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

Product Rule

When two functions are multiplied — like x² sin x, e^x ln x, or (3x − 1)² e^x — chain rule alone won’t do. The product rule says (uv)′ = u′v + uv′. Identify the two factors, differentiate each, then plug into the formula. Often you’ll need chain rule alongside, since one of the factors may itself be composite.

📘 What you need to know

Product rule formula: if y = uv where u, v are functions of x, then dy/dx = u · (dv/dx) + v · (du/dx) — given in the formula booklet.
Function notation: (fg)′ = f g′ + g f′ — also written y′ = uv′ + vu′.
When to use: a product of TWO functions of x, like x sin x, e^x cos x, x² ln x.
Don’t confuse with chain rule: sin x · cos x = product (use product rule); sin(cos x) = composite (use chain rule).
Both factors get differentiated, then summed: NOT u′ × v′ — it’s u′v + uv′ (two terms, not one).
Chain rule may be nested: when u or v is itself composite (like (3x − 1)² or sin(2x)), use chain rule to get u′ or v′.
Factor the answer when convenient — e^x, common brackets often factor cleanly. Check exam wording for required form.
Products of three or more: pair them up (apply product rule to (uv)·w treating uv as one factor).

The rule and the 2×2 setup

Product rule (formula booklet) dydx = u dvdx + v dudx

Lay out u, u′, v, v′ in a 2×2 square so that the diagonal pairs you need (u·v′ and v·u′) are on opposite diagonals — it removes a whole class of pairing mistakes.

Place u and v on top, u′ and v′ underneath. The two products that go into the answer are on the green and red diagonals.

Spotting it: if you see x² sin(x), e^x ln(x), or (2x + 1) cos(3x) — that’s two functions of x multiplied. Product rule. If only ONE factor depends on x (like 5 · sin x), it’s NOT a product — just a constant times a function. Differentiate the function and keep the constant.

🧭 Recipe — apply the product rule

Identify u and v — the two factors of y. Write them in a 2×2 layout.
Differentiate each: find u′ and v′ (use chain rule for either if needed).
Apply the formula: dy/dx = uv′ + vu′ — pair on the diagonals.
Simplify if natural — factor common terms (e^x, common brackets) for a cleaner form.
Evaluate or rearrange if a numeric gradient or specific form is requested.

Worked examples

WE 1

Algebra × trig — basic product rule

Find the derivative of y = x³ sin(2x).

Identify factors u = x³ v = sin(2x) u′ = 3x² v′ = 2 cos(2x) (chain rule on v) Apply uv′ + vu′ dy/dx = x³ · 2 cos(2x) + sin(2x) · 3x² = 2x³ cos(2x) + 3x² sin(2x) Factor x² for tidiness = x²(2x cos(2x) + 3 sin(2x)) dy/dx = x²(2x cos(2x) + 3 sin(2x)) v = sin(2x), so v′ needs chain rule: derivative of sin × derivative of 2x = 2 cos(2x)

WE 2

Linear × exponential — factor cleanly

Find the derivative of y = (2x + 1) e^3x. Give your answer in factored form.

Identify factors u = 2x + 1 v = e^(3x) u′ = 2 v′ = 3 e^(3x) (chain rule on v) Apply uv′ + vu′ dy/dx = (2x + 1) · 3 e^(3x) + e^(3x) · 2 Factor out e^(3x) = e^(3x) [3(2x + 1) + 2] = e^(3x) [6x + 3 + 2] = e^(3x) (6x + 5) dy/dx = (6x + 5) e^(3x) always factor out the exp — e^(3x) is never zero, so it always appears in the final form

WE 3

Algebra × log

Find the derivative of y = x² ln(x). Give your answer in factored form.

Identify factors u = x² v = ln(x) u′ = 2x v′ = 1/x Apply uv′ + vu′ dy/dx = x² · (1/x) + ln(x) · 2x = x + 2x ln(x) Factor x = x(1 + 2 ln(x)) dy/dx = x(2 ln(x) + 1) x²·(1/x) simplifies to x — don’t leave it as x²/x in your final answer

WE 4

Chain rule INSIDE one factor

Find the derivative of y = x cos(x²).

Identify factors u = x v = cos(x²) u′ = 1 v′ = −2x sin(x²) (chain rule: f′(x) = 2x) Apply uv′ + vu′ dy/dx = x · [−2x sin(x²)] + cos(x²) · 1 = cos(x²) − 2x² sin(x²) dy/dx = cos(x²) − 2x² sin(x²) v′ needs chain rule: d/dx [cos(x²)] = −sin(x²) · 2x — the minus sign carries through

WE 5

(Bracket)² × exponential — gradient at x = 1

The curve y = (3x − 1)² · e^x. Find the gradient at x = 1.

Identify factors u = (3x − 1)² v = e^x u′ = 2(3x − 1)·3 = 6(3x − 1) (chain rule) v′ = e^x Apply uv′ + vu′ dy/dx = (3x − 1)² · e^x + e^x · 6(3x − 1) Factor e^x and (3x − 1) = e^x (3x − 1) [(3x − 1) + 6] = e^x (3x − 1)(3x + 5) Substitute x = 1 m = e¹ · (3 − 1)(3 + 5) = e · 2 · 8 = 16e Gradient at x = 1 is 16e factoring before substituting saves arithmetic — much cleaner than expanding first

WE 6

Equation of tangent at x = e

Find the equation of the tangent to the curve y = x² ln(x) at the point where x = e.

Find y at x = e y = e² · ln(e) = e² · 1 = e² point: (e, e²) Differentiate using product rule (from WE 3) dy/dx = x(2 ln(x) + 1) Gradient at x = e m = e · (2 · 1 + 1) = 3e Apply point-slope form y − e² = 3e(x − e) y = 3e·x − 3e² + e² y = 3e·x − 2e² Tangent: y = 3e x − 2e² at x = e, ln(e) = 1 — that’s why x = e gives clean exact answers in log problems

💡 Top tips

2×2 layout — write u, v on top and u′, v′ underneath. The pairs you multiply are on the DIAGONALS, not the same column.
Factor as you go — common factors (like e^x, or a bracket) should come out before substituting numerical values.
Chain rule lives inside: when u or v is composite (like sin(2x), (3x − 1)², ln(2x + 5)), apply chain rule to find that factor’s derivative first.
Match exam notation — if the question uses f(x) g(x), give your answer in f, g notation: f g′ + g f′. If it uses u, v, use that.
Triple products (like x e^x sin x) — group two factors together: let u = x e^x, v = sin x; apply product rule again inside u′.

⚠ Common mistakes

Writing u′v′ as the answer — that’s NOT the product rule. Correct: uv′ + vu′ (two terms summed).
Forgetting chain rule when u or v is itself composite — e.g. for y = x sin(2x), v′ is 2 cos(2x), not cos(2x).
Mixing product with chain: sin x · cos x = product (use product rule); sin(cos x) = composite (use chain rule). Read the bracket structure carefully.
Forgetting to differentiate one factor — the rule has TWO terms; both u′v and uv′ must appear.
Sign slip on cos’s chain — d/dx [cos(x²)] = −2x sin(x²); don’t lose the minus sign when it lands inside a product.

Up next: Quotient Rule. For functions divided rather than multiplied — like (cos 2x) / (3x + 2). The formula is similar but with a minus sign and a denominator squared: (u/v)′ = (vu′ − uv′)/v². Order matters because of the minus, so the 2×2 layout is even more useful.

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