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

Quotient Rule

For two functions divided rather than multiplied — like (2x − 5)/(x² + 1) or e^2x/(x + 1) — the rule is (u/v)′ = (vu′ − uv′)/v². Same 2×2 layout as product rule, but with a critical minus sign in the numerator (so order matters), and a denominator squared at the end.

📘 What you need to know

Quotient rule formula: if y = u/v where u, v are functions of x, then dy/dx = (v · du/dx − u · dv/dx) / v² — given in the formula booklet.
Order matters: the minus sign means swapping u and v changes the sign of the answer. Always: (denominator × derivative of numerator) MINUS (numerator × derivative of denominator).
Don’t forget v² at the bottom: square the original denominator.
When to use: when BOTH numerator and denominator are functions of x.
Constant numerator: rewrite k/f(x) as k·(f(x))⁻¹ and use chain rule — usually faster.
Constant denominator: treat 1/k as a factor — just k times the derivative of the numerator.
Chain rule may nest inside — when u or v is itself composite (like sin(2x) or (3x + 2)²).
Quotient as a product: u/v = u · v⁻¹ — you CAN solve quotient questions with product+chain rule (good practice, but quotient rule is usually cleaner).

The rule and the diagonal trap

Quotient rule (formula booklet) dydx = v · dudx − u · dvdxv²

Same diagonal pairs as product rule, but the v·u′ pair comes FIRST (positive) and u·v′ comes second (with a minus). And don’t forget v² underneath.

Memory device: “low d-high minus high d-low, over the square of what’s below” — read top to bottom: low (denominator) times derivative of high (numerator), minus high times derivative of low, all over the bottom squared. The order of the two terms matters because of the minus.

When NOT to use quotient rule

Constant numerator

2(3x − 7)² = 2(3x − 7)⁻²

use chain rule on the negative power instead — much faster

Constant denominator

(3x − 7)²2 = ½ (3x − 7)²

treat 1/2 as a factor and use chain rule

Quotient rule still works in both cases — it just adds extra terms that always cancel. Save effort: rewrite as a power or a constant multiple first.

🧭 Recipe — apply the quotient rule

Identify u (numerator) and v (denominator). Lay them out 2×2 with u′ and v′ underneath.
Differentiate each: find u′ and v′ (chain rule inside if needed).
Apply the formula: numerator = vu′ − uv′ (positive term first); denominator = v².
Simplify the numerator — expand and collect like terms; factor common terms across v² if it helps.
Verify or evaluate — substitute the given x-value if a numeric gradient is asked for.

Worked examples

WE 1

Algebraic over algebraic — basic quotient rule

Find the derivative of y = (2x − 5)/(x² + 1).

Identify u and v u = 2x − 5 v = x² + 1 u′ = 2 v′ = 2x Apply (vu′ − uv′)/v² numerator = (x² + 1)·2 − (2x − 5)·2x = 2x² + 2 − 4x² + 10x = −2x² + 10x + 2 denominator = (x² + 1)² Final form (factor −2 from numerator) dy/dx = −2(x² − 5x − 1) / (x² + 1)² dy/dx = −2(x² − 5x − 1)/(x² + 1)² careful with −(2x − 5)·2x = −4x² + 10x — distribute the minus AND the 2x

WE 2

Trig over algebra — simplify the result

Find the derivative of y = sin(x)/x².

Identify u and v u = sin x v = x² u′ = cos x v′ = 2x Apply (vu′ − uv′)/v² dy/dx = [x² · cos x − sin x · 2x] / (x²)² = [x² cos x − 2x sin x] / x⁴ Factor x from the numerator and divide out = x(x cos x − 2 sin x) / x⁴ = (x cos x − 2 sin x) / x³ dy/dx = (x cos x − 2 sin x)/x³ v² = (x²)² = x⁴; cancelling one x with the numerator’s common factor leaves x³

WE 3

Exponential over linear — factor cleanly

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

Identify u and v u = e^(2x) v = x + 1 u′ = 2e^(2x) v′ = 1 Apply (vu′ − uv′)/v² dy/dx = [(x + 1)·2e^(2x) − e^(2x)·1] / (x + 1)² Factor e^(2x) from the numerator = e^(2x) [2(x + 1) − 1] / (x + 1)² = e^(2x) (2x + 2 − 1) / (x + 1)² = e^(2x) (2x + 1) / (x + 1)² dy/dx = e^(2x)(2x + 1)/(x + 1)² e^(2x) is never zero, so always factor it out — the (2x + 1) factor reveals where dy/dx = 0

WE 4

Show that d/dx(tan x) = sec²x

Use the quotient rule to show that the derivative of tan(x) = sin(x)/cos(x) is sec²(x).

Set u = sin x, v = cos x u′ = cos x v′ = −sin x Apply (vu′ − uv′)/v² dy/dx = [cos x · cos x − sin x · (−sin x)] / cos²x = [cos²x + sin²x] / cos²x Apply identity cos²x + sin²x = 1 = 1 / cos²x = sec²x ∎ d/dx [tan x] = sec²x ✓ double minus: −sin x · (−sin x) = +sin²x — the secret of the cos²+sin² identity appearing

WE 5

Gradient at a specific point — exact answer

Find the gradient of the curve y = (3x + 2)/(x² − 4) at the point where x = 1.

Identify u and v u = 3x + 2 v = x² − 4 u′ = 3 v′ = 2x Apply (vu′ − uv′)/v² numerator = (x² − 4)·3 − (3x + 2)·2x = 3x² − 12 − 6x² − 4x = −3x² − 4x − 12 denominator = (x² − 4)² Substitute x = 1 numerator: −3 − 4 − 12 = −19 denominator: (1 − 4)² = 9 m = −19/9 Gradient at x = 1 is −19/9 substituting BEFORE simplifying often saves time when only a numeric gradient is asked

WE 6

Real-world: drug concentration peaks

The concentration (in mg/L) of a drug in the bloodstream t hours after injection is modelled by C(t) = 5t/(t² + 4), t ≥ 0.

(a) Find dC/dt. (b) Find the time at which the concentration is at a maximum. (c) Find the maximum concentration.

(a) Apply quotient rule u = 5t v = t² + 4 u′ = 5 v′ = 2t dC/dt = [(t² + 4)·5 − 5t·2t] / (t² + 4)² = [5t² + 20 − 10t²] / (t² + 4)² = (20 − 5t²) / (t² + 4)² = 5(4 − t²) / (t² + 4)² (b) Maximum where dC/dt = 0 5(4 − t²) = 0 → t² = 4 → t = 2 (t ≥ 0) (c) Substitute t = 2 into C(t) C(2) = 5·2/(4 + 4) = 10/8 = 5/4 (a) 5(4 − t²)/(t² + 4)²; (b) t = 2 hours; (c) C_max = 5/4 mg/L denominator (t² + 4)² is always positive; sign of dC/dt comes from numerator (4 − t²) — confirms max

💡 Top tips

“Low d-high minus high d-low, over square of what’s below” — chant it before each quotient question; it locks in the order.
2×2 layout — write u, v on top and u′, v′ underneath; the diagonal pairs you need are the same as in product rule, but with a minus between them.
Constant numerator? Skip the quotient rule — rewrite as k·(f(x))⁻¹ and use chain rule.
Substitute before fully simplifying when a numeric gradient is asked — saves algebra.
For finding stationary points: dy/dx = 0 only when the NUMERATOR = 0 (the v² denominator can’t be zero on the curve’s domain).

⚠ Common mistakes

Order swap on the minus sign — writing (uv′ − vu′)/v² instead of (vu′ − uv′)/v². This flips the sign of every answer.
Forgetting to square the denominator — v² at the bottom, NOT just v.
Sign slips in expansion — −(3x + 2)·2x = −6x² − 4x; distribute the minus carefully.
Differentiating cos as +sin — d/dx(cos x) = −sin x. The double-minus in WE 4 (−sin x · −sin x = +sin²x) is the right answer; missing one minus breaks it.
Using quotient rule when chain is faster — k/(f(x))² is just k(f(x))⁻², a one-step chain rule problem.

Up next: Higher Order Derivatives. Differentiating again gives the second derivative f″(x) = d²y/dx² — used to test whether stationary points are local maxima or minima, and to determine concavity. Keep going and you get f‴, f⁽⁴⁾, f⁽⁵⁾… for many functions, the higher derivatives follow a predictable pattern.

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