01 · Explore
Intuitive Idea of Derivatives
Calculus is the branch of mathematics dealing with the study of change. We begin with a physical example to understand how rates of change are calculated at a specific instant.
Consider a body dropped from a cliff. Its distance s (in metres) after t seconds is given by s = 4.9t². To find the velocity at exactly t = 2 seconds, we can calculate the average velocity over smaller and smaller time intervals ending at 2 seconds.
Average velocity is the change in distance divided by the change in time. By observing the average velocity as the time interval approaches zero, we find that the values approach a specific limit. This limit is known as the instantaneous velocity.
In the example of s = 4.9t², as we take intervals closer to t = 2 (like 1.99 to 2 seconds or 2 to 2.01 seconds), the average velocity is found to be between 19.551 m/s and 19.649 m/s. This value is the derivative of the distance function at t = 2.
| Time Interval (seconds) | Average Velocity (m/s) |
|---|---|
| t = 0 to t = 2 | 9.8 |
| t = 1 to t = 2 | 14.7 |
| t = 1.9 to t = 2 | 19.11 |
| t = 1.99 to t = 2 | 19.551 |
| t = 2 to t = 2.01 | 19.649 |
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
What does the derivative of a distance function represent physically?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 1, 2, 3.
02 · Explore
The Concept of Limits
A limit describes the behavior of a function as its input approaches a specific value. It is the value the function 'should' assume at that point based on nearby values.
Symbolically, we write limₓ→ₐ f(x) = l. This means as x gets closer and closer to a, the value of f(x) gets closer and closer to l. Note that the function does not necessarily have to be defined at x = a for the limit to exist.
Consider f(x) = (x² - 4) / (x - 2). At x = 2, the function is undefined (0/0). However, for values very close to 2, such as 1.999 or 2.001, the function value is very close to 4. Therefore, the limit as x approaches 2 is 4.
Limits are essential because they allow us to discuss the 'expected' value of a function even where the function has a hole or a jump in its graph.
The Limiting Process
- 1
Input Approach
Variable x approaches a value 'a' from either the left (x < a) or the right (x > a).
- 2
Function Response
The function values f(x) move toward a specific value 'l'.
- 3
Limit Definition
If the values from both sides approach the same 'l', then 'l' is the limit.
The process of finding a limit involves observing function behavior as inputs converge on a point.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
If f(2) = 5, does limₓ→2 f(x) always equal 5?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 4, 5.
03 · Explore
Left and Right Hand Limits
A limit exists only if the function approaches the same value from both the left and the right sides of the point.
The Left Hand Limit (LHL) is the value f(x) approaches as x approaches 'a' from values less than 'a', denoted as limₓ→ₐ⁻ f(x). The Right Hand Limit (RHL) is the value approached from values greater than 'a', denoted as limₓ→ₐ⁺ f(x).
If LHL = RHL = l, then the limit exists and limₓ→ₐ f(x) = l. If the LHL and RHL are different, we say the limit does not exist at that point.
For example, consider a step function where f(x) = 1 for x ≤ 0 and f(x) = 2 for x > 0. At x = 0, the LHL is 1 and the RHL is 2. Since 1 ≠ 2, the limit at x = 0 does not exist.
Limit of a Piecewise Function
f(x) = {x - 2 if x < 0; x + 2 if x > 0}
To find the limit at x = 0, we check both sides. LHL: As x approaches 0 from the left, f(x) = x - 2 approaches -2. RHL: As x approaches 0 from the right, f(x) = x + 2 approaches 2. Since -2 ≠ 2, the limit does not exist.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
What is the necessary condition for limₓ→ₐ f(x) to exist?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 5, 6, 10.
04 · Explore
Algebra of Limits
Limits follow specific algebraic rules that allow us to evaluate complex functions by breaking them down into simpler parts.
Theorem 1 states that if the limits of f(x) and g(x) exist at x = a, then the limit of their sum, difference, product, and quotient also exists (provided the denominator's limit is non-zero).
Specifically: 1. lim [f(x) + g(x)] = lim f(x) + lim g(x). 2. lim [f(x) - g(x)] = lim f(x) - lim g(x). 3. lim [f(x) . g(x)] = lim f(x) . lim g(x). 4. lim [f(x) / g(x)] = [lim f(x)] / [lim g(x)].
Additionally, for any constant λ, lim [λ . f(x)] = λ . lim f(x). These rules make it possible to evaluate limits of polynomials and rational functions by direct substitution in most cases.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
Evaluate limₓ→1 (x² + 3x + 2).
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 12, 13.
05 · Explore
Limits of Polynomials and Rational Functions
Polynomials are continuous, meaning their limit at any point is simply the value of the polynomial at that point.
For any polynomial f(x), limₓ→ₐ f(x) = f(a). This is because polynomials are built from sums and products of x, and we know limₓ→ₐ x = a.
For a rational function f(x) = g(x)/h(x), the limit is g(a)/h(a) provided h(a) ≠ 0. If h(a) = 0 and g(a) = 0, we have an 'indeterminate form' 0/0. In this case, we must factorize the polynomials and cancel the common factor (x - a) before evaluating.
Theorem 2 provides a standard result for a common limit form: limₓ→ₐ (xⁿ - aⁿ) / (x - a) = naⁿ⁻¹. This is valid for any positive integer n, and even for rational n when a > 0.
Evaluating Indeterminate Form
limₓ→2 (x² - 4) / (x² - 5x + 6)
Direct substitution gives (4-4)/(4-10+6) = 0/0. Factorizing: (x-2)(x+2) / [(x-2)(x-3)]. Canceling (x-2) for x ≠ 2: limₓ→2 (x+2)/(x-3). Now substitute x=2: (2+2)/(2-3) = 4/(-1) = -4.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
Calculate limₓ→3 (x² - 9) / (x - 3).
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 13, 14, 15, 16.
06 · Explore
Limits of Trigonometric Functions
Trigonometric limits often require special theorems, such as the Sandwich Theorem, to evaluate.
The Sandwich Theorem (or Squeeze Theorem) states that if f(x) ≤ g(x) ≤ h(x) and both f(x) and h(x) approach the same limit 'l' as x approaches 'a', then g(x) must also approach 'l'.
Using geometric proofs and the Sandwich Theorem, we establish two fundamental trigonometric limits: 1. limₓ→₀ (sin x) / x = 1. 2. limₓ→₀ (1 - cos x) / x = 0. Note that x must be measured in radians.
These limits allow us to solve more complex trigonometric problems. For example, limₓ→₀ (tan x) / x = limₓ→₀ (sin x / x) . (1 / cos x) = 1 . 1 = 1.
Limit of sin(4x)/sin(2x)
limₓ→₀ sin(4x) / sin(2x)
Rewrite as [sin(4x)/4x] * [2x/sin(2x)] * [4x/2x]. As x → 0, 4x → 0 and 2x → 0. The first two terms approach 1. The remaining term is 4x/2x = 2. Thus, the limit is 1 * 1 * 2 = 2.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
What is the value of limₓ→₀ (sin 5x) / x?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 18, 19, 20.
07 · Explore
Derivatives from First Principles
The derivative measures the instantaneous rate of change of a function. It is defined formally as a limit.
The derivative of a function f at a point 'a' is defined as f'(a) = limₕ→₀ [f(a + h) - f(a)] / h. This is often called the 'First Principle' of derivatives.
Geometrically, the derivative f'(a) represents the slope of the tangent line to the curve y = f(x) at the point (a, f(a)). As the distance 'h' between two points on a curve approaches zero, the secant line connecting them becomes the tangent line.
If a derivative exists at every point in the domain, we can define a new function f'(x) = limₕ→₀ [f(x + h) - f(x)] / h. Common notations include f'(x), dy/dx, or d/dx[f(x)].
Derivative of f(x) = x²
f'(x) = limₕ→₀ [(x + h)² - x²] / h
Expand the numerator: [x² + 2xh + h² - x²] / h = [2xh + h²] / h. Simplify by dividing by h: 2x + h. As h approaches 0, the limit is 2x. Thus, the derivative of x² is 2x.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
Find the derivative of f(x) = 3x at x = 2 using the definition.
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 24, 25, 26.
08 · Explore
Algebra of Derivatives
Similar to limits, derivatives have rules that simplify the differentiation of combined functions.
Let u and v be functions of x. The following rules apply: 1. Sum/Difference Rule: (u ± v)' = u' ± v'. 2. Product Rule (Leibnitz Rule): (uv)' = u'v + uv'. 3. Quotient Rule: (u/v)′ = (u′v − uv′) / v², provided v ≠ 0.
The Product Rule tells us that the derivative of a product is not simply the product of the derivatives. Instead, it is the derivative of the first times the second, plus the first times the derivative of the second.
The Quotient Rule is used for fractions. A helpful mnemonic is 'Low d-High minus High d-Low, over the square of what's below', where 'High' is the numerator and 'Low' is the denominator.
| Rule Name | Formula |
|---|---|
| Sum Rule | d/dx [f + g] = f' + g' |
| Product Rule | d/dx [f . g] = f'g + fg' |
| Quotient Rule | d/dx [f / g] = (f'g - fg') / g² |
| Constant Multiple | d/dx [λf] = λf' |
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
Find the derivative of x . sin x.
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 28, 29.
09 · Explore
Derivatives of Standard Functions
We can derive general formulas for the derivatives of common mathematical functions.
Power Rule: The derivative of xⁿ is nxⁿ⁻¹ for any real number n. For example, the derivative of x³ is 3x², and the derivative of x¹ is 1.
Trigonometric Derivatives: 1. d/dx (sin x) = cos x. 2. d/dx (cos x) = -sin x. 3. d/dx (tan x) = sec² x. These are derived using trigonometric identities and the first principle definition.
Derivative of a Constant: The derivative of any constant function f(x) = k is 0, because a constant value does not change, so its rate of change is zero.
Derivative of a Polynomial
f(x) = 6x¹⁰⁰ - x⁵⁵ + x
Apply the power rule to each term: d/dx(6x¹⁰⁰) = 6 * 100x⁹⁹ = 600x⁹⁹. d/dx(-x⁵⁵) = -55x⁵⁴. d/dx(x) = 1. The total derivative is 600x⁹⁹ - 55x⁵⁴ + 1.
Pause & Try
Think it through first. Writing and checking your answer is free with an account.
Question
What is the derivative of f(x) = 1/x?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 29, 30, 31, 32.