IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 🎯 Skill ~4 min practice

AA SL Chain Rule skills

When a function lives inside another function, you can’t differentiate it directly — the chain rule glues two derivatives together. Learn to spot the “outside” and “inside” pieces and chain rule becomes a 10-second move on autopilot.

The Method

dydx  =  dydu  ×  dudx differentiate outer × differentiate inner · ✓ in formula booklet
  1. Spot the inner function — the bit inside brackets, square root, sin/cos, exp, etc. Call it u.
  2. Differentiate the outer — pretend u is one variable. Keep the inner function unchanged inside.
  3. Multiply by the derivative of the inner — that’s u‘(x).

Outside × Inside — three moves

Example: differentiate y = (3x + 1)⁵

Step 1 — let u = u = 3x + 1
Step 2 — diff outer 5u⁴ = 5(3x+1)⁴
×
Step 3 — diff inner u‘(x) = 3

Final answer: dy/dx = 5(3x+1)⁴ × 3 = 15(3x+1)⁴

Common patterns to recognise on sight

Function formDerivative
(f(x))nn (f(x))n−1 · f'(x)
sin(f(x))cos(f(x)) · f'(x)
cos(f(x))−sin(f(x)) · f'(x)
ef(x)ef(x) · f'(x)
ln(f(x))f'(x) / f(x)
√f(x)f'(x) / (2√f(x))

Worked examples

WE 1 EASY

Differentiate y = (2x − 5)⁴.

step 1 — identify inner u = 2x − 5 → u’ = 2step 2 — differentiate the outer d/du(u⁴) = 4u³ = 4(2x − 5)³step 3 — multiply dy/dx = 4(2x − 5)³ × 2dy/dx = 8(2x − 5)³ power on outside, expression inside — pure chain rule!
WE 2 MEDIUM

Differentiate y = sin(3x + π).

step 1 — identify inner u = 3x + π → u’ = 3step 2 — outer derivative d/du(sin u) = cos u = cos(3x + π)step 3 — multiply dy/dx = cos(3x + π) × 3dy/dx = 3 cos(3x + π) trig outer keeps the inner unchanged inside — only differentiate the outer wrapper!
WE 3 HARD

Differentiate y = √(x² + 7).

step 1 — rewrite as power y = (x² + 7)1/2step 2 — identify inner u = x² + 7 → u’ = 2xstep 3 — differentiate outer × inner dy/dx = ½(x² + 7)−1/2 × 2x simplify: ½ × 2 = 1 dy/dx = x / (x² + 7)1/2dy/dx = x / √(x² + 7) always rewrite roots as powers BEFORE applying chain rule!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Identify u first — that’s the move.
Q1 EASY Differentiate y = (5x − 2)³. Show answer ▼Hide answer ▲
u = 5x − 2 → u’ = 5 dy/dx = 3(5x − 2)² × 5 dy/dx = 15(5x − 2)²
Q2 EASY Differentiate y = cos(4x). Show answer ▼Hide answer ▲
u = 4x → u’ = 4 dy/dx = −sin(4x) × 4 dy/dx = −4 sin(4x)
Q3 MEDIUM Differentiate y = e2x² − 1. Show answer ▼Hide answer ▲
u = 2x² − 1 → u’ = 4x dy/dx = e2x² − 1 × 4x dy/dx = 4x · e2x² − 1 eu derivative keeps the same exponential — multiply by u’ on the outside!
Q4 MEDIUM Differentiate y = ln(x² + 3x). Show answer ▼Hide answer ▲
u = x² + 3x → u’ = 2x + 3 dy/dx = (2x + 3) / (x² + 3x) dy/dx = (2x + 3) / (x² + 3x) ln outer → derivative is u’/u — straight from the pattern table!
Q5 HARD Differentiate y = sin²(x). Show answer ▼Hide answer ▲
interpret sin²(x) = (sin x)² u = sin x → u’ = cos x dy/dx = 2(sin x)¹ × cos x = 2 sin x · cos x dy/dx = 2 sin x cos x  (or sin 2x) sin²(x) is a power on the outside, sin(x) on the inside — chain rule, twice if you write it out fully!

⚠ Common mistakes

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Want the theory?

Read the full Chain Rule notes for the conceptual explanation, the link to composite functions, and how chain rule combines with product/quotient rules in tougher questions.

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