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

Spot the inner function — the bit inside brackets, square root, sin/cos, exp, etc. Call it u.
Differentiate the outer — pretend u is one variable. Keep the inner function unchanged inside.
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 form	Derivative
(f(x))ⁿ	n (f(x))ⁿ⁻¹ · f'(x)
sin(f(x))	cos(f(x)) · f'(x)
cos(f(x))	−sin(f(x)) · f'(x)
e^f(x)	e^f(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 = e^{2x² − 1}. Show answer ▼Hide answer ▲

u = 2x² − 1 → u’ = 4x dy/dx = e^{2x² − 1} × 4x dy/dx = 4x · e^{2x² − 1} e^u 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

Forgetting to multiply by u‘. The biggest chain rule error — students differentiate the outer and stop. Always finish with × u‘.
Differentiating both inner and outer at once — the rule keeps the inner unchanged in the outer derivative, then multiplies by the inner’s derivative separately.
Sign error with cos. d/dx(cos f(x)) = −sin(f(x)) · f'(x) — the minus stays.
Not rewriting roots as powers. √(f(x)) = (f(x))^1/2 — easier to apply chain rule from this form.
Confusing chain rule with product rule. Chain is for f(g(x)) (one inside another). Product is for f(x) · g(x) (two multiplied).
Stopping after the outer derivative when the inner is also composite (rare in SL but possible) — chain rule can apply more than once.

📖

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.

Need help with the Chain Rule?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →