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Exercise: Chain Rule · Q13

Q.Differentiate y=cos⁡(5x−3)y=\cos(5x-3) with respect to xx.

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Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

Imagine you're assembling a toy. First you put part A into part B, then you put that combined piece into part C. The final toy's position depends on how you moved A, which then affected B, which then affected C. That's exactly what the chain rule captures — how a change in the first variable ripples through a sequence of functions to affect the final output.

Let's make this concrete. Suppose you have a function ff that depends on gg, and gg itself depends on xx:

y=f(g(x))y = f(g(x))

You want to know: if xx changes by a tiny amount, how much does yy change? The answer isn't just f′(g(x))f'(g(x)) — because g(x)g(x) itself changes when xx changes. You have to multiply the two rates:

  • How fast does gg change with respect to xx? That's g′(x)g'(x).
  • How fast does ff change with respect to its input gg? That's f′(g(x))f'(g(x)).

The total effect is the product:

dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

Note

In Leibniz notation, this looks even more natural: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}, where u=g(x)u = g(x). The dudu's "cancel" like fractions — though this is just a helpful memory aid, not a rigorous proof.

The Precise Statement

Chain Rule (single variable): If gg is differentiable at xx and ff is differentiable at g(x)g(x), then the composite function h(x)=f(g(x))h(x) = f(g(x)) is differentiable at xx, and

h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)

That's it. One multiplication. But the power is enormous — it lets you differentiate almost any nested function.

A Simple Example

Differentiate h(x)=sin⁡(3x2)h(x) = \sin(3x^2).

Here f(u)=sin⁡uf(u) = \sin u and g(x)=3x2g(x) = 3x^2. Then:

  • f′(u)=cos⁡uf'(u) = \cos u, so f′(g(x))=cos⁡(3x2)f'(g(x)) = \cos(3x^2)
  • g′(x)=6xg'(x) = 6x

Multiply: h′(x)=cos⁡(3x2)⋅6x=6xcos⁡(3x2)h'(x) = \cos(3x^2) \cdot 6x = 6x \cos(3x^2)

Watch out

The most common mistake is forgetting to multiply by the inner derivative. Students often write ddxsin⁡(3x2)=cos⁡(3x2)\frac{d}{dx} \sin(3x^2) = \cos(3x^2) and stop — that's wrong. The chain rule demands you also multiply by 6x6x.

Why It's Called a "Chain"

Think of a chain of links: x→g→fx \to g \to f. Each link has its own rate of change. To find the total rate from xx to ff, you multiply the rates of each link. If you had three functions — say h(x)=f(g(k(x)))h(x) = f(g(k(x))) — you'd multiply three derivatives:

h′(x)=f′(g(k(x)))⋅g′(k(x))⋅k′(x)h'(x) = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x)

The chain can be as long as you like. Each new function adds one more factor.

The Intuition in One Sentence

The chain rule says: the rate of change of the whole is the product of the rates of change of the parts, evaluated at the right places.

Important

The chain rule is not optional — it's the backbone of calculus. Every derivative of a trigonometric, exponential, logarithmic, or power function that isn't just xnx^n uses it. Master this, and you master differentiation.

The chain rule is one of the most heavily tested formulas in the NCERT Class 12 Continuity and Differentiability chapter, and it underlies nearly every differentiation problem in CBSE boards, JEE Main and JEE Advanced. Whether you're searching 'chain rule differentiation class 12 examples' or 'chain rule important questions for JEE', this f'(g(x))·g'(x) pattern is the formula every subsequent derivative rule in the syllabus builds on.

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