Chain Rule | JEE Main Calculus Crash Course | Mathematicon

Chain Rule: Differentiating Composite Functions for JEE Main

Hello JEE aspirants! The chain rule is your key to unlocking differentiation of composite functions, which frequently appear in JEE Main. Mastering this rule will significantly boost your problem-solving speed and accuracy. Let's dive in!

What is the Chain Rule?

Imagine you have a function within a function – a composite function. Think of it like peeling an onion. The chain rule helps you differentiate such functions layer by layer. It essentially states that the derivative of a composite function is the product of the derivative of the outer function (evaluated at the inner function) and the derivative of the inner function.

Intuition: Consider $y = (x^2 + 1)^3$ . Here, we have an inner function $u = x^2 + 1$ and an outer function $y = u^3$ . When $x$ changes, it affects $u$ , and then the change in $u$ affects $y$ . The chain rule helps us quantify how a small change in $x$ propagates through these layers to ultimately affect $y$ .

The Core Formula

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Explanation: This formula tells us that the rate of change of $y$ with respect to $x$ ( $dy/dx$ ) is equal to the rate of change of $y$ with respect to an intermediate variable $u$ ( $dy/du$ ) multiplied by the rate of change of $u$ with respect to $x$ ( $du/dx$ ). This intermediate variable $u$ is often the "inner function".

Another Common Notation

\text{If } y = f(g(x)), \text{ then } \frac{dy}{dx} = f'(g(x)) \cdot g'(x)

Explanation: This is perhaps the most frequently used form of the chain rule. Let's break it down. If we have a function $y$ that is the result of applying function $f$ to function $g(x)$ , then the derivative of $y$ with respect to $x$ is the derivative of $f$ (evaluated at $g(x)$ ) multiplied by the derivative of $g(x)$ .

Example: Let $y = \sin(x^2)$ . Here, $f(u) = \sin(u)$ and $g(x) = x^2$ . So, $f'(u) = \cos(u)$ and $g'(x) = 2x$ . Applying the chain rule:

\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2)

Yet Another Way to Write It

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Explanation: This is essentially the same as the previous formula, just written differently. It emphasizes the operation of taking the derivative of the composite function $f(g(x))$ .

Example: Let's use the same example, $y = \sin(x^2)$ . Applying this formula directly:

\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x = 2x \cos(x^2)

Differentiating Composite Functions: A Step-by-Step Approach

Identify the outer and inner functions: In $y = \cos(e^x)$ , the outer function is $\cos(u)$ and the inner function is $e^x$ .
Differentiate the outer function: Find $f'(u)$ . In our example, $f'(u) = -\sin(u)$ .
Differentiate the inner function: Find $g'(x)$ . In our example, $g'(x) = e^x$ .
Apply the chain rule: Multiply $f'(g(x))$ by $g'(x)$ . $\frac{dy}{dx} = -\sin(e^x) \cdot e^x = -e^x \sin(e^x)$

Multiple Compositions (Nested Chain Rule)

Sometimes, you'll encounter functions with multiple layers of composition, like $y = \tan(\sin(x^3))$ . The chain rule extends naturally to handle these situations. You simply apply the chain rule repeatedly, working from the outermost layer inwards.

Example: $y = \tan(\sin(x^3))$

Let $u = \sin(x^3)$ , so $y = \tan(u)$ . Then $\frac{dy}{du} = \sec^2(u)$ .
Now, let $v = x^3$ , so $u = \sin(v)$ . Then $\frac{du}{dv} = \cos(v)$ .
Finally, $\frac{dv}{dx} = 3x^2$ .

Applying the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} = \sec^2(\sin(x^3)) \cdot \cos(x^3) \cdot 3x^2 = 3x^2 \cos(x^3) \sec^2(\sin(x^3))

Tip: When dealing with multiple compositions, break down the problem into smaller, manageable steps. Introduce intermediate variables to keep track of the layers.

Tips for Solving Problems

Practice, practice, practice: The more you apply the chain rule, the more comfortable you'll become.
Start simple: Begin with basic composite functions and gradually work your way up to more complex ones.
Write it out: Explicitly identify the outer and inner functions, and their derivatives, before applying the formula.
Check your work: Make sure you've differentiated all the layers of the composite function.

Common Mistake: Forgetting to differentiate the inner function. Remember, it's a product:

f'(g(x)) \cdot g'(x)

, not just

f'(g(x))

Common Mistake: Incorrectly identifying the outer and inner functions, especially in complex expressions. Take your time to decompose the function step by step.

JEE-Specific Tricks

While there aren't specific "tricks" unique to JEE for the chain rule, a strong understanding of trigonometric identities and algebraic manipulation can significantly simplify problems.

Example: Differentiate $y = \sqrt{\frac{1 - \cos(2x)}{1 + \cos(2x)}}$

Directly applying the chain rule would be messy. However, using the identities $1 - \cos(2x) = 2\sin^2(x)$ and $1 + \cos(2x) = 2\cos^2(x)$ , we can simplify:

y = \sqrt{\frac{2\sin^2(x)}{2\cos^2(x)}} = \sqrt{\tan^2(x)} = |\tan(x)|

Now, depending on the interval of $x$ , you can differentiate either $\tan(x)$ or $-\tan(x)$ , making the problem much easier.

Tip: Look for opportunities to simplify the function *before* applying the chain rule. Trigonometric identities, algebraic manipulations, and logarithmic properties can often make differentiation much easier.

Mastering the chain rule is crucial for success in JEE Main. Understand the underlying concepts, practice diligently, and watch out for common mistakes. Good luck!