Applied Mathematics for Class 11th & 12th (Concepts and Questions) | ||
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11th | Concepts | Questions |
12th | Concepts | Questions |
Chapter 8 Calculus (Concepts)
Welcome to this essential introduction to the fundamental concepts of Calculus, a branch of mathematics specifically designed to analyze change and accumulation. In the realm of Applied Mathematics, calculus provides indispensable tools for modeling dynamic systems, optimizing processes, and understanding rates of change in diverse fields like economics, physics, engineering, and biology. This chapter will guide you through the essential building blocks: the concept of Limits, the power of Derivatives to describe instantaneous change, and the fundamentals of Integration for accumulating quantities. While we will touch upon the theoretical underpinnings, our primary focus will be on developing an intuitive understanding and showcasing the practical applications relevant to solving real-world problems, making calculus a versatile instrument in your quantitative toolkit. We begin with the foundational idea of a Limit, describing how a function $f(x)$ behaves as $x$ approaches a point 'a' ($\lim\limits_{x \to a} f(x) = L$). This idea underpins calculus. We'll cover evaluation techniques like factorization or rationalization for indeterminate forms like $\frac{0}{0}$ and introduce standard limits like $\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$ and $\lim\limits_{x \to 0} \frac{\log(1+x)}{x} = 1$, relevant for growth and financial models.
Next, we introduce the Derivative, $f'(x)$, representing the instantaneous rate of change of a function or the slope of its tangent line. While defined using the first principle, $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$, our practical focus will be on mastering the rules of differentiation: the power rule, sum/difference, product, quotient, and the vital chain rule. We will apply these to differentiate key functions including polynomials, exponential functions ($e^x, a^x$), and logarithmic functions ($\ln x, \log_a x$). A key application in economics is the interpretation of derivatives as marginal quantities. For example, Marginal Cost ($C'(x)$) and Marginal Revenue ($R'(x)$) represent the rate of change in total cost or revenue per additional unit produced or sold, perhaps expressed in $\textsf{₹}$ per item. Understanding these marginal concepts is crucial for making optimal business decisions based on incremental changes.
Finally, we touch upon the fundamentals of Integration, viewed primarily as the inverse operation of differentiation – finding the anti-derivative. We will cover basic integration formulas necessary for reversing the differentiation process for common functions like polynomials and exponential functions, such as $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), $\int e^x dx = e^x + C$, and $\int \frac{1}{x} dx = \ln|x| + C$. We may also briefly introduce the concept of the definite integral, $\int\limits_{a}^{b} f(x) dx$, representing the accumulation of a quantity (like area under a curve) between specified limits $a$ and $b$. This chapter aims to provide a solid grounding in these essential calculus tools, preparing you for more advanced quantitative analysis and modeling tasks encountered throughout Applied Mathematics and related disciplines.
Functions
In mathematics, a Function is a fundamental concept used to describe a relationship between two sets. Specifically, it is a special type of relation where every element from the first set (the input) is associated with exactly one element from the second set (the output). Functions are essential tools for modelling real-world phenomena, representing dependencies between variables, and are the cornerstone of calculus.
Definition of a Function
A function, denoted by $f$, from a set $A$ to a set $B$, is a rule or a mapping that assigns to each element $x$ in set $A$ exactly one element $y$ in set $B$.
We write this relationship as $f: A \to B$, which is read as "$f$ is a function from $A$ to $B$".
- The set $A$ is called the Domain of the function $f$. It is the set of all possible input values for which the function is defined.
- The set $B$ is called the Codomain of the function $f$. It is the set of all possible output values that the function *could* map to.
- For an element $x \in A$, the unique element $y \in B$ that the function $f$ assigns to $x$ is denoted by $f(x)$. This element $f(x)$ is called the image of $x$ under $f$, or the value of the function at $x$.
- The set of all images of the elements of $A$ under $f$ is called the Range of the function $f$. The Range is always a subset of the Codomain ($Range \subseteq B$). We will discuss Domain and Range in more detail in the next section.
Illustrative Examples of Functions:
- Consider the function $f(x) = x^2$ where the domain and codomain are both the set of real numbers, $\mathbb{R}$. So, $f: \mathbb{R} \to \mathbb{R}$. This function assigns to each real number its square. For example, $f(2) = 2^2 = 4$, and $f(-3) = (-3)^2 = 9$. Notice that for each input ($2$ or $-3$), there is only one unique output ($4$ or $9$). Even though both $2$ and $-2$ map to the same output $4$, this is allowed in a function; the key is that each *input* has only one output.
- Let $A$ be the set of students in a class, and $B$ be the set of their unique roll numbers. The relationship that assigns each student in set $A$ to their roll number in set $B$ is a function, provided each student has exactly one roll number.
Examples of Relations that are NOT Functions:
- Let $A$ be the set of students in a class, and $B$ be the set of subjects taught in the school. The relation that links each student to the subjects they study is generally NOT a function. This is because a single student (an element in set $A$) typically studies multiple subjects (multiple elements in set $B$), violating the "exactly one output" rule for functions.
- Consider the relation defined by the equation $y^2 = x$ where $x, y \in \mathbb{R}$. Let's take an input value, say $x=4$. The equation becomes $y^2 = 4$, which means $y$ can be either $2$ or $-2$. Since the input $x=4$ is associated with two different output values ($y=2$ and $y=-2$), this relation does not satisfy the condition of a function.
Notation for Functions
Functions are typically represented by symbols such as $f, g, h$, or other letters. The notation $f(x)$ is the most common way to denote the output of a function $f$ for a given input $x$. It is read as "$f$ of $x$" or "$f$ at $x$".
A function can be described in various ways:
- By Formula/Equation: e.g., $f(x) = 2x+1$.
- By Graph: A visual representation in the coordinate plane.
- By Table of Values: A list of input-output pairs.
- By Description: A rule explained in words.
Example 1. Let a function $f$ be defined by the formula $f(x) = 2x + 3$. Find the value of the function at $x=5$.
Answer:
The given function is $f(x) = 2x + 3$.
To find the value of $f$ at $x=5$, we substitute $x=5$ into the formula:
$f(5) = 2(5) + 3$
$f(5) = 10 + 3$
$f(5) = 13$
Thus, the value of the function $f$ at $x=5$ is $13$. The image of $5$ under $f$ is $13$.
Vertical Line Test (Graphical Representation)
For a relation represented by a graph in the Cartesian coordinate system, we can visually determine if it is a function using the Vertical Line Test.
A graph represents a function if and only if no vertical line intersects the graph at more than one point.
Explanation: A vertical line corresponds to a single input value ($x$). If a vertical line intersects the graph at more than one point, it means that a single input value ($x$) is associated with multiple output values ($y$), which violates the definition of a function.
Domain and Range of a function
As introduced in the definition of a function, the Domain and Range are fundamental concepts that precisely define the set of permissible inputs and the set of actual outputs of a function, respectively. Understanding these sets is crucial for analysing the behaviour of a function.
Domain of a Function
The Domain of a function $f: A \to B$ is the set of all possible input values for which the function is defined and produces a meaningful output. If the function is explicitly defined as $f: A \to B$, then the domain is the set $A$.
However, when a function is given by a formula, particularly in calculus where we often deal with real-valued functions of a real variable (i.e., $f: \mathbb{R} \to \mathbb{R}$ or a subset thereof), the domain is generally not stated explicitly. In such cases, we determine the domain as the largest possible subset of the set of real numbers $\mathbb{R}$ for which the formula yields a real number output. This is often referred to as the natural domain or the implied domain.
Common Restrictions on the Domain (for real-valued functions):
When determining the natural domain of a function defined by a formula, we must identify values of the input variable that would make the expression undefined in the real number system. The most common restrictions arise from:
1. Division by Zero: The denominator of a fraction cannot be zero. If a function is of the form $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are expressions involving $x$, the domain excludes all values of $x$ for which $Q(x) = 0$.
2. Even Roots of Negative Numbers: We cannot take the square root, fourth root, sixth root, etc. (any even root) of a negative number within the real number system. If a function involves $\sqrt[n]{g(x)}$ where $n$ is a positive even integer, the domain is restricted to values of $x$ for which $g(x) \geq 0$.
3. Logarithms of Non-positive Numbers: The argument of a logarithm function must be strictly positive. If a function is of the form $f(x) = \log_b(g(x))$ (where $b > 0$ and $b \neq 1$), the domain is restricted to values of $x$ for which $g(x) > 0$.
Combinations of these restrictions must also be considered for more complex functions.
Example 1. Find the domain of the function $f(x) = \frac{1}{x-3}$.
Answer:
The function $f(x) = \frac{1}{x-3}$ involves a fraction. For the function to be defined, the denominator cannot be equal to zero.
We must have $x - 3 \neq 0$.
Solving for $x$: $x \neq 3$.
Thus, the domain of the function $f$ is the set of all real numbers except $3$.
In set-builder notation, the domain is $\{x \in \mathbb{R} : x \neq 3\}$.
In interval notation, the domain is $(-\infty, 3) \cup (3, \infty)$.
Example 2. Find the domain of the function $g(x) = \sqrt{x+2}$.
Answer:
The function $g(x) = \sqrt{x+2}$ involves a square root (an even root). For the function to yield a real number, the expression under the square root must be greater than or equal to zero.
We must have $x + 2 \geq 0$.
Solving for $x$: $x \geq -2$.
Thus, the domain of the function $g$ is the set of all real numbers greater than or equal to $-2$.
In set-builder notation, the domain is $\{x \in \mathbb{R} : x \geq -2\}$.
In interval notation, the domain is $[-2, \infty)$.
Range of a Function
The Range of a function $f: A \to B$ is the set of all actual output values ($y$ values) that the function produces when taking inputs ($x$ values) from its domain. It is the collection of all images of the elements of the domain under the function $f$.
Formally, Range$(f) = \{y \in B : y = f(x) \text{ for some } x \in A\}$.
The Range is always a subset of the Codomain ($Range \subseteq B$). While the codomain is the set of all *possible* outputs, the range is the set of all *actual* outputs.
Methods for Finding the Range:
Determining the range of a function can be more complex than finding the domain and often requires careful analysis of the function's behaviour. Some common approaches include:
1. Analysing the Function's Behaviour: Based on the properties of the specific type of function (e.g., quadratic, linear, rational, exponential), we can determine the possible set of output values. For instance, the output of $f(x) = x^2$ is always non-negative.
2. Using the Graph: If the graph of the function is available, the range is the set of all $y$-values that are included in the graph. This can be found by projecting the graph onto the y-axis.
3. Algebraic Method: For a function $y = f(x)$, we can try to solve for $x$ in terms of $y$. If we can express $x$ as $x = g(y)$, then the range of $f$ is the domain of $g$ (considering the constraints imposed by the original function $f$). This method is particularly useful for simpler functions like linear or rational functions.
Example 3. Find the range of the function $f(x) = x^2$. Assume the domain is $\mathbb{R}$.
Answer:
The function is $f(x) = x^2$, and the domain is all real numbers ($\mathbb{R}$).
We know that the square of any real number ($x$) is always non-negative, i.e., $x^2 \geq 0$ for all $x \in \mathbb{R}$.
So, the output values $f(x)$ must be greater than or equal to $0$. This means the range must be a subset of $[0, \infty)$.
To show that the range is exactly $[0, \infty)$, we need to confirm that *any* non-negative number $y \geq 0$ can be an output of the function. If $y \geq 0$, we want to find an $x$ such that $f(x) = y$, i.e., $x^2 = y$. This equation has real solutions $x = \sqrt{y}$ and $x = -\sqrt{y}$ for any $y \geq 0$. Since these $x$ values are real numbers, they are in the domain $\mathbb{R}$.
Thus, for any $y \geq 0$, there exists an $x$ in the domain such that $f(x) = y$.
The range of $f(x) = x^2$ is the set of all non-negative real numbers.
In set-builder notation, the range is $\{y \in \mathbb{R} : y \geq 0\}$.
In interval notation, the range is $[0, \infty)$.
Example 4. Find the range of the function $g(x) = \frac{1}{x-3}$. (The domain is $\{x \in \mathbb{R} : x \neq 3\}$).
Answer:
The function is $g(x) = \frac{1}{x-3}$. Let $y$ be an output value, so $y = g(x)$. We want to find the set of all possible values for $y$.
Consider the equation $y = \frac{1}{x-3}$. We observe that for any value of $x$ in the domain ($x \neq 3$), the denominator $x-3$ will never be zero. This implies that the fraction $\frac{1}{x-3}$ can never be equal to zero.
So, $y$ cannot be $0$. This means $0$ is not in the range.
Now, let's use the algebraic method to see what other values $y$ can take. Assuming $y \neq 0$, we try to solve for $x$ in terms of $y$:
$y = \frac{1}{x-3}$
Multiply both sides by $(x-3)$ (which is non-zero since $x \neq 3$):
$y(x-3) = 1$
Distribute $y$:
$yx - 3y = 1$
Add $3y$ to both sides:
$yx = 1 + 3y$
Since we already established $y \neq 0$, we can divide by $y$:
$x = \frac{1 + 3y}{y}$
This expression for $x$ is defined for all real values of $y$ except $y=0$. For any real number $y$ where $y \neq 0$, we can find a corresponding real number $x$ given by $x = \frac{1+3y}{y}$.
We must also check if this value of $x$ is in the domain of $g(x)$, which requires $x \neq 3$. Let's see if $x = \frac{1+3y}{y}$ can ever be equal to $3$ (assuming $y \neq 0$):
$\frac{1+3y}{y} = 3$
Multiply both sides by $y$:
$1 + 3y = 3y$
Subtract $3y$ from both sides:
$1 = 0$
This result is a contradiction, which means that $\frac{1+3y}{y}$ can never be equal to $3$ for any value of $y$. Therefore, for any $y \neq 0$, the corresponding $x$ value we found will always be in the domain of $g(x)$.
Thus, the set of all possible output values $y$ is all real numbers except $0$.
In set-builder notation, the range is $\{y \in \mathbb{R} : y \neq 0\}$.
In interval notation, the range is $(-\infty, 0) \cup (0, \infty)$.
Types of Functions
Functions are fundamental mathematical objects, and they can be classified into different types based on how elements from the domain are mapped to elements in the codomain, as well as based on the form of their defining expression. Understanding these classifications helps in analysing function properties and behaviour.
Classification Based on Mapping Properties
Let $f: A \to B$ be a function, where $A$ is the domain and $B$ is the codomain.
1. One-to-One Function (Injective Function or Injection)
A function $f: A \to B$ is called a one-to-one function or an injection if distinct elements in the domain $A$ are mapped to distinct images in the codomain $B$. In simpler terms, no two different elements in the domain have the same image in the codomain.
Mathematically, $f$ is one-to-one if for any two elements $x_1, x_2 \in A$:
If $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
If $f(x_1) = f(x_2)$, then $x_1 = x_2$.
Example: Consider the function $f(x) = 2x$ from $\mathbb{R}$ to $\mathbb{R}$. Let $f(x_1) = f(x_2)$. Then $2x_1 = 2x_2$, which implies $x_1 = x_2$. Since $f(x_1) = f(x_2)$ only if $x_1 = x_2$, the function $f(x) = 2x$ is one-to-one.
Example: Consider the function $f(x) = x^2$ from $\mathbb{R}$ to $\mathbb{R}$. This function is NOT one-to-one because, for example, $f(2) = 2^2 = 4$ and $f(-2) = (-2)^2 = 4$. Here, $2 \neq -2$, but $f(2) = f(-2)$. Distinct inputs (2 and -2) have the same output (4).
2. Onto Function (Surjective Function or Surjection)
A function $f: A \to B$ is called an onto function or a surjection if every element in the codomain $B$ is the image of at least one element in the domain $A$. In other words, the range of the function is equal to its codomain.
Mathematically, $f$ is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.
Equivalently, Range$(f) = B$.
Example: Consider the function $f(x) = x+1$ from $\mathbb{R}$ to $\mathbb{R}$. Let $y$ be any element in the codomain $\mathbb{R}$. We need to check if there exists an $x \in \mathbb{R}$ (domain) such that $f(x) = y$. Setting $x+1 = y$, we solve for $x$: $x = y-1$. Since $y$ is a real number, $y-1$ is also a real number, meaning $x$ is in the domain $\mathbb{R}$. Thus, for every $y$ in the codomain, there is an $x$ in the domain such that $f(x) = y$. The function $f(x) = x+1$ is onto $\mathbb{R}$.
Example: Consider the function $f(x) = x^2$ from $\mathbb{R}$ to $\mathbb{R}$. The codomain is $\mathbb{R}$. The range of this function is $[0, \infty)$ (all non-negative real numbers), as the square of any real number is always non-negative. Since the range $[0, \infty)$ is a proper subset of the codomain $\mathbb{R}$ (e.g., $-5 \in \mathbb{R}$ but $-5$ is not in the range), the function $f(x) = x^2: \mathbb{R} \to \mathbb{R}$ is NOT onto.
However, if we define the function as $g(x) = x^2$ from $\mathbb{R}$ to $[0, \infty)$, then its codomain is $[0, \infty)$. Since the range of $g(x) = x^2$ is $[0, \infty)$, which equals this new codomain, the function $g: \mathbb{R} \to [0, \infty)$ IS onto. The codomain plays a vital role in determining if a function is onto.
3. One-to-One and Onto Function (Bijective Function or Bijection)
A function $f: A \to B$ is called a bijective function or a bijection if it is both one-to-one and onto.
A bijective function establishes a perfect pairing between the elements of the domain and the codomain, where each element in the domain is paired with exactly one element in the codomain, and every element in the codomain is paired with exactly one element in the domain.
Example: Consider the function $f(x) = x^3$ from $\mathbb{R}$ to $\mathbb{R}$.
- One-to-one check: Let $f(x_1) = f(x_2)$. Then $x_1^3 = x_2^3$. Taking the cube root of both sides, we get $x_1 = x_2$. So, it is one-to-one.
- Onto check: Let $y$ be any element in the codomain $\mathbb{R}$. We need to find an $x \in \mathbb{R}$ such that $f(x) = y$, i.e., $x^3 = y$. Solving for $x$, we get $x = \sqrt[3]{y}$. For any real number $y$, $\sqrt[3]{y}$ is also a real number. Thus, for every $y$ in the codomain, there is an $x$ in the domain such that $f(x)=y$. So, it is onto.
Since $f(x) = x^3$ is both one-to-one and onto from $\mathbb{R}$ to $\mathbb{R}$, it is a bijective function. Bijective functions are particularly important because they are precisely the functions that have inverse functions.
Other Important Types of Functions
Beyond the classification based on mapping properties, functions can also be categorized based on their algebraic structure or form:
1. Constant Function
A function $f: A \to B$ is a constant function if every element in the domain $A$ is mapped to the same single element $c$ in the codomain $B$.
The defining equation is $f(x) = c$ for all $x \in A$, where $c$ is a fixed constant.
The range of a constant function contains only one element, $\{c\}$.
Example: $f(x) = 5$ for all $x \in \mathbb{R}$. No matter what $x$ is, the output is always 5. The graph of a constant function is a horizontal line.
2. Identity Function
The identity function on a set $A$ is a function $f: A \to A$ that maps each element of $A$ to itself.
The defining equation is $f(x) = x$ for all $x \in A$.
The identity function is always a bijection (one-to-one and onto) from $A$ to $A$.
Example: The identity function on $\mathbb{R}$ is $f(x) = x$. Its graph is the line $y=x$, passing through the origin with a slope of 1.
3. Polynomial Function
A function $f(x)$ is a polynomial function if it can be written in the general form:
$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 $
The domain of a polynomial function is typically all real numbers, $\mathbb{R}$, unless otherwise specified.
Examples:
- $f(x) = 3x^2 - 5x + 2$ (Polynomial of degree 2, quadratic function)
- $g(x) = x^4 - 7$ (Polynomial of degree 4)
- $h(x) = 2x$ (Polynomial of degree 1, linear function)
- $k(x) = -10$ (Polynomial of degree 0, constant function)
4. Rational Function
A function $f(x)$ is a rational function if it can be expressed as the ratio of two polynomial functions, $P(x)$ and $Q(x)$, where $Q(x)$ is not the zero polynomial.
The form is $f(x) = \frac{P(x)}{Q(x)}$.
The domain of a rational function is all real numbers $x$ for which the denominator $Q(x)$ is not equal to zero.
Example: $f(x) = \frac{x+1}{x-2}$. This is a rational function. Its domain is $\{x \in \mathbb{R} : x \neq 2\}$.
5. Algebraic Function
An algebraic function is a function that can be constructed using a finite number of algebraic operations (addition, subtraction, multiplication, division, exponentiation by a rational exponent, and taking an $n$-th root) on a polynomial.
Polynomial functions and rational functions are special cases of algebraic functions.
Example: $f(x) = \sqrt{x^2 + 1}$ is an algebraic function involving a square root.
6. Transcendental Function
A transcendental function is any function that is NOT an algebraic function. These functions "transcend" algebraic methods.
Important classes of transcendental functions include:
- Trigonometric functions (e.g., $\sin x, \cos x, \tan x$) and their inverses (e.g., $\sin^{-1} x, \cos^{-1} x$)
- Exponential functions (e.g., $a^x$, especially $e^x$)
- Logarithmic functions (e.g., $\log_b x$, especially $\ln x$)
7. Piecewise Function
A piecewise function (also called a piecewise-defined function) is a function defined by multiple sub-functions, where each sub-function applies to a different interval of the domain.
The definition provides different formulas for different parts of the domain.
Example: The absolute value function $f(x) = |x|$ is often defined as a piecewise function:
Another example:
$g(x) = \begin{cases} x^2 + 1 & , & x \leq 1 \\ 2x & , & x > 1 \end{cases}$Graphical Representation of Functions
The graph of a function $f$ is a visual representation of the relationship between its input and output values. For a function $f: A \to B$, its graph is formally defined as the set of all ordered pairs $(x, f(x))$ where $x$ belongs to the domain of $f$.
When we consider real-valued functions of a real variable (i.e., where the domain and codomain are subsets of the set of real numbers $\mathbb{R}$), we can plot these ordered pairs $(x, y)$ on a Cartesian coordinate plane, where $x$ represents the input (plotted on the horizontal axis) and $y = f(x)$ represents the output (plotted on the vertical axis).
The graph of $f$ is the set:
Graph of $f = \{(x, y) \mid x \in \text{Domain}(f) \text{ and } y = f(x)\}$.
Interpreting Information from a Function's Graph
A function's graph provides a wealth of information about its properties and behaviour. We can visually extract several key aspects:
- Domain: The domain of the function is the set of all possible $x$-values for which the graph exists. To determine the domain from a graph, observe the horizontal extent of the graph along the x-axis.
- Range: The range of the function is the set of all actual $y$-values that the function produces. To determine the range from a graph, observe the vertical extent of the graph along the y-axis.
- Function Value: To find the value of the function for a specific input, say $x=a$, locate $a$ on the x-axis. Move vertically from $a$ until you reach the graph. Then, move horizontally from that point on the graph to the y-axis. The value where you intersect the y-axis is $f(a)$.
- Roots or Zeros: The roots or zeros of a function are the values of $x$ for which $f(x) = 0$. Graphically, these are the points where the graph intersects or touches the x-axis (the x-intercepts).
- Increasing and Decreasing Intervals: A function is considered increasing over an interval of its domain if, as $x$ increases within that interval, the corresponding $y$-values ($f(x)$) also increase. On the graph, this is where the curve rises as you move from left to right. Conversely, a function is decreasing over an interval if its graph falls as you move from left to right.
- Maximum and Minimum Values (Extrema): The highest points on the graph within a certain region are local maximums, and the lowest points are local minimums. The overall highest or lowest points on the entire graph (if they exist) are the global maximum or minimum values, respectively.
- Continuity: A function is visually continuous over an interval if its graph can be drawn without lifting the pen, meaning there are no breaks, jumps, or holes in the graph over that interval.
- Asymptotes: These are lines that the graph of a function approaches but never quite reaches as the input or output values tend towards certain limits (often infinity). They can be vertical, horizontal, or oblique.
- Vertical Line Test: As discussed in the "Functions" section (I1), a graph represents a function if and only if every vertical line intersects the graph at most once. This confirms that for each input $x$, there is only one output $y$.
(This graph represents a function)
(This graph does NOT represent a function)
Example 1. Sketch the graph of the function $f(x) = 2x - 1$ and use the graph to determine its domain and range.
Answer:
The function $f(x) = 2x - 1$ is a linear function, which is of the form $y = mx + c$. Here, the slope $m=2$ and the y-intercept $c=-1$. A linear function's graph is always a straight line.
To sketch the graph, we can find the coordinates of at least two points that lie on the line and then draw a straight line passing through them. A simple way is to find the intercepts:
- Y-intercept: Set $x=0$. $f(0) = 2(0) - 1 = -1$. The graph crosses the y-axis at $(0, -1)$.
- X-intercept: Set $f(x)=0$. $2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$. The graph crosses the x-axis at $(\frac{1}{2}, 0)$.
We can also find another point, for instance, when $x=2$:
$f(2) = 2(2) - 1 = 4 - 1 = 3$. The point $(2, 3)$ is on the graph.
Plot the points $(0, -1)$, $(\frac{1}{2}, 0)$, and $(2, 3)$ on a coordinate plane and draw a straight line through them.
From the graph, we can interpret the domain and range:
Domain: The straight line extends indefinitely to the left and right along the x-axis. This visually confirms that the function is defined for all real numbers.
Domain $= \mathbb{R}$ or $(-\infty, \infty)$.
Range: The straight line extends indefinitely downwards and upwards along the y-axis. This shows that the function can take on any real number as an output value.
Range $= \mathbb{R}$ or $(-\infty, \infty)$.
Example 2. The graph of a function is described as a parabola that opens upwards. Its vertex (the lowest point) is located at $(1, -4)$. The graph intersects the x-axis at the points $(-1, 0)$ and $(3, 0)$, and it intersects the y-axis at the point $(0, -3)$. Based on this graphical description, determine the domain, range, and zeros of the function.
Answer:
We are given key points and the general shape of the graph. Let's visualize the graph:
Now, let's find the required information from this graph:
Domain: The parabola opens upwards and extends infinitely outwards horizontally. For a standard parabola defined by a polynomial like $f(x) = a(x-h)^2 + k$ (where $a > 0$ for opening upwards), the domain is always all real numbers.
Domain $= \mathbb{R}$ or $(-\infty, \infty)$.
Range: The lowest point on the graph is the vertex, which is at $(1, -4)$. The $y$-coordinate of the vertex, $-4$, is the minimum value of the function. Since the parabola opens upwards, all other $y$-values on the graph are greater than or equal to $-4$. The graph extends infinitely upwards.
Range $= \{y \in \mathbb{R} \mid y \geq -4\}$ or $[-4, \infty)$.
Zeros: The zeros of the function are the x-values where the graph crosses the x-axis. These are the x-intercepts given as $(-1, 0)$ and $(3, 0)$.
The zeros are $x = -1$ and $x = 3$. (These are the solutions to $f(x)=0$).
Concepts of Limits and Continuity of a Function
Limits and Continuity are two of the most fundamental and interconnected concepts in calculus. They provide the tools to understand the behaviour of functions at specific points and over intervals, forming the basis for differentiation and integration. The concept of a limit describes the value a function approaches as its input gets arbitrarily close to a certain point, while continuity describes functions whose graphs are unbroken curves.
Limit of a Function
The concept of a limit allows us to analyse the behaviour of a function $f(x)$ as the input variable $x$ gets closer and closer to a particular value, say $a$, without necessarily reaching $a$. We are interested in the value that the function's output, $f(x)$, approaches.
The statement "the limit of $f(x)$ as $x$ approaches $a$ is $L$" is written notationally as:
$\lim\limits_{x \to a} f(x) = L$
Formal Definition of a Limit (Epsilon-Delta Definition):
The statement $\lim\limits_{x \to a} f(x) = L$ is formally defined as follows:
For every number $\epsilon > 0$ (epsilon, an arbitrarily small positive number), there exists a corresponding number $\delta > 0$ (delta, another positive number, which usually depends on $\epsilon$) such that if $x$ is within a distance of $\delta$ from $a$ (but not equal to $a$), then $f(x)$ is within a distance of $\epsilon$ from $L$.
Mathematically: For every $\epsilon > 0$, there exists $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$.
This rigorous definition precisely captures the idea of "arbitrarily close". We can make $f(x)$ as close as we like to $L$ (by choosing $\epsilon$ small) by ensuring $x$ is sufficiently close to $a$ (within the corresponding $\delta$).
Left-Hand Limit and Right-Hand Limit
To determine if the limit of $f(x)$ as $x$ approaches $a$ exists, we must consider the behaviour of the function as $x$ approaches $a$ from both sides:
- Left-Hand Limit (LHL): This is the value that $f(x)$ approaches as $x$ approaches $a$ from values less than $a$ (i.e., from the left side on the number line).
Notation: $\lim\limits_{x \to a^-} f(x)$ - Right-Hand Limit (RHL): This is the value that $f(x)$ approaches as $x$ approaches $a$ from values greater than $a$ (i.e., from the right side on the number line).
Notation: $\lim\limits_{x \to a^+} f(x)$
For the overall limit $\lim\limits_{x \to a} f(x)$ to exist, the left-hand limit and the right-hand limit must both exist and be equal to the same value $L$.
$\lim\limits_{x \to a} f(x) = L \iff \lim\limits_{x \to a^-} f(x) = L \text{ and } \lim\limits_{x \to a^+} f(x) = L$
... (1)
Evaluating Limits
For many basic functions, evaluating limits is straightforward.
1. Direct Substitution: If $f(x)$ is a "well-behaved" function at $x=a$, such as a polynomial function, or a rational function where the denominator is non-zero at $x=a$, the limit can be found by simply substituting $a$ into the function:
If $f$ is a polynomial function, or a rational function $f(x) = \frac{P(x)}{Q(x)}$ with $Q(a) \neq 0$, then $\lim\limits_{x \to a} f(x) = f(a)$
... (2)
2. Algebraic Manipulation: If direct substitution results in an indeterminate form, such as $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \times \infty$, $\infty - \infty$, $0^0$, $1^\infty$, or $\infty^0$, it means the limit cannot be determined by substitution alone. In such cases, algebraic techniques are often employed to simplify the expression before evaluating the limit. Common techniques include:
- Factoring the numerator and denominator to cancel common factors (useful for $\frac{0}{0}$).
- Rationalizing the numerator or denominator (especially when square roots are involved).
- Finding a common denominator.
Other techniques, such as using standard limits, trigonometric identities, or L'Hopital's Rule (a more advanced technique typically covered later), may also be necessary.
Example 1. Evaluate $\lim\limits_{x \to 2} (3x^2 - 5x + 1)$.
Answer:
The function $f(x) = 3x^2 - 5x + 1$ is a polynomial function. Polynomial functions are defined and continuous for all real numbers. Therefore, we can evaluate the limit by direct substitution.
Substitute $x=2$ into the function:
$\lim\limits_{x \to 2} (3x^2 - 5x + 1) = 3(2)^2 - 5(2) + 1$
$= 3(4) - 10 + 1$
$= 12 - 10 + 1$
$= 2 + 1$
$= 3$
The limit of the function as $x$ approaches $2$ is $3$.
Example 2. Evaluate $\lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1}$.
Answer:
Let the function be $f(x) = \frac{x^2 - 1}{x - 1}$. If we try direct substitution with $x=1$, we get $\frac{1^2 - 1}{1 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$. This is an indeterminate form, indicating that we need to simplify the expression.
We can factor the numerator, which is a difference of squares ($a^2 - b^2 = (a-b)(a+b)$):
$x^2 - 1 = (x-1)(x+1)$
Now substitute this back into the limit expression:
$\lim\limits_{x \to 1} \frac{(x-1)(x+1)}{x - 1}$
Since we are considering the limit as $x \to 1$, $x$ is approaching $1$ but $x \neq 1$. This means $x-1 \neq 0$. Therefore, we can cancel out the common factor $(x-1)$ from the numerator and the denominator.
$= \lim\limits_{x \to 1} (x+1)$
Now the expression is a polynomial $(x+1)$, which is defined and continuous at $x=1$. We can use direct substitution:
$= 1 + 1$
$= 2$
The limit of the function as $x$ approaches $1$ is $2$. Note that the original function $f(x) = \frac{x^2 - 1}{x - 1}$ is undefined at $x=1$, but its limit as $x \to 1$ exists and is $2$.
Continuity of a Function
Intuitively, a function is continuous at a point if its graph does not have any break, hole, or jump at that point. You could trace the graph at that point without lifting your pen.
Definition of Continuity at a Point
A function $f(x)$ is said to be continuous at a point $a$ if the following three conditions are satisfied:
- The function $f$ is defined at $a$. This means $f(a)$ exists and is a finite real number. The point $a$ must be in the domain of $f$.
- The limit of the function $f(x)$ as $x$ approaches $a$ exists. This means $\lim\limits_{x \to a} f(x)$ exists and is a finite real number (which implies the left-hand limit and right-hand limit at $a$ are equal and finite).
- The value of the function at $a$ is equal to the limit of the function as $x$ approaches $a$.
$\lim\limits_{x \to a} f(x) = f(a)$
... (3)
If any one of these three conditions is not met, the function is said to be discontinuous at point $a$.
Continuity over an Interval
The concept of continuity can be extended from a single point to an interval:
- A function $f$ is continuous over an open interval $(a, b)$ if it is continuous at every point $c$ within that interval ($a < c < b$).
- A function $f$ is continuous over a closed interval $[a, b]$ if it is continuous over the open interval $(a, b)$, and it is continuous from the right at $a$ ($\lim\limits_{x \to a^+} f(x) = f(a)$), and it is continuous from the left at $b$ ($\lim\limits_{x \to b^-} f(x) = f(b)$). The points $a$ and $b$ must be in the domain of $f$.
Properties of Continuous Functions
Understanding the continuity of basic functions helps in determining the continuity of more complex functions:
- Polynomial Functions: All polynomial functions are continuous everywhere on the set of real numbers, $\mathbb{R}$.
- Rational Functions: A rational function $f(x) = \frac{P(x)}{Q(x)}$ is continuous at every point in its domain. Its domain is all real numbers except where the denominator $Q(x)$ is zero. So, rational functions are continuous everywhere except possibly at the roots of the denominator.
- Root Functions: For an odd integer $n > 1$, $f(x) = \sqrt[n]{x}$ is continuous on $\mathbb{R}$. For an even integer $n > 1$, $f(x) = \sqrt[n]{x}$ is continuous on $[0, \infty)$. More generally, $\sqrt[n]{g(x)}$ is continuous where $g(x)$ is continuous and $g(x) \geq 0$ (for even $n$) or $g(x)$ is continuous (for odd $n$).
- Trigonometric Functions: $\sin x$ and $\cos x$ are continuous everywhere on $\mathbb{R}$. $\tan x$, $\cot x$, $\sec x$, and $\text{cosec} x$ are continuous on their respective domains (everywhere except where their denominators are zero).
- Exponential Functions: $f(x) = a^x$ (for $a > 0$) is continuous everywhere on $\mathbb{R}$.
- Logarithmic Functions: $f(x) = \log_b x$ (for $b > 0, b \neq 1$) is continuous everywhere on its domain, which is $(0, \infty)$. More generally, $\log_b(g(x))$ is continuous where $g(x)$ is continuous and $g(x) > 0$.
- Algebra of Continuous Functions: If $f$ and $g$ are continuous at a point $a$, then the following functions are also continuous at $a$:
- Sum: $(f+g)(x) = f(x) + g(x)$
- Difference: $(f-g)(x) = f(x) - g(x)$
- Product: $(f \cdot g)(x) = f(x) \cdot g(x)$
- Quotient: $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$, provided $g(a) \neq 0$.
- Scalar Multiplication: $(c \cdot f)(x) = c \cdot f(x)$, where $c$ is a constant.
- Composition of Continuous Functions: If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then the composite function $(g \circ f)(x) = g(f(x))$ is continuous at $a$.
Example 3. Check if the function $f(x) = x^2 + 1$ is continuous at $x = 2$.
Answer:
We check the three conditions for continuity at $x=2$:
Condition 1: Does $f(2)$ exist?
Substitute $x=2$ into the function: $f(2) = (2)^2 + 1 = 4 + 1 = 5$.
$f(2)$ exists and is equal to $5$. Condition 1 is met.
Condition 2: Does $\lim\limits_{x \to 2} f(x)$ exist?
The function $f(x) = x^2 + 1$ is a polynomial. For polynomial functions, the limit as $x$ approaches any point $a$ is equal to the function's value at that point, i.e., $\lim\limits_{x \to a} f(x) = f(a)$.
$\lim\limits_{x \to 2} (x^2 + 1) = 2^2 + 1 = 5$.
The limit exists and is equal to $5$. Condition 2 is met.
Condition 3: Is $\lim\limits_{x \to 2} f(x) = f(2)$?
From Condition 1, $f(2) = 5$.
From Condition 2, $\lim\limits_{x \to 2} f(x) = 5$.
Since $5 = 5$, the third condition is met.
As all three conditions for continuity are satisfied at $x=2$, the function $f(x) = x^2 + 1$ is continuous at $x=2$. (In fact, being a polynomial function, it is continuous at every real number).
Example 4. Check if the piecewise function $g(x) = \begin{cases} x+1 & , & x \leq 1 \\ 3-x & , & x > 1 \end{cases}$ is continuous at $x = 1$.
Answer:
We need to check the three conditions for continuity at the point $x=1$, which is where the definition of the function changes.
Condition 1: Does $g(1)$ exist?
According to the definition of $g(x)$, when $x \leq 1$, $g(x) = x+1$. Since $x=1$ falls into the case $x \leq 1$, we use the first part of the definition to find $g(1)$.
$g(1) = 1 + 1 = 2$.
$g(1)$ exists and is equal to $2$. Condition 1 is met.
Condition 2: Does $\lim\limits_{x \to 1} g(x)$ exist?
For the limit to exist at $x=1$, the left-hand limit and the right-hand limit at $x=1$ must exist and be equal.
Left-Hand Limit (LHL) at $x=1$: As $x$ approaches $1$ from the left side ($x < 1$), we use the part of the function definition for $x \leq 1$, which is $g(x) = x+1$.
$\lim\limits_{x \to 1^-} g(x) = \lim\limits_{x \to 1^-} (x+1)$
Since $x+1$ is a polynomial, we can evaluate the limit by direct substitution:
$= 1 + 1 = 2$.
Right-Hand Limit (RHL) at $x=1$: As $x$ approaches $1$ from the right side ($x > 1$), we use the part of the function definition for $x > 1$, which is $g(x) = 3-x$.
$\lim\limits_{x \to 1^+} g(x) = \lim\limits_{x \to 1^+} (3-x)$
Since $3-x$ is a polynomial, we can evaluate the limit by direct substitution:
$= 3 - 1 = 2$.
Since the LHL ($2$) is equal to the RHL ($2$), the limit $\lim\limits_{x \to 1} g(x)$ exists and is equal to $2$. Condition 2 is met.
Condition 3: Is $\lim\limits_{x \to 1} g(x) = g(1)$?
From Condition 1, $g(1) = 2$.
From Condition 2, $\lim\limits_{x \to 1} g(x) = 2$.
Since $2 = 2$, the third condition is met.
Because all three conditions for continuity are satisfied at $x=1$, the function $g(x)$ is continuous at $x=1$.
Instantaneous Rate of Change
The concept of the rate of change is fundamental to understanding how quantities vary. In calculus, we distinguish between the rate of change over an interval and the rate of change at a single point. The latter is known as the Instantaneous Rate of Change and is a core concept leading to the definition of the derivative.
Average Rate of Change
The Average Rate of Change of a quantity $y$ with respect to another quantity $x$, over an interval from $x_1$ to $x_2$, measures the total change in $y$ divided by the total change in $x$. If $y$ is given by a function $f(x)$, this can be expressed as the change in the function's value divided by the change in the input value.
For a function $y = f(x)$, the average rate of change over the interval $[x_1, x_2]$ is given by the formula:
Average Rate of Change $= \frac{\text{Change in } y}{\text{Change in } x} = \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$
... (1)
Geometrically, if we plot the graph of $y = f(x)$, the average rate of change between two points $(x_1, f(x_1))$ and $(x_2, f(x_2))$ is equal to the slope of the secant line connecting these two points. A secant line is a straight line that intersects a curve at two or more points.
From Average Rate of Change to Instantaneous Rate of Change
The Instantaneous Rate of Change measures how fast a quantity is changing at a precise moment or point. It is the limiting value of the average rate of change as the interval over which the change is measured shrinks to zero.
To define the instantaneous rate of change at a specific point, say $x=a$, we consider the average rate of change over a very small interval starting at $a$. Let this small interval be from $a$ to $a+h$, where $h = \Delta x$ represents a small change in $x$ and $h \neq 0$. The points on the graph are $(a, f(a))$ and $(a+h, f(a+h))$.
The average rate of change over the interval $[a, a+h]$ (or $[a+h, a]$ if $h < 0$) is:
Average Rate of Change on $[a, a+h] = \frac{f(a+h) - f(a)}{(a+h) - a} = \frac{f(a+h) - f(a)}{h}$
The expression $\frac{f(a+h) - f(a)}{h}$ is called the difference quotient.
To find the instantaneous rate of change at the point $x=a$, we take the limit of this average rate of change as the increment $h$ approaches zero. This captures the rate of change at the exact point $x=a$.
Instantaneous Rate of Change at $x=a = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$
... (2)
This limit, provided it exists, gives the precise rate at which $f(x)$ is changing with respect to $x$ exactly at the point $x=a$.
Connection to the Derivative and Tangent Line Slope
The limit expression in equation (2) is the fundamental definition of the Derivative of the function $f$ at the point $a$. It is denoted by $f'(a)$.
$f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$
... (3)
Geometrically, as $h$ approaches $0$, the point $(a+h, f(a+h))$ on the graph gets closer and closer to the point $(a, f(a))$. The secant line connecting these two points rotates and approaches the tangent line to the graph of $f(x)$ at the point $(a, f(a))$.
Therefore, the value of the limit $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ represents the slope of the tangent line to the curve $y = f(x)$ at the point $(a, f(a))$. The slope of the tangent line at a point provides a measure of the steepness and direction of the curve at that specific point, which is precisely what the instantaneous rate of change describes.
Example 1. Find the average rate of change of the function $f(x) = x^2$ over the interval $[1, 3]$.
Answer:
We are asked to find the average rate of change of $f(x) = x^2$ over the interval $[1, 3]$.
Using the formula for average rate of change $\frac{f(x_2) - f(x_1)}{x_2 - x_1}$, with $x_1 = 1$ and $x_2 = 3$:
First, find the function values at $x_1$ and $x_2$:
$f(x_1) = f(1) = (1)^2 = 1$.
$f(x_2) = f(3) = (3)^2 = 9$.
Now substitute these values into the formula:
Average Rate of Change $= \frac{f(3) - f(1)}{3 - 1}$
$= \frac{9 - 1}{2}$
$= \frac{8}{2}$
$= 4$
The average rate of change of $f(x) = x^2$ over the interval $[1, 3]$ is $4$. This means that, on average, for every 1-unit increase in $x$ from $1$ to $3$, the value of $f(x)$ increases by $4$ units.
Example 2. Find the instantaneous rate of change of the function $f(x) = x^2$ at $x = 1$ using the limit definition.
Answer:
We need to find the instantaneous rate of change of $f(x) = x^2$ at $x=1$. This is given by the limit definition:
Instantaneous Rate of Change at $x=1 = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$.
First, let's find the required function values:
$f(1) = (1)^2 = 1$.
$f(1+h) = (1+h)^2$. Expand $(1+h)^2$ using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$:
$f(1+h) = 1^2 + 2(1)(h) + h^2 = 1 + 2h + h^2$.
Now substitute $f(1+h)$ and $f(1)$ into the difference quotient:
$\frac{f(1+h) - f(1)}{h} = \frac{(1 + 2h + h^2) - 1}{h}$
$= \frac{1 + 2h + h^2 - 1}{h}$
$= \frac{2h + h^2}{h}$
Since we are considering the limit as $h \to 0$, $h$ is approaching $0$ but $h \neq 0$. Thus, we can factor out $h$ from the numerator and cancel it with the denominator:
$= \frac{h(2 + h)}{h}$
$= 2 + h$ (for $h \neq 0$)
Now, evaluate the limit of the simplified expression as $h \to 0$:
Instantaneous Rate of Change at $x=1 = \lim\limits_{h \to 0} (2 + h)$
Substitute $h=0$ into the expression $(2+h)$ (since $2+h$ is a polynomial in $h$, the limit is found by direct substitution):
$= 2 + 0$
$= 2$
The instantaneous rate of change of $f(x) = x^2$ at $x=1$ is $2$. This means that at the exact moment when $x=1$, the function's value is increasing at a rate of $2$ units of $f(x)$ per unit of $x$. This value $2$ is also the slope of the tangent line to the graph of $y=x^2$ at the point $(1, f(1)) = (1, 1)$.
Differentiation as a Process of Finding Derivative
Differentiation is a fundamental operation in calculus. It is the process of finding the derivative of a function. The derivative is a powerful tool that allows us to determine the instantaneous rate of change of a function at any point in its domain, and it also gives us the slope of the tangent line to the function's graph at that point.
The Derivative of a Function
The derivative of a function $f(x)$ with respect to the independent variable $x$ is formally defined using the concept of a limit, specifically the limit of the difference quotient.
The derivative of $f(x)$, denoted by $f'(x)$ (read as "f prime of x") or $\frac{dy}{dx}$ (read as "dee y dee x" or "the derivative of y with respect to x" when $y = f(x)$), is defined as:
$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$
... (1)
This definition is known as the limit definition of the derivative or finding the derivative "from first principles". The derivative $f'(x)$ is itself a new function of $x$, which gives the instantaneous rate of change of $f$ at any point $x$ where the limit exists.
Differentiability
A function $f(x)$ is said to be differentiable at a point $x=a$ if the limit in the definition of $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists and is a finite real number.
A function is said to be differentiable over an interval if it is differentiable at every point within that interval.
Relationship between Differentiability and Continuity:
A crucial theorem states that if a function is differentiable at a point, then it must be continuous at that point. In other words, differentiability implies continuity.
However, the converse is NOT true. A function can be continuous at a point but not differentiable at that point.
Common situations where a function is continuous but not differentiable at a point include:
- Sharp corners or cusps: At these points, the slope of the tangent line is not uniquely defined. The left-hand derivative (limit of the difference quotient as $h \to 0^-$) is not equal to the right-hand derivative (limit as $h \to 0^+$). The absolute value function $f(x) = |x|$ is continuous at $x=0$ but not differentiable at $x=0$ due to a sharp corner there.
- Vertical tangents: If the tangent line at a point is vertical, its slope is undefined (infinitely steep). The limit of the difference quotient will tend to $\infty$ or $-\infty$. Example: $f(x) = \sqrt[3]{x}$ at $x=0$.
- Discontinuities: Since differentiability requires continuity, any point of discontinuity (jump, hole, or vertical asymptote) is also a point where the function is not differentiable.
Basic Differentiation Rules (for Algebraic Functions)
While the limit definition always works for finding the derivative, using it for complex functions can be very laborious. Fortunately, rules have been derived from the limit definition that allow us to find derivatives much more efficiently. Here are some fundamental rules for differentiating algebraic functions:
1. The Constant Rule
The derivative of a constant function is zero. If $f(x) = c$, where $c$ is any real number, then the instantaneous rate of change of a value that never changes is always 0.
If $f(x) = c$, then $f'(x) = 0$.
$\frac{d}{dx}(c) = 0$
2. The Power Rule
The power rule is used to differentiate functions of the form $x^n$, where $n$ is any real number.
If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.
$\frac{d}{dx}(x^n) = nx^{n-1}$
Examples:
- $\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$
- $\frac{d}{dx}(x^1) = 1x^{1-1} = 1x^0 = 1(1) = 1$
- $\frac{d}{dx}(x^0) = 0x^{0-1} = 0$ (This matches the constant rule for $x^0=1$)
- $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{(1/2)-1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
- $\frac{d}{dx}(\frac{1}{x^2}) = \frac{d}{dx}(x^{-2}) = -2x^{-2-1} = -2x^{-3} = -\frac{2}{x^3}$
3. The Constant Multiple Rule
If a function $f(x)$ is multiplied by a constant $c$, the derivative of the new function is the constant multiplied by the derivative of $f(x)$.
If $g(x) = c \cdot f(x)$ and $f$ is differentiable, then $g'(x) = c \cdot f'(x)$.
$\frac{d}{dx}(c \cdot f(x)) = c \cdot \frac{d}{dx}(f(x))$
Example: $\frac{d}{dx}(5x^3) = 5 \cdot \frac{d}{dx}(x^3) = 5 \cdot (3x^2) = 15x^2$
4. The Sum and Difference Rule
The derivative of a sum or difference of two or more differentiable functions is the sum or difference of their individual derivatives.
If $h(x) = f(x) \pm g(x)$ and $f, g$ are differentiable, then $h'(x) = f'(x) \pm g'(x)$.
$\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))$
Example: $\frac{d}{dx}(x^2 + x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x) = 2x + 1$
By combining these basic rules, we can differentiate any polynomial function and many other algebraic functions term by term.
Example 1. Find the derivative of the function $f(x) = 5x^3 - 2x^2 + 7x - 4$ using the differentiation rules.
Answer:
We want to find $f'(x)$ for $f(x) = 5x^3 - 2x^2 + 7x - 4$. We can apply the Sum and Difference Rule to differentiate each term separately, and then use the Constant Multiple Rule and the Power Rule for each term.
$f'(x) = \frac{d}{dx}(5x^3 - 2x^2 + 7x - 4)$
Applying the Sum and Difference Rule:
$f'(x) = \frac{d}{dx}(5x^3) - \frac{d}{dx}(2x^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(4)$
Now, apply the Constant Multiple Rule to the first three terms:
$f'(x) = 5 \cdot \frac{d}{dx}(x^3) - 2 \cdot \frac{d}{dx}(x^2) + 7 \cdot \frac{d}{dx}(x) - \frac{d}{dx}(4)$
Next, apply the Power Rule ($\frac{d}{dx}(x^n) = nx^{n-1}$) to the terms involving $x$, and the Constant Rule ($\frac{d}{dx}(c) = 0$) to the last term:
$f'(x) = 5 \cdot (3x^{3-1}) - 2 \cdot (2x^{2-1}) + 7 \cdot (1x^{1-1}) - 0$
$f'(x) = 5 \cdot (3x^2) - 2 \cdot (2x^1) + 7 \cdot (1x^0) - 0$
Simplify the terms:
$f'(x) = 15x^2 - 4x + 7(1)$
$f'(x) = 15x^2 - 4x + 7$
The derivative of the function $f(x) = 5x^3 - 2x^2 + 7x - 4$ is $f'(x) = 15x^2 - 4x + 7$.
Derivatives of Algebraic Functions using Chain Rule
The Chain Rule is one of the most crucial differentiation rules in calculus, particularly useful for differentiating composite functions. Many functions encountered in calculus, especially algebraic ones raised to a power or square roots of expressions, are composite functions. The Chain Rule provides a systematic way to find their derivatives.
Composite Function
A function $h(x)$ is called a composite function if it can be formed by combining two functions, say $f$ and $g$, such that the output of one function becomes the input of the other. Specifically, if we have two functions $f$ and $g$, the composite function $(f \circ g)(x)$ is defined as $f(g(x))$.
In the composite function $h(x) = f(g(x))$, $g$ is often referred to as the inner function, and $f$ is the outer function. The domain of $f(g(x))$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.
Example: Let $f(u) = u^2$ and $g(x) = x+1$.
The composite function $h(x) = f(g(x))$ is obtained by substituting $g(x)$ into $f(u)$ for $u$.
$h(x) = f(g(x)) = f(x+1) = (x+1)^2$.
Here, $g(x) = x+1$ is the inner function, and $f(u) = u^2$ is the outer function.
Example: Let $y = \sqrt{2x+5}$. This can be viewed as a composite function.
Let the inner function be $u = g(x) = 2x+5$.
Let the outer function be $y = f(u) = \sqrt{u} = u^{1/2}$.
Then $y = f(g(x)) = f(2x+5) = \sqrt{2x+5}$.
The Chain Rule
The Chain Rule states how to find the derivative of a composite function. If $y = f(u)$ is a differentiable function of $u$, and $u = g(x)$ is a differentiable function of $x$, then the composite function $y = f(g(x))$ is a differentiable function of $x$. Its derivative is given by the product of the derivative of $y$ with respect to $u$ and the derivative of $u$ with respect to $x$.
In Leibniz notation:
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
... (1)
In Prime notation, if $h(x) = f(g(x))$, then the derivative $h'(x)$ is given by:
$h'(x) = f'(g(x)) \cdot g'(x)$
... (2)
In simpler terms, to differentiate $f(g(x))$:
- Differentiate the outer function $f$, evaluating its derivative at the inner function $g(x)$ (i.e., find $f'(g(x))$).
- Multiply the result by the derivative of the inner function $g(x)$ (i.e., find $g'(x)$ and multiply).
Applying Chain Rule to Algebraic Functions
The Chain Rule is frequently applied in conjunction with the Power Rule to differentiate expressions where a function (instead of just $x$) is raised to a power.
Power of a Function Rule (derived using Chain Rule)
Consider a function of the form $y = [g(x)]^n$, where $g(x)$ is a differentiable function of $x$, and $n$ is a real number. We can think of this as a composite function where the inner function is $u = g(x)$ and the outer function is $y = f(u) = u^n$.
Using the Power Rule for the outer function $f(u) = u^n$, the derivative with respect to $u$ is:
$\frac{dy}{du} = \frac{d}{du}(u^n) = nu^{n-1}$
The derivative of the inner function $u = g(x)$ with respect to $x$ is:
$\frac{du}{dx} = g'(x)$
Now, apply the Chain Rule $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$:
$\frac{dy}{dx} = (nu^{n-1}) \times g'(x)$
Substitute back $u = g(x)$:
$\frac{d}{dx}([g(x)]^n) = n[g(x)]^{n-1} \cdot g'(x)$
... (3)
Example 1. Find the derivative of $h(x) = (x^2 + 3x - 1)^5$ using the chain rule.
Answer:
The function is $h(x) = (x^2 + 3x - 1)^5$. This is in the form $[g(x)]^n$ where:
- Inner function $g(x) = x^2 + 3x - 1$
- Outer function is raising to the power of $5$, i.e., $f(u) = u^5$ with $u = g(x)$.
- The power $n=5$.
According to the Chain Rule (Power of a Function Rule), the derivative is $n[g(x)]^{n-1} \cdot g'(x)$.
First, find the derivative of the inner function $g(x)$:
$g'(x) = \frac{d}{dx}(x^2 + 3x - 1)$
Using the Sum/Difference, Constant Multiple, and Power Rules:
$g'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(1)$
$g'(x) = 2x^{2-1} + 3 \cdot \frac{d}{dx}(x^1) - 0$
$g'(x) = 2x + 3 \cdot (1x^{1-1})$
$g'(x) = 2x + 3 \cdot (1)$
$g'(x) = 2x + 3$
Now, apply the Chain Rule formula $\frac{d}{dx}([g(x)]^n) = n[g(x)]^{n-1} \cdot g'(x)$:
Here, $n=5$, $g(x) = x^2 + 3x - 1$, and $g'(x) = 2x+3$.
$h'(x) = 5 \cdot (x^2 + 3x - 1)^{5-1} \cdot (2x + 3)$
$h'(x) = 5 (x^2 + 3x - 1)^4 (2x + 3)$
The derivative of $h(x) = (x^2 + 3x - 1)^5$ is $h'(x) = 5(2x+3)(x^2 + 3x - 1)^4$.
Example 2. Find the derivative of $y = \sqrt{2x + 5}$ using the chain rule.
Answer:
The function is $y = \sqrt{2x + 5}$. We can rewrite the square root using a fractional exponent: $y = (2x + 5)^{1/2}$.
This is in the form $[g(x)]^n$ where:
- Inner function $g(x) = 2x + 5$
- Outer function is raising to the power of $1/2$, i.e., $f(u) = u^{1/2}$ with $u = g(x)$.
- The power $n=1/2$.
We apply the Chain Rule (Power of a Function Rule): $\frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$.
First, find the derivative of the inner function $g(x)$:
$g'(x) = \frac{d}{dx}(2x + 5)$
Using the Sum/Difference, Constant Multiple, and Power/Constant Rules:
$g'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(5)$
$g'(x) = 2 \cdot \frac{d}{dx}(x) + 0$
$g'(x) = 2 \cdot (1)$
$g'(x) = 2$
Now, apply the Chain Rule formula $\frac{d}{dx}([g(x)]^n) = n[g(x)]^{n-1} \cdot g'(x)$:
Here, $n=1/2$, $g(x) = 2x + 5$, and $g'(x) = 2$.
$\frac{dy}{dx} = \frac{1}{2} \cdot (2x + 5)^{\frac{1}{2}-1} \cdot (2)$
Simplify the exponent:
$\frac{dy}{dx} = \frac{1}{2} \cdot (2x + 5)^{-\frac{1}{2}} \cdot (2)$
Multiply the constant terms $\frac{1}{2}$ and $2$ (they cancel out):
$\frac{dy}{dx} = \cancel{\frac{1}{2}} \cdot (2x + 5)^{-\frac{1}{2}} \cdot \cancel{2}$
$\frac{dy}{dx} = (2x + 5)^{-\frac{1}{2}}$
Rewrite using radical notation (optional, but standard form):
$\frac{dy}{dx} = \frac{1}{(2x + 5)^{1/2}} = \frac{1}{\sqrt{2x + 5}}$
The derivative of $y = \sqrt{2x + 5}$ is $\frac{dy}{dx} = \frac{1}{\sqrt{2x + 5}}$.