(\frac{d}{dx} y\right)$

$\frac{d}{dx}(xy) = (1) y + x \frac{dy}{dx} = y + x \frac{dy}{dx}$

For the right side, use the chain rule $\frac{d}{dx}(e^w) = e^w \frac{dw}{dx}$ with $w = x-y$.

$\frac{d}{dx}(e^{x-y}) = e^{x-y} \cdot \frac{d}{dx}(x-y)$

Differentiate the exponent:

$\frac{d}{dx}(x-y) = \frac{d}{dx}(x) - \frac{d}{dx}(y) = 1 - \frac{dy}{dx}$

Substitute the exponent's derivative back:

$\frac{d}{dx}(e^{x-y}) = e^{x-y} (1 - \frac{dy}{dx})$

Now, equate the derivatives of both sides:

$y + x \frac{dy}{dx} = e^{x-y} (1 - \frac{dy}{dx})$

$y + x \frac{dy}{dx} = e^{x-y} - e^{x-y} \frac{dy}{dx}$

Group terms containing $\frac{dy}{dx}$ on one side and terms without $\frac{dy}{dx}$ on the other side:

$x \frac{dy}{dx} + e^{x-y} \frac{dy}{dx} = e^{x-y} - y$

Factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx} (x + e^{x-y}) = e^{x-y} - y$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{e^{x-y} - y}{x + e^{x-y}}$

From the original equation (1), we know that $e^{x-y} = xy$. We can substitute this into the expression for $\frac{dy}{dx}$ to simplify it:

$\frac{dy}{dx} = \frac{xy - y}{x + xy}$

Factor out common terms in the numerator and denominator:

$\frac{dy}{dx} = \frac{y(x - 1)}{x(1 + y)}$

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

$\frac{dy}{dx} = \frac{y(x - 1)}{x(1 + y)}$

or equivalently,

$\frac{dy}{dx} = \frac{e^{x-y} - y}{x + e^{x-y}}$

Given:

The function is $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$.

Solution:

To find the derivative of $f(x)$, we can use logarithmic differentiation, which is convenient for functions involving products.

Let $y = f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$.

Taking the natural logarithm of both sides:

$\log y = \log[(1 + x) (1 + x^2) (1 + x^4) (1 + x^8)]$

Using the logarithm property $\log(abcd) = \log a + \log b + \log c + \log d$:

$\log y = \log(1 + x) + \log(1 + x^2) + \log(1 + x^4) + \log(1 + x^8)$

Now, differentiate both sides with respect to $x$. Recall that $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(1+x^n) = nx^{n-1}$.

$\frac{d}{dx}(\log y) = \frac{d}{dx}(\log(1 + x)) + \frac{d}{dx}(\log(1 + x^2)) + \frac{d}{dx}(\log(1 + x^4)) $$ + \frac{d}{dx}(\log(1 + x^8))$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x) + \frac{1}{1+x^2} \cdot \frac{d}{dx}(1+x^2) + \frac{1}{1+x^4} \cdot \frac{d}{dx}(1+x^4) $$ + \frac{1}{1+x^8} \cdot \frac{d}{dx}(1+x^8)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} \cdot (1) + \frac{1}{1+x^2} \cdot (2x) + \frac{1}{1+x^4} \cdot (4x^3) + \frac{1}{1+x^8} \cdot (8x^7)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$

Substitute the original expression for $y = f(x)$ back into the equation to get $f'(x)$:

$f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$

Now, we need to find $f'(1)$. Substitute $x=1$ into the expression for $f'(x)$:

$f'(1) = (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8) \left( \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} \right)$

$f'(1) = (2) (1 + 1) (1 + 1) (1 + 1) \left( \frac{1}{2} + \frac{2}{1+1} + \frac{4}{1+1} + \frac{8}{1+1} \right)$

$f'(1) = (2) (2) (2) (2) \left( \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right)$

$f'(1) = 16 \left( \frac{1}{2} + 1 + 2 + 4 \right)$

$f'(1) = 16 \left( \frac{1}{2} + 7 \right)$

$f'(1) = 16 \left( \frac{1}{2} + \frac{14}{2} \right)$

$f'(1) = 16 \left( \frac{1 + 14}{2} \right)$

$f'(1) = 16 \left( \frac{15}{2} \right)$

$f'(1) = \cancel{16}^{8} \cdot \frac{15}{\cancel{2}_{1}}$

$f'(1) = 8 \cdot 15$

$f'(1) = 120$

Final Answer:

The derivative of the function $f(x)$ is:

$f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$

The value of the derivative at $x=1$ is:

$f'(1) = 120$

Given:

The function to differentiate is $f(x) = (x^2 – 5x + 8) (x^3 + 7x + 9)$.

Solution - Method (i): Using Product Rule

Let $u(x) = x^2 - 5x + 8$ and $v(x) = x^3 + 7x + 9$.

Then $f(x) = u(x)v(x)$.

The derivative of $u(x)$ is $u'(x) = \frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$.

The derivative of $v(x)$ is $v'(x) = \frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$.

Using the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (2x - 5)(x^3 + 7x + 9) + (x^2 - 5x + 8)(3x^2 + 7)$

Expand the first term:

$(2x - 5)(x^3 + 7x + 9) = 2x(x^3 + 7x + 9) - 5(x^3 + 7x + 9)$

$= 2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45$

$= 2x^4 - 5x^3 + 14x^2 - 17x - 45$

Expand the second term:

$(x^2 - 5x + 8)(3x^2 + 7) = x^2(3x^2 + 7) - 5x(3x^2 + 7) + 8(3x^2 + 7)$

$= 3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56$

$= 3x^4 - 15x^3 + 31x^2 - 35x + 56$

Combine the expanded terms:

$f'(x) = (2x^4 - 5x^3 + 14x^2 - 17x - 45) + (3x^4 - 15x^3 + 31x^2 $$ - 35x + 56)$

$f'(x) = (2x^4 + 3x^4) + (-5x^3 - 15x^3) + (14x^2 + 31x^2) $$ + (-17x - 35x) $$ + (-45 + 56)$

$f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Solution - Method (ii): By Expanding the Product

First, expand the product $f(x) = (x^2 - 5x + 8)(x^3 + 7x + 9)$:

$f(x) = x^2(x^3 + 7x + 9) - 5x(x^3 + 7x + 9) + 8(x^3 + 7x + 9)$

$f(x) = (x^5 + 7x^3 + 9x^2) - (5x^4 + 35x^2 + 45x) + (8x^3 + 56x + 72)$

$f(x) = x^5 + 7x^3 + 9x^2 - 5x^4 - 35x^2 - 45x + 8x^3 + 56x + 72$

Combine like terms:

$f(x) = x^5 - 5x^4 + (7x^3 + 8x^3) + (9x^2 - 35x^2) + (-45x + 56x) + 72$

$f(x) = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72$

Now, differentiate this polynomial term by term:

$f'(x) = \frac{d}{dx}(x^5) - \frac{d}{dx}(5x^4) + \frac{d}{dx}(15x^3) - \frac{d}{dx}(26x^2) + \frac{d}{dx}(11x) + \frac{d}{dx}(72)$

$f'(x) = 5x^{5-1} - 5(4x^{4-1}) + 15(3x^{3-1}) - 26(2x^{2-1}) + 11(1) + 0$

$f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Solution - Method (iii): By Logarithmic Differentiation

Let $y = f(x) = (x^2 - 5x + 8)(x^3 + 7x + 9)$.

Take the natural logarithm of both sides:

$\log y = \log[(x^2 - 5x + 8)(x^3 + 7x + 9)]$

Using the logarithm property $\log(ab) = \log a + \log b$:

$\log y = \log(x^2 - 5x + 8) + \log(x^3 + 7x + 9)$

Differentiate both sides with respect to $x$. Use the chain rule $\frac{d}{dx}(\log w) = \frac{1}{w} \frac{dw}{dx}$.

$\frac{d}{dx}(\log y) = \frac{d}{dx}(\log(x^2 - 5x + 8)) + \frac{d}{dx}(\log(x^3 + 7x + 9))$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2 - 5x + 8} \cdot \frac{d}{dx}(x^2 - 5x + 8) + \frac{1}{x^3 + 7x + 9} \cdot \frac{d}{dx}(x^3 + 7x + 9)$

$\frac{1}{y} \frac{dy}{dx} = \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right)$

Substitute the original expression for $y$ back:

$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \left( \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right)$

Distribute the term $(x^2 - 5x + 8)(x^3 + 7x + 9)$ into the parenthesis:

$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \frac{2x - 5}{x^2 - 5x + 8} $$ + (x^2 - 5x + 8)(x^3 + 7x + 9) \frac{3x^2 + 7}{x^3 + 7x + 9}$

Cancel the common factors:

$\frac{dy}{dx} = (x^3 + 7x + 9) (2x - 5) + (x^2 - 5x + 8) (3x^2 + 7)$

This is the same expression obtained in Method (i) before combining terms. Expanding this leads to the same polynomial:

$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Comparison of Results:

From Method (i), we got $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

From Method (ii), we got $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

From Method (iii), we got $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

All three methods yield the same derivative for the function.

Final Answer:

Yes, all three methods give the same answer. The derivative is $5x^4 - 20x^3 + 45x^2 - 52x + 11$.

Given:

u, v, and w are functions of x.

We need to show that $\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$.

Solution - Method 1: By Repeated Application of Product Rule

Let $y = uvw$. We can treat $uv$ as a single function first. Let $z = uv$. Then $y = zw$.

Using the product rule $\frac{d}{dx}(zw) = \frac{dz}{dx} w + z \frac{dw}{dx}$:

Now, we need to find $\frac{d}{dx}(uv)$ using the product rule for two functions:

Substitute the expression for $\frac{d}{dx}(uv)$ from equation (2) into equation (1):

$\frac{dy}{dx} = \left(\frac{du}{dx} v + u \frac{dv}{dx}\right) w + uv \frac{dw}{dx}$

Distribute the term $w$ into the parentheses:

$\frac{dy}{dx} = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Thus, by repeated application of the product rule:

$\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Solution - Method 2: By Logarithmic Differentiation

Let $y = uvw$.

Take the natural logarithm on both sides:

Using the logarithm property $\log(abc) = \log a + \log b + \log c$ for the right side of equation (3):

Now, differentiate both sides of equation (4) with respect to $x$. Remember to use the chain rule for $\log u$, $\log v$, and $\log w$ since $u$, $v$, and $w$ are functions of $x$.

$\frac{d}{dx}(\log y) = \frac{d}{dx}(\log u) + \frac{d}{dx}(\log v) + \frac{d}{dx}(\log w)$

Using $\frac{d}{dx}(\log f(x)) = \frac{1}{f(x)} f'(x)$, we get:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}$

To solve for $\frac{dy}{dx}$, multiply both sides by $y$:

$\frac{dy}{dx} = y \left(\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}\right)$

Substitute $y = uvw$ back into the equation:

$\frac{dy}{dx} = (uvw) \left(\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}\right)$

Distribute the term $(uvw)$ into the parentheses:

$\frac{dy}{dx} = (uvw) \frac{1}{u} \frac{du}{dx} + (uvw) \frac{1}{v} \frac{dv}{dx} + (uvw) \frac{1}{w} \frac{dw}{dx}$

Cancel the common terms in each part:

$\frac{dy}{dx} = \cancel{u}vw \frac{1}{\cancel{u}} \frac{du}{dx} + u\cancel{v}w \frac{1}{\cancel{v}} \frac{dv}{dx} + uv\cancel{w} \frac{1}{\cancel{w}} \frac{dw}{dx}$

$\frac{dy}{dx} = vw \frac{du}{dx} + uw \frac{dv}{dx} + uv \frac{dw}{dx}$

This can be written in the desired form:

$\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Conclusion:

Both methods, repeated application of the product rule and logarithmic differentiation, yield the same formula for the derivative of the product of three functions:

$\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Given:

The parametric equations are $x = a \cos \theta$ and $y = a \sin \theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find the derivative of $x$ with respect to $\theta$:

Next, find the derivative of $y$ with respect to $\theta$:

Now, divide equation (ii) by equation (i):

Simplify the expression:

Using the trigonometric identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is $-\cot \theta$.

Given:

The parametric equations are $x = at^2$ and $y = 2at$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, we find the derivative of $x$ with respect to the parameter $t$:

Next, we find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$:

Simplify the expression by cancelling the common terms:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{1}{t}$

Given:

The parametric equations are $x = a (\theta + \sin \theta)$ and $y = a (1 – \cos \theta)$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, we find the derivative of $x$ with respect to the parameter $\theta$:

Next, we find the derivative of $y$ with respect to the parameter $\theta$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Cancel the common factor $a$:

Use the trigonometric half-angle identities: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$:

Simplify the expression:

Using the identity $\tan \phi = \frac{\sin \phi}{\cos \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \tan(\theta/2)$

Given:

The implicit equation is $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Differentiating both sides of equation (1) with respect to $x$:

$\frac{d}{dx}(x^{\frac{2}{3}} + y^{\frac{2}{3}}) = \frac{d}{dx}(a^{\frac{2}{3}})$

$\frac{d}{dx}(x^{\frac{2}{3}}) + \frac{d}{dx}(y^{\frac{2}{3}}) = \frac{d}{dx}(a^{\frac{2}{3}})$

For the term $\frac{d}{dx}(x^{\frac{2}{3}})$, use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

For the term $\frac{d}{dx}(y^{\frac{2}{3}})$, use the power rule and the chain rule (since $y$ is a function of $x$):

For the term $\frac{d}{dx}(a^{\frac{2}{3}})$, since $a$ is a constant, $a^{\frac{2}{3}}$ is also a constant:

Substitute these derivatives back into the differentiated equation:

Now, we solve equation (2) for $\frac{dy}{dx}$. Subtract $\frac{2}{3} x^{-\frac{1}{3}}$ from both sides:

Divide both sides by $\frac{2}{3} y^{-\frac{1}{3}}$:

Cancel the common factor $\frac{2}{3}$:

Using the property of exponents $\frac{a^{-n}}{b^{-m}} = \frac{b^m}{a^n}$:

This can also be written using radicals:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = - \left(\frac{y}{x}\right)^{\frac{1}{3}}$

or equivalently,

$\frac{dy}{dx} = - \sqrt[3]{\frac{y}{x}}$

Given:

The parametric equations are $x = 2at^2$ and $y = at^4$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling the common terms:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = t^2$

Given:

The parametric equations are $x = a \cos \theta$ and $y = b \cos \theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find the derivative of $x$ with respect to the parameter $\theta$:

Next, find the derivative of $y$ with respect to the parameter $\theta$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling the common terms:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{b}{a}$

Given:

The parametric equations are $x = \sin t$ and $y = \cos 2t$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Using the chain rule, let $u = 2t$. Then $\frac{d}{dt}(\cos u) = -\sin u \frac{du}{dt}$.

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Using the double angle identity $\sin(2t) = 2 \sin t \cos t$:

Cancel the common $\cos t$ term (assuming $\cos t \neq 0$):

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -4 \sin t$

Given:

The parametric equations are $x = 4t$ and $y = \frac{4}{t} = 4t^{-1}$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Simplify the expression:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\frac{1}{t^2}$

Given:

The parametric equations are $x = \cos \theta – \cos 2\theta$ and $y = \sin \theta – \sin 2\theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find the derivative of $x$ with respect to the parameter $\theta$:

Using the derivative rule $\frac{d}{d\theta}(\cos u) = -\sin u \frac{du}{d\theta}$:

$\frac{d}{d\theta}(\cos \theta) = -\sin \theta \cdot \frac{d}{d\theta}(\theta) = -\sin \theta \cdot 1 = -\sin \theta$

$\frac{d}{d\theta}(\cos 2\theta) = -\sin(2\theta) \cdot \frac{d}{d\theta}(2\theta) = -\sin(2\theta) \cdot 2 = -2\sin 2\theta$

Substitute these back into the expression for $\frac{dx}{d\theta}$:

Next, find the derivative of $y$ with respect to the parameter $\theta$:

Using the derivative rule $\frac{d}{d\theta}(\sin u) = \cos u \frac{du}{d\theta}$:

$\frac{d}{d\theta}(\sin \theta) = \cos \theta \cdot \frac{d}{d\theta}(\theta) = \cos \theta \cdot 1 = \cos \theta$

$\frac{d}{d\theta}(\sin 2\theta) = \cos(2\theta) \cdot \frac{d}{d\theta}(2\theta) = \cos(2\theta) \cdot 2 = 2\cos 2\theta$

Substitute these back into the expression for $\frac{dy}{d\theta}$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

The expression can be left in terms of $\theta$ as requested.

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{\cos \theta - 2\cos 2\theta}{2\sin 2\theta - \sin \theta}$

Given:

The parametric equations are $x = a (\theta – \sin \theta)$ and $y = a (1 + \cos \theta)$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find the derivative of $x$ with respect to the parameter $\theta$:

Next, find the derivative of $y$ with respect to the parameter $\theta$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Cancel the common factor $a$:

Use the trigonometric half-angle identities: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$:

Simplify the expression:

Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\cot(\theta/2)$

Given:

The parametric equations are $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

$x = \frac{\sin^3 t}{(\cos 2t)^{1/2}}$

Using the quotient rule $\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = \sin^3 t$ and $v = (\cos 2t)^{1/2}$.

$u' = \frac{d}{dt}(\sin^3 t) = 3\sin^2 t \cos t$

$v' = \frac{d}{dt}((\cos 2t)^{1/2}) = \frac{1}{2}(\cos 2t)^{-1/2} (-2\sin 2t) = -\frac{\sin 2t}{\sqrt{\cos 2t}}$

$\frac{dx}{dt} = \frac{(3\sin^2 t \cos t)(\cos 2t)^{1/2} - (\sin^3 t)(-\frac{\sin 2t}{\sqrt{\cos 2t}})}{(\sqrt{\cos 2t})^2}$

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t \sqrt{\cos 2t} + \frac{\sin^3 t \sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$

Multiply numerator and denominator by $\sqrt{\cos 2t}$ to simplify:

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t (\cos 2t) + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Simplify the numerator using $\sin 2t = 2\sin t \cos t$:

Numerator $= 3\sin^2 t \cos t \cos 2t + \sin^3 t (2\sin t \cos t)$

Numerator $= 3\sin^2 t \cos t \cos 2t + 2\sin^4 t \cos t$

Factor out $\sin^2 t \cos t$:

Numerator $= \sin^2 t \cos t (3\cos 2t + 2\sin^2 t)$

Using $\cos 2t = 1 - 2\sin^2 t$:

Numerator $= \sin^2 t \cos t (3(1 - 2\sin^2 t) + 2\sin^2 t)$

Numerator $= \sin^2 t \cos t (3 - 6\sin^2 t + 2\sin^2 t)$

Numerator $= \sin^2 t \cos t (3 - 4\sin^2 t)$

Next, find the derivative of $y$ with respect to the parameter $t$:

$y = \frac{\cos^3 t}{(\cos 2t)^{1/2}}$

Using the quotient rule, where $p = \cos^3 t$ and $q = (\cos 2t)^{1/2}$.

$p' = \frac{d}{dt}(\cos^3 t) = 3\cos^2 t (-\sin t) = -3\cos^2 t \sin t$

$q' = -\frac{\sin 2t}{\sqrt{\cos 2t}}$

$\frac{dy}{dt} = \frac{(-3\cos^2 t \sin t)(\cos 2t)^{1/2} - (\cos^3 t)(-\frac{\sin 2t}{\sqrt{\cos 2t}})}{(\sqrt{\cos 2t})^2}$

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t \sqrt{\cos 2t} + \frac{\cos^3 t \sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$

Multiply numerator and denominator by $\sqrt{\cos 2t}$ to simplify:

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t (\cos 2t) + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Simplify the numerator using $\sin 2t = 2\sin t \cos t$:

Numerator $= -3\cos^2 t \sin t \cos 2t + \cos^3 t (2\sin t \cos t)$

Numerator $= -3\cos^2 t \sin t \cos 2t + 2\cos^4 t \sin t$

Factor out $\cos^2 t \sin t$:

Numerator $= \cos^2 t \sin t (-3\cos 2t + 2\cos^2 t)$

Using $\cos 2t = 2\cos^2 t - 1$:

Numerator $= \cos^2 t \sin t (-3(2\cos^2 t - 1) + 2\cos^2 t)$

Numerator $= \cos^2 t \sin t (-6\cos^2 t + 3 + 2\cos^2 t)$

Numerator $= \cos^2 t \sin t (3 - 4\cos^2 t)$

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Cancel the common denominator $(\cos 2t)^{3/2}$:

Cancel common factors $\sin t$ and $\cos t$ (assuming $\sin t \neq 0$ and $\cos t \neq 0$):

Using the triple angle identities $\cos 3t = \cos t (4\cos^2 t - 3)$ and $\sin 3t = \sin t (3 - 4\sin^2 t)$:

$3 - 4\cos^2 t = -(4\cos^2 t - 3) = -\frac{\cos 3t}{\cos t}$

$3 - 4\sin^2 t = \frac{\sin 3t}{\sin t}$

Substitute these into the expression for $\frac{dy}{dx}$:

Cancel $\cos t$ from the numerator and $\sin t$ from the denominator:

Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\cot 3t$

Given:

The parametric equations are $x = a \left( \cos t + \log \tan \frac{t}{2} \right)$ and $y = a \sin t$, where $t$ is the parameter. We assume $\log$ denotes the natural logarithm $\log$.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

We know $\frac{d}{dt}(\cos t) = -\sin t$.

For the term $\frac{d}{dt}\left(\log \tan \frac{t}{2}\right)$, use the chain rule. Let $u = \tan \frac{t}{2}$.

$\frac{d}{dt}\left(\log u\right) = \frac{1}{u} \frac{du}{dt} = \frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{dt}\left(\tan \frac{t}{2}\right)$

Now, differentiate $\tan \frac{t}{2}$ using the chain rule. Let $v = \frac{t}{2}$.

$\frac{d}{dt}\left(\tan v\right) = \sec^2 v \frac{dv}{dt} = \sec^2 \frac{t}{2} \cdot \frac{d}{dt}\left(\frac{t}{2}\right)$

$\frac{d}{dt}\left(\frac{t}{2}\right) = \frac{1}{2}$

So, $\frac{d}{dt}\left(\tan \frac{t}{2}\right) = \sec^2 \frac{t}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{t}{2}$.

Substitute this back into the derivative of $\log \tan \frac{t}{2}$:

$\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}$

Rewrite in terms of $\sin$ and $\cos$: $\tan \frac{t}{2} = \frac{\sin(t/2)}{\cos(t/2)}$ and $\sec^2 \frac{t}{2} = \frac{1}{\cos^2(t/2)}$.

$\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{2} \cdot \frac{1}{\cos^2(t/2)} = \frac{1}{2 \sin(t/2) \cos(t/2)}$

Using the identity $\sin t = 2 \sin(t/2) \cos(t/2)$:

$\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{1}{\sin t} = \text{cosec } t$

Now substitute the derivatives of $\cos t$ and $\log \tan \frac{t}{2}$ back into the expression for $\frac{dx}{dt}$:

Rewrite $\text{cosec } t$ as $\frac{1}{\sin t}$:

Combine the terms inside the parenthesis:

Using the identity $1 - \sin^2 t = \cos^2 t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Simplify the expression:

Using the identity $\tan t = \frac{\sin t}{\cos t}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \tan t$

Given:

The parametric equations are $x = a \sec \theta$ and $y = b \tan \theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, find the derivative of $x$ with respect to the parameter $\theta$:

Next, find the derivative of $y$ with respect to the parameter $\theta$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling common terms and using identities $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

Using the identity $\frac{1}{\sin \theta} = \text{cosec } \theta$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{b}{a} \text{ cosec } \theta$

Given:

The parametric equations are $x = a (\cos \theta + \theta \sin \theta)$ and $y = a (\sin \theta – \theta \cos \theta)$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, we find the derivative of $x$ with respect to the parameter $\theta$:

The derivative of $\cos \theta$ is $-\sin \theta$.

To differentiate $\theta \sin \theta$, we use the product rule $\frac{d}{d\theta}(uv) = u'v + uv'$ with $u = \theta$ and $v = \sin \theta$:

$\frac{d}{d\theta}(\theta \sin \theta) = (1)(\sin \theta) + (\theta)(\cos \theta) = \sin \theta + \theta \cos \theta$

Substitute these derivatives back into the expression for $\frac{dx}{d\theta}$:

Next, we find the derivative of $y$ with respect to the parameter $\theta$:

The derivative of $\sin \theta$ is $\cos \theta$.

To differentiate $\theta \cos \theta$, use the product rule $\frac{d}{d\theta}(uv) = u'v + uv'$ with $u = \theta$ and $v = \cos \theta$:

$\frac{d}{d\theta}(\theta \cos \theta) = (1)(\cos \theta) + (\theta)(-\sin \theta) = \cos \theta - \theta \sin \theta$

Substitute these derivatives back into the expression for $\frac{dy}{d\theta}$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling the common terms $a$ and $\theta$ (assuming $a \neq 0$ and $\theta \neq 0$):

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \tan \theta$

Given:

The parametric equations are $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$, where $t$ is the parameter and $a$ is a constant (presumably $a > 0$ for the expressions to be defined). We need to find $\frac{dy}{dx}$ and show that $\frac{dy}{dx} = -\frac{y}{x}$.

To Show:

$\frac{dy}{dx} = -\frac{y}{x}$

Solution - Method 1: Using Parametric Differentiation

We are given $x = \sqrt{a^{\sin^{-1} t}} = (a^{\sin^{-1} t})^{1/2} = a^{\frac{1}{2}\sin^{-1} t}$.

To find $\frac{dx}{dt}$, we use the derivative rule $\frac{d}{du}(a^u) = a^u \log a \cdot \frac{du}{dt}$. Here $u = \frac{1}{2}\sin^{-1} t$.

Substituting back $x = a^{\frac{1}{2}\sin^{-1} t}$:

Similarly, for $y = \sqrt{a^{\cos^{-1} t}} = (a^{\cos^{-1} t})^{1/2} = a^{\frac{1}{2}\cos^{-1} t}$.

To find $\frac{dy}{dt}$, we use the derivative rule $\frac{d}{du}(a^u) = a^u \log a \cdot \frac{du}{dt}$. Here $u = \frac{1}{2}\cos^{-1} t$.

Substituting back $y = a^{\frac{1}{2}\cos^{-1} t}$:

Now, find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

Cancel the common factor $\frac{\log a}{2\sqrt{1-t^2}}$ (assuming $\log a \neq 0$ and $\sqrt{1-t^2} \neq 0$):

Solution - Method 2: Eliminating the Parameter (Implicit Differentiation)

We are given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.

Square both equations:

Multiply these two equations:

Using the property of exponents $a^m \cdot a^n = a^{m+n}$:

Using the inverse trigonometric identity $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$ (for $t \in [-1, 1]$):

Since $a$ is a constant, $a^{\pi/2}$ is also a constant. Let $C = a^{\pi/2}$.

Now, differentiate this implicit equation with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ on the left side, where $u=x^2$ and $v=y^2$. Remember that $y$ is a function of $x$, so use the chain rule for the term involving $y^2$.

$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(C)$

$\frac{d}{dx}(x^2) y^2 + x^2 \frac{d}{dx}(y^2) = 0$

$(2x) y^2 + x^2 (2y \frac{dy}{dx}) = 0$

$2xy^2 + 2x^2y \frac{dy}{dx} = 0$

Divide the entire equation by $2xy$ (assuming $x \neq 0$ and $y \neq 0$):

$\frac{2xy^2}{2xy} + \frac{2x^2y}{2xy} \frac{dy}{dx} = 0$

$y + x \frac{dy}{dx} = 0$

Subtract $y$ from both sides:

$x \frac{dy}{dx} = -y$

Divide both sides by $x$ (assuming $x \neq 0$):

Conclusion:

Both methods demonstrate that the derivative $\frac{dy}{dx}$ of the given parametric equations is equal to $-\frac{y}{x}$.

Given:

The function is $y = x^3 + \tan x$.

To Find:

We need to find the second derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$.

Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative with respect to $x$.

For the first term, $\frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x$ (using the constant multiple rule and power rule).

For the second term, $\frac{d}{dx}(\sec^2 x)$, use the chain rule. Let $u = \sec x$. Then $\sec^2 x = u^2$.

$\frac{d}{dx}(\sec^2 x) = \frac{d}{du}(u^2) \cdot \frac{du}{dx}$

Combine the derivatives of the terms:

Final Answer:

The second derivative of $y$ with respect to $x$ is:

$\frac{d^2y}{dx^2} = 6x + 2 \sec^2 x \tan x$

Given:

$y = A \sin x + B \cos x$, where $A$ and $B$ are constants.

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 of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{d}{dx}(A \sin x + B \cos x)$

Using the properties of differentiation (linearity):

$\frac{dy}{dx} = A \frac{d}{dx}(\sin x) + B \frac{d}{dx}(\cos x)$

We know that $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$.

Substituting these derivatives:

$\frac{dy}{dx} = A (\cos x) + B (-\sin x)$

$\frac{dy}{dx} = A \cos x - B \sin x$

Next, we find the second derivative of $y$ with respect to $x$, denoted as $\frac{d^2y}{dx^2}$, by differentiating the first derivative $\frac{dy}{dx}$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(A \cos x - B \sin x)$

Using the properties of differentiation:

$\frac{d^2y}{dx^2} = A \frac{d}{dx}(\cos x) - B \frac{d}{dx}(\sin x)$

Substituting the derivatives of $\cos x$ and $\sin x$ again:

$\frac{d^2y}{dx^2} = A (-\sin x) - B (\cos x)$

$\frac{d^2y}{dx^2} = -A \sin x - B \cos x$

Now, consider the expression $\frac{d^2y}{dx^2} + y$. Substitute the expressions we found for $\frac{d^2y}{dx^2}$ and the original expression for $y$:

$\frac{d^2y}{dx^2} + y = (-A \sin x - B \cos x) + (A \sin x + B \cos x)$

Rearranging the terms:

$\frac{d^2y}{dx^2} + y = (-A \sin x + A \sin x) + (-B \cos x + B \cos x)$

$\frac{d^2y}{dx^2} + y = 0 + 0$

$\frac{d^2y}{dx^2} + y = 0$

This shows that the given relation holds true.

Conclusion:

We have shown that if $y = A \sin x + B \cos x$, then $\frac{d^2y}{dx^2} + y = 0$.

Given:

$y = 3e^{2x} + 2e^{3x}$

To Prove:

$\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$

Proof:

We are given the function:

First, find the first derivative $\frac{dy}{dx}$ by differentiating equation (1) with respect to $x$:

Using the chain rule $\frac{d}{dx}(e^{kx}) = ke^{kx}$:

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (2) with respect to $x$:

Using the chain rule again:

Now, substitute the expressions for $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ from equations (1), (2), and (3) into the left side of the equation to be proven:

LHS = $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$

Distribute the constants $-5$ and $6$:

Group the terms with $e^{2x}$ and $e^{3x}$:

Combine the coefficients for each exponential term:

Since the Left Hand Side equals 0, which is the Right Hand Side of the equation to be proven, the statement is true.

Conclusion:

We have shown that if $y = 3e^{2x} + 2e^{3x}$, then $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$.

Given:

The function is $y = \sin^{-1} x$.

To Prove:

$(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0$

Proof:

We are given the function $y = \sin^{-1} x$.

First, find the first derivative $\frac{dy}{dx}$ with respect to $x$.

We can rewrite this using a negative exponent:

Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (1) with respect to $x$. Use the chain rule.

Let $u = 1 - x^2$. Then $\frac{d}{dx}(u^{-\frac{1}{2}}) = -\frac{1}{2} u^{-\frac{3}{2}} \frac{du}{dx}$.

$\frac{du}{dx} = \frac{d}{dx}(1 - x^2) = 0 - 2x = -2x$.

Substitute $u$ and $\frac{du}{dx}$ back:

Simplify the expression:

Rewrite using positive exponents:

Now, substitute the expressions for $\frac{dy}{dx}$ from equation (1) and $\frac{d^2y}{dx^2}$ from equation (2) into the left-hand side of the equation to be proven: $(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx}$.

In the first term, the $(1 - x^2)$ in the numerator and denominator cancel out (assuming $x^2 \neq 1$).

Since the Left Hand Side equals 0, which is the Right Hand Side of the equation to be proven, the statement is true.

Conclusion:

We have shown that if $y = \sin^{-1} x$, then $(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0$.

Given:

Let the given function be $y = x^2 + 3x + 2$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$.

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = 2$

Given:

Let the given function be $y = x^{20}$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$. Use the power rule again, along with the constant multiple rule $\frac{d}{dx}(cf(x)) = c \frac{d}{dx}(f(x))$.

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = 380 x^{18}$

Given:

Let the given function be $y = x \cos x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x$ and $v=\cos x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$.

The derivative of $\cos x$ is $-\sin x$.

To differentiate $x \sin x$, use the product rule again. Let $u=x$ and $v=\sin x$.

$\frac{d}{dx}(x \sin x) = \left(\frac{d}{dx} x\right)(\sin x) + (x)\left(\frac{d}{dx} \sin x\right) = (1)(\sin x) + (x)(\cos x) $$ = \sin x + x \cos x$.

Substitute these derivatives back:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -2\sin x - x \cos x$

Given:

Let the given function be $y = \log x$. We assume $\log x$ represents the natural logarithm, i.e., $y = \log x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y = \log x$ with respect to $x$.

We can write $\frac{1}{x}$ as $x^{-1}$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = x^{-1}$ with respect to $x$. Use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

Rewrite using a positive exponent:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -\frac{1}{x^2}$

Given:

Let the given function be $y = x^3 \log x$. We assume $\log x$ represents the natural logarithm, i.e., $y = x^3 \log x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x^3$ and $v=\log x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = 3x^2 \log x + x^2$ with respect to $x$. Use the sum rule and the product rule again for the term $3x^2 \log x$.

For the first term, $\frac{d}{dx}(3x^2 \log x)$, use the product rule with $u=3x^2$ and $v=\log x$.

$\frac{d}{dx}(3x^2 \log x) = \left(\frac{d}{dx} 3x^2\right) (\log x) + (3x^2) \left(\frac{d}{dx} \log x\right)$

$= (6x)(\log x) + (3x^2)\left(\frac{1}{x}\right)$

$= 6x \log x + 3x$

For the second term, $\frac{d}{dx}(x^2) = 2x$.

Substitute these derivatives back into the expression for $\frac{d^2y}{dx^2}$:

Factor out $x$:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = x(6 \log x + 5)$

Given:

Let the given function be $y = e^x \sin 5x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=e^x$ and $v=\sin 5x$.

The derivative of $u = e^x$ is $u' = \frac{d}{dx}(e^x) = e^x$.

The derivative of $v = \sin 5x$ requires the chain rule. Let $w=5x$, then $\frac{d}{dx}(\sin w) = \cos w \frac{dw}{dx} = \cos(5x) \cdot 5 = 5\cos 5x$. So $v' = 5\cos 5x$.

Factor out $e^x$:

$\frac{dy}{dx} = e^x (\sin 5x + 5 \cos 5x)$

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$. Use the product rule again for the expression $e^x (\sin 5x + 5 \cos 5x)$. Let $A=e^x$ and $B=(\sin 5x + 5 \cos 5x)$.

The derivative of $A = e^x$ is $A' = e^x$.

The derivative of $B = \sin 5x + 5 \cos 5x$ is:

$B' = \frac{d}{dx}(\sin 5x) + \frac{d}{dx}(5 \cos 5x)$

$B' = (5 \cos 5x) + 5(-\sin 5x \cdot 5)$

$B' = 5 \cos 5x - 25 \sin 5x$

Apply the product rule for $\frac{d^2y}{dx^2} = \frac{d}{dx}(AB) = A'B + AB'$:

Factor out $e^x$:

Combine like terms inside the bracket:

Rearrange the terms:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = e^x (10 \cos 5x - 24 \sin 5x)$

Given:

Let the given function be $y = e^{6x} \cos 3x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=e^{6x}$ and $v=\cos 3x$.

The derivative of $u = e^{6x}$ is $u' = \frac{d}{dx}(e^{6x})$. Using the chain rule, let $w=6x$. Then $\frac{d}{dx}(e^w) = e^w \frac{dw}{dx} = e^{6x} \cdot 6 = 6e^{6x}$. So $u' = 6e^{6x}$.

The derivative of $v = \cos 3x$ is $v' = \frac{d}{dx}(\cos 3x)$. Using the chain rule, let $z=3x$. Then $\frac{d}{dx}(\cos z) = -\sin z \frac{dz}{dx} = -\sin(3x) \cdot 3 = -3\sin 3x$. So $v' = -3\sin 3x$.

Using the product rule for $\frac{dy}{dx} = u'v + uv'$:

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$. Differentiate each term in equation (1) using the product rule and chain rule.

Differentiate the first term $6e^{6x} \cos 3x$. Use product rule with $A=6e^{6x}$ and $B=\cos 3x$. $A'=36e^{6x}$ and $B'=-3\sin 3x$.

$\frac{d}{dx}(6e^{6x} \cos 3x) = A'B + AB' = (36e^{6x})(\cos 3x) + (6e^{6x})(-3\sin 3x)$

$= 36e^{6x} \cos 3x - 18e^{6x} \sin 3x$

Differentiate the second term $3e^{6x} \sin 3x$. Use product rule with $C=3e^{6x}$ and $D=\sin 3x$. $C'=18e^{6x}$ and $D'=3\cos 3x$.

$\frac{d}{dx}(3e^{6x} \sin 3x) = C'D + CD' = (18e^{6x})(\sin 3x) + (3e^{6x})(3\cos 3x)$

$= 18e^{6x} \sin 3x + 9e^{6x} \cos 3x$

Substitute these results back into the expression for $\frac{d^2y}{dx^2}$:

Remove the parenthesis and combine like terms:

Factor out $e^{6x}$:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = e^{6x} (27 \cos 3x - 36 \sin 3x)$

Given:

Let the given function be $y = \tan^{-1} x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$.

We can rewrite the first derivative using a negative exponent for easier differentiation:

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = (1 + x^2)^{-1}$ with respect to $x$. Use the chain rule.

Let $u = 1 + x^2$. Then $\frac{d}{dx}(u^{-1}) = -1 \cdot u^{-1-1} \cdot \frac{du}{dx} = -u^{-2} \frac{du}{dx}$.

Find the derivative of $u$ with respect to $x$:

Substitute $u = 1 + x^2$ and $\frac{du}{dx} = 2x$ back into the chain rule expression:

Rearrange the terms:

Rewrite the expression using a positive exponent:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -\frac{2x}{(1 + x^2)^2}$

Given:

Let the given function be $y = \log (\log x)$. We assume $\log$ denotes the natural logarithm $\log$, so $y = \log (\log x)$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y = \log (\log x)$ with respect to $x$. Use the chain rule $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$. Let $u = \log x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = \frac{1}{x \log x}$ with respect to $x$. We can use either the quotient rule or rewrite the expression using negative exponents and use the product rule.

Method A: Using Quotient Rule

Rewrite $\frac{dy}{dx} = \frac{1}{x \log x}$. Let $p = 1$ and $q = x \log x$. The derivative is $\frac{d}{dx}\left(\frac{p}{q}\right) = \frac{p'q - pq'}{q^2}$.

$p' = \frac{d}{dx}(1) = 0$.

$q' = \frac{d}{dx}(x \log x)$. Use the product rule on $x \log x$: $(1)(\log x) + (x)(\frac{1}{x}) = \log x + 1$. So $q' = \log x + 1$.

Substitute into the quotient rule formula:

Method B: Using Product Rule (with negative exponent)

Rewrite $\frac{dy}{dx} = (x \log x)^{-1}$. Use the chain rule, with $u = x \log x$. The derivative is $\frac{d}{dx}(u^{-1}) = -1 \cdot u^{-2} \cdot \frac{du}{dx}$.

$\frac{du}{dx} = \frac{d}{dx}(x \log x) = \log x + 1$ (as calculated in Method A).

Substitute $u = x \log x$ and $\frac{du}{dx} = \log x + 1$ back into the chain rule expression:

Rewrite using positive exponents:

Both methods give the same result.

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x \log x)^2}$

Given:

Let the given function be $y = \sin(\log x)$. We assume $\log x$ represents the natural logarithm, i.e., $y = \sin(\log x)$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y = \sin(\log x)$ with respect to $x$. Use the chain rule.

Using the chain rule, $\frac{d}{dx}(\sin u) = \cos u \frac{du}{dx}$, where $u = \log x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = \frac{\cos(\log x)}{x}$ with respect to $x$. Use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

Let $u = \cos(\log x)$ and $v = x$.

Find the derivative of $u = \cos(\log x)$. Use the chain rule $\frac{d}{dx}(\cos w) = -\sin w \frac{dw}{dx}$, where $w = \log x$.

Find the derivative of $v = x$:

Apply the quotient rule:

Simplify the numerator:

Factor out the negative sign from the numerator:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -\frac{\sin(\log x) + \cos(\log x)}{x^2}$

Given:

The function is $y = 5 \cos x – 3 \sin x$.

To Prove:

$\frac{d^2y}{dx^2} + y = 0$

Proof:

We are given the function:

First, find the first derivative $\frac{dy}{dx}$ by differentiating equation (1) with respect to $x$:

We know that $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\sin x) = \cos x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (2) with respect to $x$:

Substituting the derivatives of $\sin x$ and $\cos x$ again:

Now, substitute the expressions for $\frac{d^2y}{dx^2}$ from equation (3) and $y$ from equation (1) into the left side of the equation to be proven:

LHS = $\frac{d^2y}{dx^2} + y$

Rearranging the terms:

Since the Left Hand Side equals 0, which is the Right Hand Side of the equation to be proven, the statement is true.

Conclusion:

We have shown that if $y = 5 \cos x – 3 \sin x$, then $\frac{d^2y}{dx^2} + y = 0$.

Given:

The function is $y = \cos^{-1} x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$ in terms of $y$ alone.

Solution:

We are given $y = \cos^{-1} x$. This implies that $x = \cos y$.

First, find the first derivative $\frac{dy}{dx}$ with respect to $x$.

Since we need the answer in terms of $y$, substitute $x = \cos y$ into the expression for $\frac{dy}{dx}$.

Using the identity $\sin^2 y + \cos^2 y = 1$, we have $\sqrt{1 - \cos^2 y} = \sqrt{\sin^2 y} = |\sin y|$. Assuming the principal value branch for $\cos^{-1} x$, $y \in [0, \pi]$, where $\sin y \geq 0$. Thus, $\sqrt{1 - \cos^2 y} = \sin y$.

Using the identity $\frac{1}{\sin y} = \text{cosec } y$:

Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = -\text{cosec } y$ with respect to $x$. Remember that $y$ is a function of $x$, so use the chain rule.

Using the chain rule $\frac{d}{dx}(\text{cosec } y) = \frac{d}{dy}(\text{cosec } y) \cdot \frac{dy}{dx}$.

The derivative of $\text{cosec } y$ with respect to $y$ is $-\text{cosec } y \cot y$.

Substitute the expression for $\frac{dy}{dx}$ from equation (1):

Simplify the expression:

This expression is in terms of $y$ alone.

Final Answer:

The second order derivative $\frac{d^2y}{dx^2}$ in terms of $y$ alone is:

$\frac{d^2y}{dx^2} = -\text{cosec}^2 y \cot y$

Given:

The function is $y = 3 \cos (\log x) + 4 \sin (\log x)$. We assume $\log x$ represents the natural logarithm, i.e., $y = 3 \cos (\log x) + 4 \sin (\log x)$.

We use the notation $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$.

To Prove:

$x^2 y_2 + xy_1 + y = 0$

Proof:

We are given the function:

First, find the first derivative $y_1 = \frac{dy}{dx}$ by differentiating equation (1) with respect to $x$. Use the chain rule.

For $\frac{d}{dx}(\cos (\log x))$, use chain rule: $\frac{d}{du}(\cos u) \frac{du}{dx}$ with $u = \log x$. $\frac{du}{dx} = \frac{1}{x}$.

$\frac{d}{dx}(\cos (\log x)) = -\sin(\log x) \cdot \frac{1}{x} = -\frac{\sin(\log x)}{x}$.

For $\frac{d}{dx}(\sin (\log x))$, use chain rule: $\frac{d}{du}(\sin u) \frac{du}{dx}$ with $u = \log x$. $\frac{du}{dx} = \frac{1}{x}$.

$\frac{d}{dx}(\sin (\log x)) = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x}$.

Substitute these derivatives back:

Multiply both sides of equation (2) by $x$ to simplify the next differentiation step:

Now, find the second derivative $y_2 = \frac{d^2y}{dx^2}$ by differentiating equation (3) with respect to $x$. Use the product rule on the left side and chain rule on the right side.

$\frac{d}{dx}\left(x \frac{dy}{dx}\right) = \frac{d}{dx}(-3 \sin(\log x) + 4 \cos(\log x))$

Left side (product rule): $\frac{d}{dx}\left(x y_1\right) = (1)y_1 + x\frac{dy_1}{dx} = y_1 + x y_2$.

Right side (linearity and chain rule):

$\frac{d}{dx}(-3 \sin(\log x)) = -3 \frac{d}{dx}(\sin(\log x)) = -3 \left(\cos(\log x) \cdot \frac{1}{x}\right) $$ = -\frac{3 \cos(\log x)}{x}$

$\frac{d}{dx}(4 \cos(\log x)) = 4 \frac{d}{dx}(\cos(\log x)) = 4 \left(-\sin(\log x) \cdot \frac{1}{x}\right) = -\frac{4 \sin(\log x)}{x}$

Equating the derivatives of both sides of equation (3):

Multiply both sides by $x$:

From equation (1), we know that $y = 3 \cos (\log x) + 4 \sin (\log x)$. Substitute this into the equation:

Rearrange the terms to match the required form:

This proves the required equation.

Conclusion:

We have shown that if $y = 3 \cos (\log x) + 4 \sin (\log x)$, then $x^2 y_2 + xy_1 + y = 0$.

Given:

$y = Ae^{mx} + Be^{nx}$, where A, B, m, and n are constants.

To Prove:

$\frac{d^2y}{dx^2} - (m + n) \frac{dy}{dx} + mny = 0$

Proof:

We are given the function:

First, find the first derivative $\frac{dy}{dx}$ by differentiating equation (1) with respect to $x$. Use the chain rule $\frac{d}{dx}(e^{kx}) = ke^{kx}$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (2) with respect to $x$:

Using the chain rule again:

Now, substitute the expressions for $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ from equations (1), (2), and (3) into the left side of the equation to be proven:

LHS = $\frac{d^2y}{dx^2} - (m + n) \frac{dy}{dx} + mny$

Expand the term $-(m + n)(Ame^{mx} + Bne^{nx})$:

$-(m + n)(Ame^{mx} + Bne^{nx}) = -(m(Ame^{mx} + Bne^{nx}) $$ + n(Ame^{mx} + Bne^{nx}))$

$= -(Am^2e^{mx} + Bmne^{nx} + Amne^{mx} + Bn^2e^{nx})$

$= -Am^2e^{mx} - Bmne^{nx} - Amne^{mx} - Bn^2e^{nx}$

Expand the term $mn(Ae^{mx} + Be^{nx})$:

$mn(Ae^{mx} + Be^{nx}) = Amne^{mx} + Bmne^{nx}$

Substitute these expanded terms back into the expression for LHS:

Group the terms with $e^{mx}$ and $e^{nx}$:

Combine the coefficients for each exponential term:

Since the Left Hand Side equals 0, which is the Right Hand Side of the equation to be proven, the statement is true.

Conclusion:

We have shown that if $y = Ae^{mx} + Be^{nx}$, then $\frac{d^2y}{dx^2} - (m + n) \frac{dy}{dx} + mny = 0$.

Given:

The function is $y = 500e^{7x} + 600e^{–7x}$.

To Prove:

$\frac{d^2y}{dx^2} = 49y$

Proof:

We are given the function:

First, find the first derivative $\frac{dy}{dx}$ by differentiating equation (1) with respect to $x$. Use the chain rule $\frac{d}{dx}(e^{kx}) = ke^{kx}$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (2) with respect to $x$:

Using the chain rule again:

Now consider the right-hand side of the equation to be proven, which is $49y$. Substitute the expression for $y$ from equation (1):

Distribute the constant 49:

Calculate the products:

So, the right-hand side becomes:

Comparing equation (3) and equation (4), we see that $\frac{d^2y}{dx^2}$ is equal to $49y$.

This proves the required equation.

Conclusion:

We have shown that if $y = 500e^{7x} + 600e^{–7x}$, then $\frac{d^2y}{dx^2} = 49y$.

Given:

The implicit equation is $e^y(x + 1) = 1$.

To Show:

$\frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2$

Proof:

We are given the equation:

To find the derivatives, it is convenient to isolate one variable. From equation (1), we can write $e^y = \frac{1}{x+1}$.

Taking the natural logarithm of both sides:

Using logarithm properties $\log(e^y) = y$ and $\log(1/a) = -\log a$:

Now, find the first derivative $\frac{dy}{dx}$ by differentiating equation (2) with respect to $x$. Use the chain rule $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$, where $u = x+1$.

Now find the square of the first derivative:

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = -\frac{1}{x+1} = -(x+1)^{-1}$ with respect to $x$. Use the chain rule.

Let $u = x+1$. Then $\frac{d}{dx}(u^{-1}) = -1 \cdot u^{-2} \cdot \frac{du}{dx}$.

Rewrite using a positive exponent:

Comparing equation (4) and equation (5), we see that $\frac{d^2y}{dx^2}$ is equal to $\left(\frac{dy}{dx}\right)^2$.

This proves the required equation.

Alternate Proof: Starting from $\frac{dy}{dx}$ in terms of $y$

From equation (1), $e^y(x+1) = 1$, we have $x+1 = e^{-y}$.

Differentiating this equation with respect to $x$:

$\frac{d}{dx}(x+1) = \frac{d}{dx}(e^{-y})$

$1 = e^{-y} \cdot \frac{d}{dx}(-y)$ (using chain rule)

$1 = e^{-y} \cdot (-1) \frac{dy}{dx}$

$1 = -e^{-y} \frac{dy}{dx}$

Solving for $\frac{dy}{dx}$:

Now find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (A) with respect to $x$. Use the chain rule, remembering $y$ is a function of $x$.

Substitute the expression for $\frac{dy}{dx}$ from equation (A):

Now, calculate the square of the first derivative $\left(\frac{dy}{dx}\right)^2$ using equation (A):

Comparing equation (B) and equation (C), we see that $\frac{d^2y}{dx^2}$ is equal to $\left(\frac{dy}{dx}\right)^2$.

This also proves the required equation.

Conclusion:

We have shown by two methods that if $e^y(x + 1) = 1$, then $\frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2$.

Given:

The function is $y = (\tan^{-1} x)^2$.

We use the notation $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$.

To Show:

$(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$

Proof:

We are given the function:

First, find the first derivative $y_1 = \frac{dy}{dx}$ by differentiating equation (1) with respect to $x$. Use the chain rule: $\frac{d}{dx}(u^n) = nu^{n-1} \frac{du}{dx}$, where $u = \tan^{-1} x$ and $n=2$. The derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$.

To simplify the next differentiation step, multiply both sides of equation (2) by $(1+x^2)$:

Now, find the second derivative $y_2 = \frac{d^2y}{dx^2}$ by differentiating equation (3) with respect to $x$. Use the product rule on the left side and the derivative of $\tan^{-1} x$ on the right side.

$\frac{d}{dx}[(1+x^2) y_1] = \frac{d}{dx}[2 \tan^{-1} x]$

Apply the product rule $\frac{d}{dx}(uv) = u'v + uv'$ on the left side, with $u = 1+x^2$ and $v = y_1$.

$\frac{d}{dx}(1+x^2) = 2x$ and $\frac{d}{dx}(y_1) = y_2$.

Multiply the entire equation by $(1+x^2)$ to eliminate the denominator:

Rearranging the terms, we get:

Since $1+x^2 = x^2+1$, this matches the equation we were asked to show.

Conclusion:

We have shown that if $y = (\tan^{-1} x)^2$, then $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$.

Given:

Function $y = f(x) = x^2 + 2$

Interval $[a, b] = [-2, 2]$

To Verify:

Rolle's Theorem for the given function in the given interval.

Solution:

For Rolle's Theorem to be applicable, the function must satisfy the following three conditions:

1. Continuity:

The function $f(x) = x^2 + 2$ is a polynomial function.

2. Differentiability:

Differentiating the function with respect to $x$:

Since $2x$ is a polynomial, it exists for all real numbers. Thus, $f(x)$ is differentiable on the open interval $(-2, 2)$.

3. Equality of function values at endpoints:

We evaluate $f(x)$ at $a = -2$ and $b = 2$.

Observing the values:

Since all three conditions of Rolle's Theorem are satisfied, there exists at least one $c \in (-2, 2)$ such that $f'(c) = 0$.

Using equation (i), we set $f'(c) = 0$:

Now we check if this value lies in the interval:

Thus, the value of $c$ is $0$.

Hence, Rolle's Theorem is verified.

Given:

Function $f(x) = x^2$

Interval $[a, b] = [2, 4]$

To Verify:

Mean Value Theorem (MVT) for the function on $[2, 4]$.

Solution:

The Mean Value Theorem states that if a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $c \in (a, b)$ such that:

Step 1: Check Validity of Hypotheses

Continuity: The function $f(x) = x^2$ is a polynomial function.

Differentiability: The derivative is $f'(x) = 2x$.

Step 2: Find the value of c

First, we calculate the values of the function at the endpoints:

Substituting these values into the MVT formula from equation (i):

We know that $f'(x) = 2x$, so $f'(c) = 2c$. Equating the values:

Step 3: Verification

We check if the calculated value $c$ lies within the open interval $(2, 4)$.

Thus, $c = 3$.

Hence, Mean Value Theorem is verified.

Given:

The function is $f(x) = x^2 + 2x – 8$.

The interval is $[a, b] = [-4, 2]$.

To Verify:

Rolle's Theorem for the function $f(x)$ on the interval $[-4, 2]$.

Solution (Verification of Conditions):

Rolle's Theorem requires three conditions to be satisfied:

1. Continuity on the closed interval $[a, b] = [-4, 2]$:

The function $f(x) = x^2 + 2x – 8$ is a polynomial function. Polynomial functions are continuous for all real numbers $\mathbb{R}$.

Therefore, $f(x)$ is continuous on the closed interval $[-4, 2]$.

2. Differentiability on the open interval $(a, b) = (-4, 2)$:

The derivative of $f(x)$ is: $f'(x) = \frac{d}{dx}(x^2 + 2x – 8) = 2x + 2$.

Since $f'(x)$ exists for all real numbers, $f(x)$ is differentiable on the open interval $(-4, 2)$.

3. $f(a) = f(b)$ (Value at endpoints):

We evaluate the function at the endpoints $a = -4$ and $b = 2$:

Conclusion of Rolle's Theorem:

Since all three conditions are satisfied, by Rolle's Theorem, there must exist at least one value $c \in (-4, 2)$ such that $f'(c) = 0$.

We set the derivative $f'(x) = 2x + 2$ to zero to find $c$:

The value $c = -1$ lies within the open interval $(-4, 2)$, since $-4 < -1 < 2$.

Thus, Rolle's Theorem is $\mathbf{verified}$.

Rolle’s theorem is applicable to $f(x)$ on $[a, b]$ if $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$.

Case (i): $f(x) = \lfloor x \rfloor$ for $x \in [5, 9]$

1. Continuity on $[5, 9]$: The greatest integer function $f(x) = \lfloor x \rfloor$ is discontinuous at all integer points. Since the interval $[5, 9]$ contains integers $6, 7, 8$, $f(x)$ is not continuous on $[5, 9]$.

2. Differentiability on $(5, 9)$: $f(x) = \lfloor x \rfloor$ is not differentiable at any integer point. Since the open interval $(5, 9)$ contains integers $6, 7, 8$, $f(x)$ is not differentiable on $(5, 9)$.

3. $f(5) = f(9)$: $f(5) = \lfloor 5 \rfloor = 5$. $f(9) = \lfloor 9 \rfloor = 9$. Thus, $f(5) \neq f(9)$.

Rolle's theorem is $\mathbf{not\ applicable}$ to $f(x) = \lfloor x \rfloor$ on $[5, 9]$.

Case (ii): $f(x) = \lfloor x \rfloor$ for $x \in [– 2, 2]$

1. Continuity on $[-2, 2]$: The greatest integer function $f(x) = \lfloor x \rfloor$ is discontinuous at integer points. Since the interval $[-2, 2]$ contains integers $-1, 0, 1$, $f(x)$ is not continuous on $[-2, 2]$.

2. Differentiability on $(-2, 2)$: $f(x) = \lfloor x \rfloor$ is not differentiable at any integer point. Since the open interval $(-2, 2)$ contains integers $-1, 0, 1$, $f(x)$ is not differentiable on $(-2, 2)$.

3. $f(-2) = f(2)$: $f(-2) = \lfloor -2 \rfloor = -2$. $f(2) = \lfloor 2 \rfloor = 2$. Thus, $f(-2) \neq f(2)$.

Rolle's theorem is $\mathbf{not\ applicable}$ to $f(x) = \lfloor x \rfloor$ on $[-2, 2]$.

Case (iii): $f(x) = x^2 – 1$ for $x \in [1, 2]$

1. Continuity on $[1, 2]$: $f(x)$ is a polynomial function, so it is continuous on $[1, 2]$.

2. Differentiability on $(1, 2)$: $f'(x) = 2x$, which exists for all $x$. So, $f(x)$ is differentiable on $(1, 2)$.

3. $f(1) = f(2)$: $f(1) = (1)^2 - 1 = 0$. $f(2) = (2)^2 - 1 = 3$. Thus, $f(1) \neq f(2)$.

Rolle's theorem is $\mathbf{not\ applicable}$ to $f(x) = x^2 - 1$ on $[1, 2]$.

Comment on the Converse of Rolle’s Theorem:

The converse of Rolle's theorem is: "If there exists a point $c \in (a, b)$ such that $f'(c) = 0$, then $f(x)$ must satisfy the three conditions (Continuity, Differentiability, $f(a)=f(b)$)."

Consider $f(x) = \lfloor x \rfloor$ on $x \in [5, 9]$. For any non-integer $x$, $f'(x) = 0$. Since there are many non-integer points in $(5, 9)$ (e.g., $c=5.5$), the conclusion $f'(c)=0$ is satisfied.

However, as shown above, $f(x) = \lfloor x \rfloor$ on $[5, 9]$ does $\mathbf{not}$ satisfy any of the conditions of Rolle's theorem.

Since the existence of $f'(c)=0$ does not imply the conditions of Rolle's theorem are met, the $\mathbf{converse\ of\ Rolle's\ theorem\ is\ not\ true}$.

Given:

1. $f : [-5, 5] \to \mathbb{R}$ is a differentiable function.

2. $f'(x)$ does not vanish anywhere, i.e., $f'(x) \neq 0$ for all $x \in [-5, 5]$.

To Prove:

$f(-5) \neq f(5)$.

Proof (by Contradiction using Rolle's Theorem):

We prove this by assuming the opposite of the conclusion and showing that it leads to a contradiction with the given information.

Assume that $\mathbf{f(-5) = f(5)}$.

Now, we check the conditions of Rolle's Theorem for $f(x)$ on the interval $[-5, 5]$:

1. Continuity on $[-5, 5]$: Since $f(x)$ is differentiable on $[-5, 5]$, it must be continuous on $[-5, 5]$.

2. Differentiability on $(-5, 5)$: It is given that $f(x)$ is differentiable on $[-5, 5]$, and thus on $(-5, 5)$.

3. $f(-5) = f(5)$: This is our assumption for the proof by contradiction.

Since all three conditions of Rolle's Theorem are satisfied, the theorem guarantees that there exists at least one value $c$ in the open interval $(-5, 5)$ such that:

This conclusion ($\mathbf{f'(c) = 0}$ for some $c$) directly $\mathbf{contradicts}$ the given statement that $f'(x)$ does not vanish anywhere, i.e., $f'(x) \neq 0$ for all $x$.

Since our assumption leads to a contradiction, the assumption must be false.

Therefore, the negation of the assumption is true.

Conclusion:

If $f : [– 5, 5] \to \mathbb{R}$ is a differentiable function and $f'(x)$ does not vanish anywhere, then $f(-5) \neq f(5)$. (Hence Proved).

Given:

The function is $f(x) = x^2 – 4x – 3$.

The interval is $[a, b] = [1, 4]$.

To Verify:

Mean Value Theorem (MVT) for the function $f(x)$ on the interval $[1, 4]$.

Solution (Verification of Conditions):

MVT requires two conditions to be satisfied:

1. Continuity on the closed interval $[1, 4]$:

The function $f(x) = x^2 – 4x – 3$ is a polynomial function, which is continuous for all real numbers.

Therefore, $f(x)$ is continuous on the closed interval $[1, 4]$.

2. Differentiability on the open interval $(1, 4)$:

The derivative of $f(x)$ is: $f'(x) = \frac{d}{dx}(x^2 – 4x – 3) = 2x - 4$.

Since $f'(x)$ exists for all real numbers, $f(x)$ is differentiable on the open interval $(1, 4)$.

Finding the value of $c$:

Since the conditions of MVT are satisfied, there exists $c \in (1, 4)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

First, calculate the value of the slope $\frac{f(b) - f(a)}{b - a}$:

Next, set $f'(c)$ equal to this slope and solve for $c$:

Since $c = 2.5$ lies within the open interval $(1, 4)$, the conclusion of the MVT is verified.

Conclusion:

The Mean Value Theorem is $\mathbf{verified}$ for $f(x) = x^2 – 4x – 3$ on $[1, 4]$ with $c = \frac{5}{2} \in (1, 4)$ such that $f'\left(\frac{5}{2}\right) = 1 = \frac{f(4) - f(1)}{4 - 1}$.

Given:

The function is $f(x) = x^3 – 5x^2 – 3x$.

The interval is $[a, b] = [1, 3]$.

To Verify:

Mean Value Theorem (MVT) for the function $f(x)$ on the interval $[1, 3]$.

To Find:

All values $c \in (1, 3)$ for which $f'(c) = 0$.

Solution (Verification of MVT):

1. Continuity on $[1, 3]$: $f(x)$ is a polynomial function, so it is continuous on $[1, 3]$.

2. Differentiability on $(1, 3)$: The derivative is $f'(x) = 3x^2 - 10x - 3$. This exists for all $x$. So, $f(x)$ is differentiable on $(1, 3)$.

The MVT guarantees the existence of $c \in (1, 3)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Calculate $\frac{f(b) - f(a)}{b - a}$:

Set $f'(c) = -10$ and solve for $c$:

Factor the quadratic equation:

$3c^2 - 3c - 7c + 7 = 0$

$3c(c - 1) - 7(c - 1) = 0$

$(3c - 7)(c - 1) = 0$

Possible values for $c$ are $c = \frac{7}{3}$ and $c = 1$.

We check if $c$ is in the open interval $(1, 3)$.

$c = \frac{7}{3} = 2.33\overline{3}$. Since $1 < \frac{7}{3} < 3$, this value is in the interval.

$c = 1$ is an endpoint and not in the open interval $(1, 3)$.

Thus, $c = \frac{7}{3}$ verifies the Mean Value Theorem.

Finding $c \in (1, 3)$ for which $f'(c) = 0$:

We set $f'(c) = 0$:

Using the quadratic formula, $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

The two roots are:

$c_1 = \frac{5 + \sqrt{34}}{3}$ and $c_2 = \frac{5 - \sqrt{34}}{3}$.

Since $\sqrt{34} \approx 5.83$:

$c_1 \approx \frac{5 + 5.83}{3} \approx 3.61$. This is $\mathbf{not}$ in $(1, 3)$.

$c_2 \approx \frac{5 - 5.83}{3} \approx -0.27$. This is $\mathbf{not}$ in $(1, 3)$.

Therefore, there are $\mathbf{no\ values}$ of $c$ in the open interval $(1, 3)$ for which $f'(c) = 0$.

Conclusion:

The Mean Value Theorem is $\mathbf{verified}$ for $f(x) = x^3 – 5x^2 – 3x$ on $[1, 3]$ with $c = \frac{7}{3}$. There are $\mathbf{no\ values}$ of $c \in (1, 3)$ for which $f'(c) = 0$.

The Mean Value Theorem (MVT) is applicable to $f(x)$ on $[a, b]$ if and only if:

1. $f(x)$ is continuous on the closed interval $[a, b]$.

2. $f(x)$ is differentiable on the open interval $(a, b)$.

Case (i): $f(x) = \lfloor x \rfloor$ for $x \in [5, 9]$

1. Continuity on $[5, 9]$: The function $f(x) = \lfloor x \rfloor$ is discontinuous at integers $6, 7, 8$ within the interval.

2. Differentiability on $(5, 9)$: Since $f(x)$ is discontinuous at integer points in $(5, 9)$, it is not differentiable throughout the interval.