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

Quotient Rule

Use the quotient rule when one function is divided by another — like sin xx or x2ex. Looks intimidating, but it’s just one formula. The order matters because of the minus sign.

📘 What you need to know

Product or quotient? Spot the difference

Product rule
x · sin x
two functions multiplied
Quotient rule
sin x / x
one function divided by another

The formula

Quotient rule
If   y = uv   then   y′ = vu′ − uvv2
✓ in formula booklet

The 2-step method

How to apply the quotient rule

  1. Identify u (top) and v (bottom), then differentiate to get u′ and v′.
  2. Substitute into y′ = (vu′ − uv′)/v². Be careful with the order.

The square trick

Same 2×2 layout as product rule — but now with a minus instead of a plus, and divided by v2.

The quotient rule layout

u
v
u′
v′
y′ = (v · u′u · v′) / v2
v · u′ comes FIRST — order matters!
🧠

“Lo d-Hi minus Hi d-Lo, all over Lo squared”

Lo = bottom, Hi = top, d = “derivative of”. So: bottom × derivative-of-top, minus top × derivative-of-bottom, all divided by the bottom squared. Sing it once and it sticks.

📍

If the top is just a constant, skip the quotient rule

For something like 2(3x − 7)2, rewrite as 2(3x − 7)−2 and use the chain rule. Faster, fewer mistakes.

Worked examples

WE 1

Polynomial / polynomial

Differentiate y = (x² + 1) / (x − 3).

step 1 — set up u = x² + 1,  v = x − 3 u′ = 2x,  v′ = 1step 2 — apply formula y′ = ((x − 3)(2x) − (x² + 1)(1)) / (x − 3)² = (2x² − 6x − x² − 1) / (x − 3)² y′ = (x² − 6x − 1) / (x − 3)² expand the top first, then collect like terms — keep the bottom factored!
WE 2

Trig / polynomial

Differentiate y = sin x / x.

step 1 u = sin x,  v = x u′ = cos x,  v′ = 1step 2 y′ = (x cos x − sin x · 1) / x² y′ = (x cos x − sin x) / x² classic AA SL question — no further simplification needed!
WE 3

With chain rule inside

Differentiate f(x) = cos 2x / (3x + 2).

step 1 — chain rule for u′ u = cos 2x,  v = 3x + 2 u′ = −2 sin 2x  (chain rule) v′ = 3step 2 f′(x) = ((3x+2)(−2 sin 2x) − (cos 2x)(3)) / (3x+2)² f′(x) = (−2(3x+2) sin 2x − 3 cos 2x) / (3x+2)² factor out −1 if needed: −((2(3x+2) sin 2x + 3 cos 2x)) / (3x+2)²
WE 4

Polynomial / exponential

Differentiate y = x² / ex.

step 1 u = x²,  v = ex u′ = 2x,  v′ = exstep 2 y′ = (ex(2x) − x²(ex)) / (ex = ex(2x − x²) / e2x = (2x − x²) / ex y′ = x(2 − x) / ex ex cancels: ex/e2x = 1/ex. Always look for these!
WE 5

Gradient at a specific point

Find the gradient of y = x / (x² + 1) at x = 2.

step 1 u = x,  v = x² + 1 u′ = 1,  v′ = 2xstep 2 — derivative y′ = ((x² + 1)(1) − x(2x)) / (x² + 1)² = (x² + 1 − 2x²) / (x² + 1)² = (1 − x²) / (x² + 1)²step 3 — sub x = 2 y′(2) = (1 − 4) / (4 + 1)² = −3/25 gradient = −3/25 simplify the algebra BEFORE plugging in numbers!

💡 Top tips

⚠ Common mistakes

You’ve now got chain, product, and quotient — the three big rules. Next we look at second order derivatives — what you get by differentiating twice, and why that’s so useful.

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