Chain Rule

This lesson covers the chain rule for differentiating composite functions as required by the Edexcel A-Level Mathematics specification (9MA0). The chain rule is one of the most important differentiation techniques and is used extensively throughout A-Level Mathematics.

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What Is a Composite Function?

A composite function is a "function of a function" — one function is applied inside another.

Examples:

• (3x + 1)⁵ — a power of a linear expression
• sin(2x) — sin applied to 2x
• e^(x²) — exponential of x²
• √(4x - 1) — square root of a linear expression
• ln(x² + 5) — ln of a quadratic

These cannot be differentiated using the power rule alone — they require the chain rule.

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The Chain Rule

If y = f(g(x)) — that is, y is a function of u, where u is a function of x — then:

dy/dx = (dy/du) × (du/dx)

Or equivalently, using function notation:

d/dx[f(g(x))] = f'(g(x)) × g'(x)

In words: "differentiate the outer function (keeping the inner function unchanged), then multiply by the derivative of the inner function."

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The Chain Rule for Powers

The most common application: differentiating expressions of the form (f(x))ⁿ.

d/dx[(f(x))ⁿ] = n(f(x))ⁿ⁻¹ × f'(x)

"Bring down the power, reduce the power by 1, multiply by the derivative of the inside."

Worked Example 1

Differentiate y = (3x + 1)⁵.

Solution: Let u = 3x + 1, so y = u⁵. dy/du = 5u⁴, du/dx = 3 dy/dx = 5u⁴ × 3 = 15(3x + 1)⁴ = 15(3x + 1)⁴

Worked Example 2

Differentiate y = (2x² - 3)⁴.

Solution: Outer function: u⁴ → derivative 4u³ Inner function: 2x² - 3 → derivative 4x

dy/dx = 4(2x² - 3)³ × 4x = 16x(2x² - 3)³

Worked Example 3

Differentiate y = 1/(x² + 1)³.

Solution: Rewrite: y = (x² + 1)⁻³

dy/dx = -3(x² + 1)⁻⁴ × 2x = -6x(x² + 1)⁻⁴ = -6x / (x² + 1)⁴

Worked Example 4

Differentiate y = √(5x - 2).

Solution: Rewrite: y = (5x - 2)^(1/2)

dy/dx = (1/2)(5x - 2)^(-1/2) × 5 = 5 / (2√(5x - 2))

Exam Tip: Always rewrite roots and reciprocals as powers before applying the chain rule. For example, √(f(x)) = (f(x))^(1/2) and 1/(f(x)) = (f(x))⁻¹.

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Chain Rule with Trigonometric Functions

Function	Derivative
sin(kx)	k cos(kx)
cos(kx)	-k sin(kx)
tan(kx)	k sec²(kx)
sin(f(x))	f'(x) cos(f(x))
cos(f(x))	-f'(x) sin(f(x))

Worked Example 5

Differentiate y = sin(3x).

Solution: dy/dx = 3 cos(3x)

Worked Example 6

Differentiate y = cos(x²).

Solution: Outer: cos u → derivative -sin u Inner: x² → derivative 2x

dy/dx = -sin(x²) × 2x = -2x sin(x²)

Worked Example 7

Differentiate y = 4 sin²(x) = 4(sin x)².

Solution: Rewrite as y = 4(sin x)². Outer: 4u² → derivative 8u Inner: sin x → derivative cos x

dy/dx = 8 sin x × cos x = 8 sin x cos x = 4 sin(2x)

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Chain Rule with Exponential Functions

Function	Derivative
eᵏˣ	keᵏˣ
e^(f(x))	f'(x) × e^(f(x))

Worked Example 8

Differentiate y = e^(3x²+1).

Solution: f(x) = 3x² + 1, f'(x) = 6x

dy/dx = 6x × e^(3x²+1) = 6xe^(3x²+1)

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Chain Rule with Logarithmic Functions

Function	Derivative
ln(kx)	1/x
ln(f(x))	f'(x)/f(x)

Worked Example 9

Differentiate y = ln(x³ + 2x).

Solution: f(x) = x³ + 2x, f'(x) = 3x² + 2

dy/dx = (3x² + 2)/(x³ + 2x)

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Multiple Applications of the Chain Rule

Sometimes you need to apply the chain rule more than once (nested functions).

Worked Example 10

Differentiate y = e^(sin x).

Solution: Outer: eᵘ → derivative eᵘ Inner: sin x → derivative cos x

dy/dx = e^(sin x) × cos x = cos x × e^(sin x)

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Finding Tangents and Normals Using the Chain Rule

Worked Example 11

Find the equation of the tangent to y = (2x - 1)³ at the point where x = 1.

Solution: At x = 1: y = (2(1) - 1)³ = 1³ = 1. Point is (1, 1).

dy/dx = 3(2x - 1)² × 2 = 6(2x - 1)²

At x = 1: dy/dx = 6(1)² = 6

Tangent: y - 1 = 6(x - 1) y = 6x - 5

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Common Mistakes

Mistake	Correction
d/dx(sin 2x) = cos 2x	Missing the chain rule factor: d/dx(sin 2x) = 2 cos 2x
d/dx((3x)⁴) = 4(3x)³	Missing the inner derivative: d/dx((3x)⁴) = 4(3x)³ × 3 = 12(3x)³
Forgetting to multiply by the inner derivative	Always check: "Did I multiply by the derivative of the inside?"
Applying chain rule when not needed	y = x⁵ does not need the chain rule — just use the power rule directly

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Summary

• The chain rule: dy/dx = (dy/du) × (du/dx), or d/dx[f(g(x))] = f'(g(x)) × g'(x).
• "Differentiate the outside, keep the inside, multiply by the derivative of the inside."
• For (f(x))ⁿ: bring down the power, reduce by 1, multiply by f'(x).
• The chain rule applies to powers, trig, exponential, and logarithmic functions.
• Always rewrite roots and reciprocals as powers before differentiating.

A-Level Deep Dive: Chain Rule

Spec mapping

Edexcel 9MA0 specification section 9 — Differentiation covers using the chain rule, including problems involving connected rates of change and inverse functions (refer to the official specification document for exact wording). The chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$dxdy​=dudy​⋅dxdu​ is the engine that converts every "simple" differentiation rule into a tool for composite functions $y = f(g(x))$y=f(g(x)). It is examined explicitly on Paper 1 (Pure) and Paper 2 (Pure), and arises synoptically with the product rule (when the chain output is one factor of a product), with implicit differentiation (where chain rule is what makes $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$dxd​(yn)=nyn−1dxdy​ work), with parametric differentiation ($\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​ is chain rule rearranged), and with integration by substitution (the reverse direction). The Edexcel formula booklet does not state the chain rule — it must be memorised, although the booklet does provide derivatives of $\sin$sin, $\cos$cos, $\tan$tan, $e^x$ex and $\ln x$lnx used inside the chain.

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Worked example with full mark scheme

Question (8 marks):

(a) Given $y = \bigl(\sin(x^3)\bigr)^2$y=(sin(x3))2, find $\dfrac{dy}{dx}$dxdy​, simplifying your answer. (5)

(b) Hence find the exact value of $\dfrac{dy}{dx}$dxdy​ when $x = \sqrt[3]{\pi/6}$x=3π/6​. (3)

Solution with mark scheme:

(a) Step 1 — identify the nesting. The function is a triple composition: outer $u^2$u2, middle $\sin(\cdot)$sin(⋅), inner $x^3$x3. Set $u = \sin(x^3)$u=sin(x3) and $v = x^3$v=x3, so $y = u^2$y=u2, $u = \sin(v)$u=sin(v), $v = x^3$v=x3.

M1 — recognising and writing down the composition. Awarded for any clear identification of the layers (substitutions, brackets, or words). Candidates who attempt $y = \sin^2(x^3) = \sin(x^3) \cdot \sin(x^3)$y=sin2(x3)=sin(x3)⋅sin(x3) and apply the product rule earn the same M1, although the route is longer.

Step 2 — differentiate each layer.

$\dfrac{dy}{du} = 2u, \qquad \dfrac{du}{dv} = \cos(v), \qquad \dfrac{dv}{dx} = 3x^2$dudy​=2u,dvdu​=cos(v),dxdv​=3x2

M1 — correct derivative of each layer. The middle layer $\dfrac{d}{dv}\sin(v) = \cos(v)$dvd​sin(v)=cos(v) is the one most often slipped (sign error $-\cos$−cos or differentiation in degrees not radians).

Step 3 — apply the chain rule.

$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dv} \cdot \dfrac{dv}{dx} = 2u \cdot \cos(v) \cdot 3x^2$dxdy​=dudy​⋅dvdu​⋅dxdv​=2u⋅cos(v)⋅3x2

M1 — correct triple product (or two applications of the two-layer chain rule).

Step 4 — back-substitute.

$\dfrac{dy}{dx} = 2\sin(x^3) \cdot \cos(x^3) \cdot 3x^2 = 6x^2 \sin(x^3)\cos(x^3)$dxdy​=2sin(x3)⋅cos(x3)⋅3x2=6x2sin(x3)cos(x3)

A1 — correct unsimplified expression in $x$x.

Step 5 — simplify using the double-angle identity $\sin(2A) = 2\sin A \cos A$sin(2A)=2sinAcosA:

$6x^2 \sin(x^3)\cos(x^3) = 3x^2 \cdot 2\sin(x^3)\cos(x^3) = 3x^2 \sin(2x^3)$6x2sin(x3)cos(x3)=3x2⋅2sin(x3)cos(x3)=3x2sin(2x3)

A1 — fully simplified form. The double-angle simplification is the AO2 reasoning step — examiners reward it because it demonstrates synoptic command across sections 5 and 9.

(b) Step 1 — substitute $x = \sqrt[3]{\pi/6}$x=3π/6​, so $x^3 = \pi/6$x3=π/6 and $2x^3 = \pi/3$2x3=π/3. Also $x^2 = (\pi/6)^{2/3}$x2=(π/6)2/3.

M1 — correct substitution into the simplified derivative.

Step 2 — evaluate $\sin(\pi/3) = \sqrt{3}/2$sin(π/3)=3​/2:

$\dfrac{dy}{dx} = 3 \cdot \Bigl(\dfrac{\pi}{6}\Bigr)^{2/3} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2} \Bigl(\dfrac{\pi}{6}\Bigr)^{2/3}$dxdy​=3⋅(6π​)2/3⋅23​​=233​​(6π​)2/3

A1 (AO1.1b) — correct exact substitution.

A1 (AO2.5) — exact form retained (no decimal approximation).

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A curve has equation $y = \ln(x^2 + 1) + e^{\sin x}$y=ln(x2+1)+esinx for $x \in \mathbb{R}$x∈R.

(a) Find $\dfrac{dy}{dx}$dxdy​. (4)

(b) Hence find the gradient of the curve at $x = 0$x=0. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying the chain rule to $\ln(x^2 + 1)$ln(x2+1): $\dfrac{d}{dx}\ln(u) = \dfrac{1}{u}\dfrac{du}{dx}$dxd​ln(u)=u1​dxdu​, with $u = x^2 + 1$u=x2+1, $\dfrac{du}{dx} = 2x$dxdu​=2x, giving $\dfrac{2x}{x^2 + 1}$x2+12x​.
• M1 (AO1.1a) — applying the chain rule to $e^{\sin x}$esinx: $\dfrac{d}{dx}e^{u} = e^{u}\dfrac{du}{dx}$dxd​eu=eudxdu​, with $u = \sin x$u=sinx, $\dfrac{du}{dx} = \cos x$dxdu​=cosx, giving $\cos(x) \cdot e^{\sin x}$cos(x)⋅esinx.
• A1 (AO1.1b) — sum: $\dfrac{dy}{dx} = \dfrac{2x}{x^2 + 1} + \cos(x) \cdot e^{\sin x}$dxdy​=x2+12x​+cos(x)⋅esinx.
• A1 (AO2.5) — answer left in exact, unsimplified form (no spurious tidying that loses a term).

(b)

• M1 (AO1.1b) — substituting $x = 0$x=0: $\dfrac{2 \cdot 0}{0 + 1} + \cos(0) \cdot e^{\sin 0} = 0 + 1 \cdot e^0 = 1$0+12⋅0​+cos(0)⋅esin0=0+1⋅e0=1.
• A1 (AO1.1b) — gradient $= 1$=1.

Total: 6 marks split AO1 = 5, AO2 = 1. Edexcel uses chain-rule questions to test structural recognition: can the candidate see that $\ln(x^2 + 1)$ln(x2+1) and $e^{\sin x}$esinx each require chain rule independently before the sum is taken?

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Synoptic links

Connects to:

1. Composite functions (Section 2): the chain rule is the differential analogue of function composition $(f \circ g)(x) = f(g(x))$(f∘g)(x)=f(g(x)). Every chain-rule problem is a "decompose, differentiate, recompose" exercise — the same skill that underpins inverse-function questions.

2. Product rule (Section 9): chain and product rules combine constantly. Differentiating $x^2 \sin(3x)$x2sin(3x) requires the product rule, but the $\sin(3x)$sin(3x) factor needs the chain rule first. The order matters: differentiate inner factors before assembling the product-rule formula.

3. Implicit differentiation (Section 9): when differentiating $y^3 + xy = 7$y3+xy=7 implicitly, the term $y^3$y3 becomes $3y^2 \cdot \dfrac{dy}{dx}$3y2⋅dxdy​ — that extra factor of $\dfrac{dy}{dx}$dxdy​ is the chain rule treating $y$y as a function of $x$x. Without chain rule, implicit differentiation collapses.

4. Parametric differentiation (Section 9): the formula $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​ is the chain rule rearranged: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$dtdy​=dxdy​⋅dtdx​ inverted. Recognising this turns parametric problems from rote into reasoned.

5. Integration by substitution (Section 8): $\int f(g(x)) g'(x)\,dx = \int f(u)\,du$∫f(g(x))g′(x)dx=∫f(u)du where $u = g(x)$u=g(x) is the chain rule run backwards. Every "by substitution" integral is an undone chain-rule derivative — spotting the inner function and its derivative inside the integrand is exactly the chain-rule skill in reverse. A practical consequence: students who internalise the chain rule's $f'(g(x)) g'(x)$f′(g(x))g′(x) shape will instantly recognise integrands like $2x \cos(x^2)$2xcos(x2) or $\dfrac{2x}{x^2 + 1}$x2+12x​ as "obvious substitution" — these are exact derivatives produced by $\sin(x^2)$sin(x2) and $\ln(x^2 + 1)$ln(x2+1) respectively. The skill transfers without re-learning.

6. Connected rates of change (Section 9, modelling): real-world problems such as "a spherical balloon's radius is increasing at $0.2$0.2 cm/s; how fast is its volume increasing when $r = 5$r=5?" are pure chain rule applied to $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$dtdV​=drdV​⋅dtdr​. Identifying the right two quantities to chain together is the AO3 modelling skill — and it is the same mechanical pattern as $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$dxdy​=dudy​⋅dxdu​ in disguise.

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Mark-scheme literacy

Chain-rule questions on 9MA0 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–75%	Identifying composition; differentiating each layer correctly; assembling the product of derivatives
AO2 (reasoning / interpretation)	20–30%	Choosing whether to simplify (e.g. via double-angle identity); leaving exact form; presenting answer in requested style
AO3 (problem-solving)	5–10%	Connected-rates problems where the chain rule must be set up from a real-world scenario

Examiner-rewarded phrasing: "Let $u = \ldots$u=… and $y = \ldots$y=…, so $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$dxdy​=dudy​⋅dxdu​"; "differentiating the outer function and multiplying by the derivative of the inner function"; "in exact form". Phrases that lose marks: differentiating "the outside" without ever writing the inner derivative; combining $2x$2x and $x^2 + 1$x2+1 into a single $\ln$ln argument by mistake; leaving an answer in terms of $u$u when the question asks for $\dfrac{dy}{dx}$dxdy​.

A specific Edexcel pattern: questions that ask "find $\dfrac{dy}{dx}$dxdy​" expect the answer in terms of $x$x only — back-substitution is mandatory. Leaving $2u \cdot 3x^2$2u⋅3x2 without writing $u = \sin(x^3)$u=sin(x3) explicitly costs the final A1.

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Grade-band model answers

3-mark question

Question: Differentiate $y = (2x + 1)^5$y=(2x+1)5 with respect to $x$x.

Grade C response (~210 words):

Let $u = 2x + 1$u=2x+1. Then $y = u^5$y=u5.

$\dfrac{dy}{du} = 5u^4$dudy​=5u4, and $\dfrac{du}{dx} = 2$dxdu​=2.

By the chain rule, $\dfrac{dy}{dx} = 5u^4 \cdot 2 = 10u^4 = 10(2x + 1)^4$dxdy​=5u4⋅2=10u4=10(2x+1)4.