Learn9MA0Paper 1 · Pure Mathematics 1DifferentiationThe chain rule

Differentiation · Subtopic 6 of 10

The Chain Rule

How to differentiate a function of a function, such as $(2x-5)^4$ or $\sqrt{x^2+5}$ , without multiplying out: differentiate the outside, keep the inside, and multiply by the derivative of the inside.

Cambridge 9709 Paper 1Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Recognise a composite function and identify the "inside" and "outside".
Use $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}$ .
Differentiate $(ax+b)^n$ , $(f(x))^n$ , roots and reciprocals of brackets.
Use the chain rule to find gradients at points.

Pause and think

Gear A turns $3$ times for every turn of gear B, and gear B turns $2$ times for every turn of gear C. How many times does A turn for each turn of C?

Section 2 of 14

Core idea

Start hereGears inside gears

A small gear turns $3$ times for each turn of a middle gear, and the middle gear turns $5$ times for each turn of a big one. How fast does the small gear turn compared with the big one? $3\times5=15$ times. Rates of change multiply in the same way when one function sits inside another. That's the chain rule.

The big idea

For a composite $y=f(g(x))$ , let $u=g(x)$ . Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}$ : differentiate the outside (leaving the inside alone), then multiply by the derivative of the inside.

1. Spotting the layers

Two layers: inside then outside

\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}

: the rates of the two layers multiply, like gears in a chain.

$y=(3x+1)^5$ is built in two steps: inner $u=3x+1$ , outer $y=u^5$ . Expanding to the fifth power would be painful.

Why the rates multiply

$\frac{\delta y}{\delta x}=\frac{\delta y}{\delta u}\times\frac{\delta u}{\delta x}$
Small changes: the $\delta u$ cancels (ordinary fractions, $\delta u\neq0$ ).
$\delta x\to0\Rightarrow\delta u\to0$
A small change in $x$ makes a small change in $u$ .
$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$
Take limits of each fraction.

2. Examples

Worked example $y=(3x+1)^5$

Find $\dfrac{dy}{dx}$ .

Goal: The derivative without expanding.
Strategy: Inner $u=3x+1$ ; outer $u^5$ .

$\frac{du}{dx}=3,\qquad\frac{dy}{du}=5u^4$
Differentiate each layer.
$\frac{dy}{dx}=5u^4\times3=15(3x+1)^4$
Multiply; put $u$ back.

Answer

15(3x+1)^4

✓ CheckAt

x=0

: chain rule gives

15

. Direct:

(3x+1)^5\approx1+15x+\ldots

near

0

, gradient

15

✓.

Worked example $y=\sqrt{x^2+9}$

Find $\dfrac{dy}{dx}$ and the gradient at $x=4$ .

Goal: Formula and value.
Strategy: Rewrite as a power.

$y=(x^2+9)^{1/2}$
Root as a power.
$\frac{dy}{dx}=\tfrac12(x^2+9)^{-1/2}\times2x$
Outer derivative × inner derivative.
$=\frac{x}{\sqrt{x^2+9}}$
Simplify.
$x=4:\ \frac45$
$\sqrt{25}=5$ .

Answer

\dfrac{x}{\sqrt{x^2+9}}

;

\frac45

x=4

✓ CheckFor large

x

the curve is almost

y=x

, gradient near

1

, and

\frac x{\sqrt{x^2+9}}\to1

✓.

Worked example $y=\dfrac4{(2-5x)^3}$

Differentiate.

Goal: $\frac{dy}{dx}$ .
Strategy: Negative power, chain rule; watch the inner $-5$ .

$y=4(2-5x)^{-3}$
Rewrite.
$\frac{dy}{dx}=4(-3)(2-5x)^{-4}\times(-5)$
Power down, reduce, times inner derivative.
$=\frac{60}{(2-5x)^4}$
$4\times(-3)\times(-5)=60$ .

Answer

\dfrac{60}{(2-5x)^4}

✓ CheckPositive everywhere:

y

is increasing on each branch ✓ (the denominator shrinks as

x

approaches

0.4

Common misconceptionForgetting the inner derivative

\frac{d}{dx}(3x+1)^5\neq5(3x+1)^4

. The factor

3

from the inside is essential.

Pause and think

Gradient of $y=(x^2-3)^4$ at $x=2$ ?

Summary

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ .
Outside first, times inside's derivative.

Composite function: A function inside another.
Chain rule: Rule for differentiating composites.

HistoryNotation that does the work. Leibniz wrote the chain rule in 1676 almost exactly as we do now. His notation makes it look as if the

du

s cancel in

\dfrac{dy}{du}\times\dfrac{du}{dx}

. They are not really fractions, but the notation is so well designed that it guides you to the right answer.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in