Mastering the Chain Rule for Competitive Exams

Welcome to this module on the Chain Rule, a fundamental concept in differential calculus essential for success in competitive exams like JEE. The Chain Rule allows us to differentiate composite functions – functions within functions. Understanding this rule is key to solving complex differentiation problems efficiently.

What is a Composite Function?

A composite function is formed when one function is substituted into another. For example, if $f(x) = x^2$ and $g(x) = \sin(x)$ , then a composite function could be $f(g(x)) = (\sin(x))^2$ or $g(f(x)) = \sin(x^2)$ . We denote a composite function as $(f \circ g)(x) = f(g(x))$ .

h(x) = \sqrt{x^3 + 1}

, what are the 'outer' and 'inner' functions?

The outer function is $f(u) = \sqrt{u}$ and the inner function is $g(x) = x^3 + 1$ . So, $h(x) = f(g(x))$ .

The Chain Rule: The Core Idea

The derivative of a composite function is the derivative of the outer function (evaluated at the inner function) multiplied by the derivative of the inner function.

Think of it as peeling an onion. You differentiate the outermost layer first, then multiply by the derivative of the next layer, and so on, until you reach the innermost function.

Mathematically, if $y = f(u)$ and $u = g(x)$ , then the derivative of $y$ with respect to $x$ is given by: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ . This can also be written as $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$ .

Let's visualize the Chain Rule. Imagine a machine where you input 'x'. This machine passes 'x' through an inner function $g(x)$ to produce 'u'. Then, this 'u' is fed into an outer function $f(u)$ to produce the final output 'y'. The Chain Rule tells us how a small change in 'x' affects 'y' by considering the intermediate change in 'u'. The rate of change of y with respect to x ( $\frac{dy}{dx}$ ) is the product of the rate of change of y with respect to u ( $\frac{dy}{du}$ ) and the rate of change of u with respect to x ( $\frac{du}{dx}$ ).

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Applying the Chain Rule: Examples

Let's work through a common example: Differentiate $y = (2x^3 + 5)^4$ .

Identify the outer and inner functions:
- Outer function: $f(u) = u^4$
- Inner function: $g(x) = 2x^3 + 5$
- So, $y = f(g(x))$ .

Find the derivatives of the individual functions:
- $\frac{dy}{du} = f'(u) = 4u^3$
- $\frac{du}{dx} = g'(x) = 6x^2$

Apply the Chain Rule formula:
- $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (4u^3) \cdot (6x^2)$

Substitute the inner function back for 'u':
- Since $u = 2x^3 + 5$ , substitute this back into the expression:
- $\frac{dy}{dx} = 4(2x^3 + 5)^3 \cdot (6x^2)$
- $\frac{dy}{dx} = 24x^2 (2x^3 + 5)^3$

What is the derivative of

y = \sin(x^2)

using the Chain Rule?

Outer function $f(u) = \sin(u)$ , inner function $g(x) = x^2$ . $f'(u) = \cos(u)$ , $g'(x) = 2x$ . So, $\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2)$ .

Chain Rule with Trigonometric and Exponential Functions

The Chain Rule is particularly useful when dealing with composite trigonometric, exponential, and logarithmic functions. For instance, differentiating $y = e^{\cos(x)}$ requires applying the Chain Rule.

Example: Differentiate $y = e^{\cos(x)}$

Outer function: $f(u) = e^u \implies f'(u) = e^u$
Inner function: $g(x) = \cos(x) \implies g'(x) = -\sin(x)$
$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = e^{\cos(x)} \cdot (-\sin(x)) = -\sin(x) e^{\cos(x)}$

Remember to always differentiate the 'outermost' function first, keeping the 'inner' function intact, and then multiply by the derivative of the 'inner' function.

Multiple Applications of the Chain Rule

For functions composed of more than two functions, like $y = f(g(h(x)))$ , you apply the Chain Rule iteratively: $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$ .

Example: Differentiate $y = \sin(e^{x^2})$

Let $u = e^{x^2}$ and $v = x^2$ .
Then $y = \sin(u)$ , $u = e^v$ , $v = x^2$ .
$\frac{dy}{du} = \cos(u)$
$\frac{du}{dv} = e^v$
$\frac{dv}{dx} = 2x$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} = \cos(u) \cdot e^v \cdot 2x$
Substitute back: $\frac{dy}{dx} = \cos(e^{x^2}) \cdot e^{x^2} \cdot 2x = 2x e^{x^2} \cos(e^{x^2})$

Key Takeaways for Competitive Exams

To excel in competitive exams:

Practice consistently: Solve a variety of problems involving the Chain Rule.
Identify functions: Clearly distinguish between outer and inner functions.
Systematic approach: Follow the step-by-step process of finding derivatives and multiplying them.
Watch for common pitfalls: Be careful with signs (especially with trigonometric functions) and powers.

Learning Resources

Khan Academy: The Chain Rule(video)

A clear video explanation of the chain rule with examples, perfect for building foundational understanding.

Paul's Online Math Notes: The Chain Rule(documentation)

Comprehensive notes covering the chain rule, including multiple examples and practice problems.

Brilliant.org: Chain Rule(blog)

An interactive explanation of the chain rule, focusing on intuition and application.

YouTube: Chain Rule Explained - Calculus(video)

A visual and step-by-step tutorial on how to apply the chain rule for differentiation.

Mathway: Chain Rule Examples(documentation)

Provides a collection of solved examples demonstrating the application of the chain rule to various functions.

Wikipedia: Chain Rule(wikipedia)

A detailed overview of the chain rule, its history, and generalizations in calculus.

Calculus Made Easy: Chain Rule Practice(blog)

A blog post with practical tips and examples for mastering the chain rule.

Symbolab: Chain Rule Calculator and Examples(documentation)

An online tool to practice and verify chain rule differentiation, with step-by-step solutions.

Art of Problem Solving: Chain Rule(documentation)

A resource focused on problem-solving strategies for calculus, including detailed explanations of the chain rule.

MIT OpenCourseware: Calculus - Chain Rule(video)

Lecture videos from MIT covering the chain rule, offering a rigorous academic perspective.