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NotesMath AATopic 5.6
Unit 5 · Calculus · Topic 5.6

IB Math AA — Chain, product & quotient

Topic 5.6 of IB Mathematics: Analysis and Approaches covers Chain, product & quotient, which is part of Unit 5: Calculus. Students explore key concepts including Chain rule, Product rule, Quotient rule. A strong understanding of chain, product & quotient is essential for IB Math AA exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Chain, product & quotient

Key Idea: Three rules let you differentiate anything built from simpler functions — composites, products and fractions. They're pure Paper 1 by-hand skills, and the first job in every question is to spot which rule to use.

🧭 Which rule, when?

Shape of the functionRule to useWhat you do
Composite — a function inside another, f(g(x)), e.g. (x²+3)⁴, sin(3x)ChainDifferentiate the outside, × the inside's derivative.
Product — two factors multiplied, u·v, e.g. x·eˣ, (2x+1)(x²−3)Productu′v + uv′ — both ways round, added.
Quotient — one thing over another, u/v, e.g. (x+1)/(x−2)Quotient(u′v − uv′)/v² — top derivative first, then minus.

🔗 The three rules

ddx f(g(x))=f′(g(x))⋅g′(x)\frac{d}{dx}\,f\big(g(x)\big) = f'\big(g(x)\big)\cdot g'(x)dxd​f(g(x))=f′(g(x))⋅g′(x)
fff
the outer function (the wrapper)
g(x)g(x)g(x)
the inner function — its derivative g'(x) is the extra factor
ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv'dxd​(uv)=u′v+uv′
u, vu,\ vu, v
the two factors being multiplied
u′, v′u',\ v'u′, v′
their derivatives — label all four before you substitute
ddx ⁣(uv)=u′v−uv′v2\frac{d}{dx}\!\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^{2}}dxd​(vu​)=v2u′v−uv′​
uuu
the top (numerator)
vvv
the bottom — the answer is over v²
Know these: sin x → cos x, cos x → −sin x (the minus!), eˣ → eˣ (unchanged), ln x → 1/x. For composites the inner derivative is the multiplier: (ax+b)ⁿ → n(ax+b)ⁿ⁻¹ × a, sin(ax+b) → a cos(ax+b), eᵃˣ⁺ᵇ → a·eᵃˣ⁺ᵇ. If a product/quotient factor is itself composite (e.g. e²ˣ), differentiate that part with the chain rule.

✏️ IB-style worked examples

IB-style question — differentiate using the chain rule

Differentiate y = (2x² − 5)³.

Step by step:

  1. Outer is ( )³, inner is 2x² − 5. Differentiate the outer.

    3(2x2−5)23(2x^{2}-5)^{2}3(2x2−5)2
  2. Multiply by the inner's derivative (4x).

    dydx=3(2x2−5)2⋅4x=12x(2x2−5)2\frac{dy}{dx} = 3(2x^{2}-5)^{2}\cdot 4x = 12x(2x^{2}-5)^{2}dxdy​=3(2x2−5)2⋅4x=12x(2x2−5)2
Final answer:

dy/dx = 12x(2x² − 5)².

IB-style question — differentiate a product

Differentiate y = x³·e²ˣ.

Step by step:

  1. u = x³ (u′ = 3x²), v = e²ˣ (v′ = 2e²ˣ by the chain rule).

    u′v+uv′=3x2e2x+x3⋅2e2xu'v + uv' = 3x^{2}e^{2x} + x^{3}\cdot 2e^{2x}u′v+uv′=3x2e2x+x3⋅2e2x
  2. Factor out the common e²ˣ and x².

    =x2e2x(3+2x)= x^{2}e^{2x}(3 + 2x)=x2e2x(3+2x)
Final answer:

dy/dx = x²e²ˣ(3 + 2x).

IB-style question — differentiate a quotient

Differentiate y = (3x − 1)/(x² + 2).

Step by step:

  1. u = 3x−1 (u′ = 3), v = x²+2 (v′ = 2x). Apply (u′v − uv′)/v².

    3(x2+2)−(3x−1)(2x)(x2+2)2\frac{3(x^{2}+2) - (3x-1)(2x)}{(x^{2}+2)^{2}}(x2+2)23(x2+2)−(3x−1)(2x)​
  2. Expand the numerator carefully (mind the minus).

    =3x2+6−6x2+2x(x2+2)2=−3x2+2x+6(x2+2)2= \frac{3x^{2}+6 - 6x^{2}+2x}{(x^{2}+2)^{2}} = \frac{-3x^{2}+2x+6}{(x^{2}+2)^{2}}=(x2+2)23x2+6−6x2+2x​=(x2+2)2−3x2+2x+6​
Final answer:

dy/dx = (−3x² + 2x + 6)/(x² + 2)².

Important: Chain: never drop the × (inside)′ factor — the × 4x, the × 3 in (3x−1)⁵. Quotient: it's u′v − uv′, not the other way round, all over v² — getting the order wrong flips every sign.

Tap each card to reveal the answer.

Which rule for y = (x² + 1)⁵? Chain — it's a composite (something inside a power).

Differentiate y = sin(4x) 4cos(4x) — cos of it, times the inner derivative 4.

Differentiate y = (x + 2)(3x − 1) 6x + 5 — product rule: 1(3x−1) + (x+2)(3) = 6x + 5.

Differentiate y = x·ln x ln x + 1 — u′v + uv′ = 1·ln x + x·(1/x).

Differentiate y = (x − 1)/(x + 1) 2/(x + 1)² — (1(x+1) − (x−1)(1))/(x+1)² = 2/(x+1)².

What's the derivative of e⁵ˣ? 5e⁵ˣ — chain rule: the inner derivative 5 comes out front.

Exam Tips

  • Spot the shape first: composite → chain, product → product, fraction → quotient.
  • Chain: differentiate the outside, then × the derivative of the inside (don't forget it).
  • Standard derivatives: cos x → −sin x (sign!), eˣ stays eˣ, ln x → 1/x.
  • Product u′v + uv′ and quotient (u′v − uv′)/v²: label u, v, u′, v′ before substituting.
  • Quotient order matters — top derivative first, then minus, all over v². Factorise to tidy.

What you'll learn in Topic 5.6

  • 5.6.1 Chain rule
  • 5.6.2 Product rule
  • 5.6.3 Quotient rule
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 5.6 Chain, product & quotient

5.6.1

Chain rule

Notes
5.6.2

Product rule

Notes
5.6.3

Quotient rule

Notes

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Topic 5.6 Chain, product & quotient forms a core part of Unit 5: Calculus in IB Math AA. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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