A-Level Maths / Pure Mathematics / Differentiation

Chain Rule

Differentiating composite functions using the chain rule.

Pure Mathematics A2 45 min

Learning Objectives

  • Recognise composite functions that require the chain rule
  • Apply the chain rule to differentiate composite functions
  • Use the chain rule with trigonometric, exponential, and logarithmic functions
  • Combine the chain rule with other differentiation techniques

Key Formulae

dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Key Concepts

The chain rule is used to differentiate composite functions — where one function is applied inside another. Whenever you see a “function of a function”, you need the chain rule.

When to use it

Use the chain rule when differentiating expressions like:

  • (3x+1)5(3x + 1)^5 — a polynomial raised to a power
  • sin(x2)\sin(x^2) — trig of a non-trivial argument
  • e3x+1e^{3x+1} — exponential of a non-trivial argument
  • ln(x2+1)\ln(x^2 + 1) — log of a non-trivial argument

The rule

If y=f(g(x))y = f(g(x)), let u=g(x)u = g(x) so that y=f(u)y = f(u). Then:

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

In function notation: ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Worked Examples

Example 1: Power of a linear expression

Differentiate y=(2x+3)4y = (2x + 3)^4

Let u=2x+3u = 2x + 3, so y=u4y = u^4

dudx=2,dydu=4u3\frac{du}{dx} = 2, \qquad \frac{dy}{du} = 4u^3

dydx=4u32=8(2x+3)3\frac{dy}{dx} = 4u^3 \cdot 2 = 8(2x + 3)^3

Example 2: Trig of a quadratic

Differentiate y=sin(x2)y = \sin(x^2)

Let u=x2u = x^2, so y=sinuy = \sin u

dudx=2x,dydu=cosu\frac{du}{dx} = 2x, \qquad \frac{dy}{du} = \cos u

dydx=cos(x2)2x=2xcos(x2)\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2)

Example 3: Exponential with chain rule

Differentiate y=e3x+1y = e^{3x+1}

Let u=3x+1u = 3x + 1, so y=euy = e^u

dudx=3,dydu=eu\frac{du}{dx} = 3, \qquad \frac{dy}{du} = e^u

dydx=3e3x+1\frac{dy}{dx} = 3e^{3x+1}

Example 4: Natural log (A2)

Differentiate y=ln(x2+1)y = \ln(x^2 + 1)

Let u=x2+1u = x^2 + 1, so y=lnuy = \ln u

dudx=2x,dydu=1u\frac{du}{dx} = 2x, \qquad \frac{dy}{du} = \frac{1}{u}

dydx=2xx2+1\frac{dy}{dx} = \frac{2x}{x^2 + 1}

Example 5: Nested chain rule

Differentiate y=esin(2x)y = e^{\sin(2x)}

Apply the chain rule twice. Outer: e()e^{(\cdot)}, Middle: sin()\sin(\cdot), Inner: 2x2x

dydx=esin(2x)cos(2x)2=2cos(2x)esin(2x)\frac{dy}{dx} = e^{\sin(2x)} \cdot \cos(2x) \cdot 2 = 2\cos(2x)\,e^{\sin(2x)}

Common Patterns

These results follow directly from the chain rule and are worth memorising:

yydydx\frac{dy}{dx}
(ax+b)n(ax + b)^nan(ax+b)n1an(ax + b)^{n-1}
sin(ax+b)\sin(ax + b)acos(ax+b)a\cos(ax + b)
cos(ax+b)\cos(ax + b)asin(ax+b)-a\sin(ax + b)
eax+be^{ax+b}aeax+bae^{ax+b}
ln(ax+b)\ln(ax + b)aax+b\frac{a}{ax + b}

Exam Tips

  • Always clearly define your substitution when showing working
  • Check your answer is in terms of x, not u
  • Look for the "function of a function" structure — brackets, powers of trig/exp, etc.
  • The chain rule often appears combined with product or quotient rule in A2 questions

Specification

Edexcel A Level Maths
Pure: Differentiation > Chain Rule
WJEC A Level Maths
Pure: Differentiation > Chain, Product & Quotient Rules

Resources

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