IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Chain Rule

The chain rule differentiates a “function of a function” — anything where x doesn’t appear alone, like sin(x² + 1), e^{cos x}, or (3x − 7)⁵. The recipe: differentiate the outer function, leaving the inner alone, then multiply by the derivative of the inner. Once you’ve drilled it, you’ll do it in one line without the “let u = …” substitution.

📘 What you need to know

Chain rule formula: if y = g(u) and u = f(x), then dy/dx = (dy/du) × (du/dx) — given in the formula booklet.
Function notation: if y = g(f(x)), then dy/dx = g′(f(x)) · f′(x).
When to use: composite functions — x “doesn’t appear alone”. sin(3x + 2) needs it; sin x doesn’t.
Mental shortcut: “differentiate the outer ignoring the inner, then multiply by the derivative of the inner.”
Power inside: d/dx [f(x)]ⁿ = n[f(x)]ⁿ⁻¹ · f′(x) — the most common chain-rule pattern.
Trig/exp/log inside: each standard derivative gets multiplied by the inner’s derivative — see the table below.
Nested chain: chain rule may need to be applied twice or more for things like e^sin(2x).
Composite vs product: sin(x) cos(x) is a product (next note); sin(cos x) is a composite — chain rule.

The rule and how to spot it

Chain rule (formula booklet) dydx = dydu × dudx

Function notation if y = g(f(x)) then y′ = g′(f(x)) · f′(x)

Spotting test: does x appear alone, or is it inside something? sin(3x²) — the x is wrapped in 3x², which is inside sin → composite → chain rule. (2x + 5)⁴ — the x is wrapped in 2x + 5, which is inside ()⁴ → composite → chain rule.

Standard chain-rule patterns

Memorise these five — they cover the vast majority of HL exam questions. The pattern is always the same: standard derivative of the outer × derivative of the inner.

Function y	Derivative dy/dx
(f(x))ⁿ	n · f′(x) · (f(x))^{n − 1}
e^f(x)	f′(x) · e^f(x)
ln(f(x))	f′(x) / f(x)
sin(f(x))	f′(x) · cos(f(x))
cos(f(x))	−f′(x) · sin(f(x))

🧭 Recipe — apply the chain rule

Identify outer and inner: outer is the “wrapper” (sin, e^□, ln, ()ⁿ); inner is the expression in x.
Differentiate the outer first, leaving the inner unchanged.
Multiply by f′(x) — the derivative of the inner.
Simplify if straightforward — factor common terms when factoring is asked or natural.
If nested (chain inside chain), apply the rule again to the inner derivative.

Worked examples

WE 1

Power chain rule — bracket to a power

Find the derivative of y = (3x² + 2x − 5)⁴.

Identify outer and inner outer: u⁴ inner: u = 3x² + 2x − 5 f′(x) = 6x + 2 Apply pattern: n · f′(x) · (f(x))^(n−1) dy/dx = 4 · (6x + 2) · (3x² + 2x − 5)³ Factor 2 from (6x + 2) for tidiness = 8(3x + 1)(3x² + 2x − 5)³ dy/dx = 8(3x + 1)(3x² + 2x − 5)³ power drops by 1 (4 → 3); 4 multiplies in; f′(x) multiplies in. Pattern: n · f′ · (f)^(n−1)

WE 2

Trig with polynomial inside

Find the derivative of y = sin(4x³ − x).

Identify outer and inner outer: sin(u) inner: u = 4x³ − x f′(x) = 12x² − 1 Apply pattern: f′(x) cos(f(x)) dy/dx = (12x² − 1) · cos(4x³ − x) dy/dx = (12x² − 1) cos(4x³ − x) “differentiate sin, ignore inside” → cos(4x³ − x), then “× derivative of inside” → ×(12x² − 1)

WE 3

Exponential with polynomial inside — tangent at a point

A curve has equation y = e^{x² − 3x}. Find the equation of the tangent at the point where x = 2.

Find y at x = 2 y = e^(4 − 6) = e⁻² point: (2, e⁻²) Differentiate using chain rule outer: e^u inner: u = x² − 3x f′(x) = 2x − 3 dy/dx = (2x − 3) e^(x² − 3x) Gradient at x = 2 m = (4 − 3) e⁻² = e⁻² Apply point-slope form y − e⁻² = e⁻²(x − 2) y = e⁻² · x − 2 e⁻² + e⁻² = e⁻²(x − 1) Tangent: y = e⁻²(x − 1) also written as y = (x − 1)/e². Verify: at x = 2, y = 1·e⁻² = e⁻² ✓

WE 4

Logarithm with polynomial inside

Find the derivative of y = ln(5x³ + 2x + 1).

Identify outer and inner outer: ln(u) inner: u = 5x³ + 2x + 1 f′(x) = 15x² + 2 Apply pattern: f′(x) / f(x) dy/dx = (15x² + 2) / (5x³ + 2x + 1) dy/dx = (15x² + 2)/(5x³ + 2x + 1) ln(stuff) → derivative of stuff over stuff. Numerator = derivative of inside; denominator = inside itself

WE 5

Nested — chain rule applied twice

Find the derivative of y = e^sin(2x).

Outer is e^u where u = sin(2x). Apply chain rule (1st time) dy/dx = e^(sin 2x) · d/dx [sin(2x)] Differentiate sin(2x) — chain rule again (2nd time) d/dx [sin(2x)] = 2 cos(2x) Combine dy/dx = e^(sin 2x) · 2 cos(2x) dy/dx = 2 cos(2x) · e^(sin 2x) work outside-in: peel off e^□ first, then handle sin(2x). Each layer adds a multiplication

WE 6

Multi-part: derivative, stationary points, value at x = 1

The curve y = (3x² − 5)⁴.

(a) Find dy/dx in factored form. (b) Hence find the x-coordinates of all stationary points. (c) Find the gradient at x = 1.

(a) Apply chain rule: outer = u⁴, inner = 3x² − 5, f′ = 6x dy/dx = 4 · (6x) · (3x² − 5)³ = 24x(3x² − 5)³ (b) Stationary points: dy/dx = 0 24x(3x² − 5)³ = 0 → x = 0 OR 3x² − 5 = 0 → x² = 5/3 → x = ±√(5/3) = ±√15/3 (c) Substitute x = 1 into dy/dx dy/dx = 24(1)(3 − 5)³ = 24(−8) = −192 (a) 24x(3x² − 5)³; (b) x = 0, ±√15/3; (c) gradient = −192 factored form makes (b) trivial — never expand a chain-rule answer if the question doesn’t ask

💡 Top tips

Mental mantra: “differentiate the outer ignoring the inner, then multiply by the derivative of the inner.” Say it in your head every time.
Don’t expand brackets like (3x² − 5)⁴ before differentiating — chain rule keeps it factored, which is essential for finding stationary points.
Linear inside is the easiest case: d/dx sin(ax + b) = a cos(ax + b) — just multiply by the constant a.
Nested chain: peel the layers one at a time. Each chain rule application multiplies one new derivative.
Spot it fast: if x is wrapped inside anything (brackets, sin, e^□, ln), you need chain rule.

⚠ Common mistakes

Forgetting f′(x) — the most common chain-rule slip. Differentiating sin(3x² + 1) and writing just cos(3x² + 1) with no ×6x = wrong.
Differentiating the inner instead of multiplying it: the inner stays inside the outer’s derivative — you ALSO multiply by f′(x), you don’t replace the inner.
Mixing chain with product/quotient when the problem is a chain-only composite. sin(x) cos(x) is a product; sin(cos x) is a composite. Read carefully.
Expanding brackets first on a power like (2x − 1)⁵ — chain rule is faster and keeps the answer factored.
Forgetting the second chain on nested cases like e^sin(2x): you need both layers — derivative of e^□ AND the 2 from sin(2x).

Up next: Product Rule. When two functions are multiplied — like x² sin x or e^x ln x — chain rule alone won’t work. The product rule says (uv)′ = u′v + uv′. Combined with chain rule, you’ll be able to differentiate almost any algebraic combination of the standard functions.

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