IB Maths AA SL
Topic 5 — Calculus
Paper 1 & 2
~7 min read
Product Rule
Use the product rule when two functions are multiplied together — like x2 sin x or ex ln x. The trick: (first × derivative of second) + (second × derivative of first).
📘 What you need to know
- The rule: if y = uv, then dydx = u · dvdx + v · dudx (in formula booklet).
- Use it when two functions are multiplied — not nested inside each other.
- Find u, v, u′, v′ — then plug into the formula.
- Sometimes you need chain rule inside to find u′ or v′.
- Simplify by factorising at the end if possible.
Product or chain rule? Spot the difference
Chain rule
sin(cos x)
“sin OF cos x” — nested, function of a function
Product rule
sin x · cos x
“sin x TIMES cos x” — two functions multiplied
The formula
The 2-step method
How to apply the product rule
- Identify u and v, then differentiate to get u′ and v′.
- Substitute into y′ = uv′ + vu′ and simplify if possible.
The square trick
Lay out u, v, u′, v′ in a 2×2 grid. The two pairs that multiply in the formula are on the diagonals.
The product rule layout
y′ = u · v′ + v · u′
diagonal pairs always multiply!
🧠“Each one keeps the other’s derivative”
The first function (u) keeps the derivative of the second (v′). Then the second function (v) keeps the derivative of the first (u′). Add them. Done.
📍Chain rule may sneak in
If u or v is itself a composite function (like cos 3x2), use the chain rule to find its derivative — then plug into the product rule.
Worked examples
Differentiate y = x² sin x.
step 1 — set up the square
u = x², v = sin x
u′ = 2x, v′ = cos xstep 2 — apply formula
y′ = uv′ + vu′
= x²(cos x) + sin x(2x)
y′ = x² cos x + 2x sin x
factor out x: y′ = x(x cos x + 2 sin x) ✓
Find the derivative of y = ex sin x.
step 1
u = ex, v = sin x
u′ = ex, v′ = cos xstep 2
y′ = ex(cos x) + sin x(ex)
= ex(cos x + sin x)
y′ = ex(cos x + sin x)
ex is a common factor — always factor it out!
WE 3With chain rule inside
Find the derivative of y = 5x² cos(3x²).
step 1 — chain rule for v′
u = 5x², v = cos(3x²)
u′ = 10x
v′: chain rule on cos(3x²) → −sin(3x²) × 6x = −6x sin(3x²)step 2
y′ = 5x²(−6x sin 3x²) + cos(3x²)(10x)
= −30x³ sin(3x²) + 10x cos(3x²)
y′ = 10x[cos(3x²) − 3x² sin(3x²)]
factor out 10x — always check what’s common!
Differentiate y = x³ ln x.
step 1
u = x³, v = ln x
u′ = 3x², v′ = 1/xstep 2
y′ = x³(1/x) + ln x(3x²)
= x² + 3x² ln x
y′ = x²(1 + 3 ln x)
x³/x simplified to x² — always tidy up!
WE 5Gradient at a specific point
Find the gradient of y = (2x + 1) ex at x = 0.
step 1
u = 2x + 1, v = ex
u′ = 2, v′ = exstep 2 — derivative
y′ = (2x + 1)ex + ex(2)
= ex(2x + 3)step 3 — sub x = 0
y′(0) = e⁰(2(0) + 3) = 1 × 3
gradient = 3
e⁰ = 1 — handy!
💡 Top tips
- Lay out u, v, u′, v′ in a square. Diagonal pairs multiply in the formula.
- Use chain rule when needed to find u′ or v′. The product rule applies on top.
- Always factorise at the end. Most product rule answers can be simplified — usually the question expects it.
- Check for hidden simplifications like x³ × (1/x) = x².
- Match the question’s notation — use dy/dx if the question does, y′ if it does.
⚠ Common mistakes
- Multiplying derivatives instead of using the rule. d/dx(uv) ≠ u′v′.
- Confusing product with composite. sin x · cos x = product. sin(cos x) = composite.
- Forgetting chain rule when u or v is itself a composite (like cos(3x²)).
- Sign errors when v involves cos. cos differentiates to −sin.
- Skipping the factorisation step — examiners often want a tidy form.
Two functions multiplied → product rule. Two functions divided → quotient rule, which is the next note. Same idea, slightly different formula.
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