IB Math AA SL Topic 5.6: Chain, product & quotient | Notes & Flashcards | Aimnova

IB Math AA SL — 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 SL exams and builds the foundation for connected topics across the syllabus.

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?

🔗 The three rules

\frac{d}{dx}\,f\big(g(x)\big) = f'\big(g(x)\big)\cdot g'(x)

$f$: the outer function (the wrapper)
$g(x)$: the inner function — its derivative g'(x) is the extra factor

\frac{d}{dx}(uv) = u'v + uv'

$u,\ v$: the two factors being multiplied
$u',\ v'$: their derivatives — label all four before you substitute

\frac{d}{dx}\!\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^{2}}

$u$: the top (numerator)
$v$: 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:

Outer is ( )³, inner is 2x² − 5. Differentiate the outer.
$3(2x^{2}-5)^{2}$
Multiply by the inner's derivative (4x).
$\frac{dy}{dx} = 3(2x^{2}-5)^{2}\cdot 4x = 12x(2x^{2}-5)^{2}$

Final answer:

dy/dx = 12x(2x² − 5)².

IB-style question — differentiate a product

Differentiate y = x³·e²ˣ.

Step by step:

u = x³ (u′ = 3x²), v = e²ˣ (v′ = 2e²ˣ by the chain rule).
$u'v + uv' = 3x^{2}e^{2x} + x^{3}\cdot 2e^{2x}$
Factor out the common e²ˣ and x².
$= x^{2}e^{2x}(3 + 2x)$

Final answer:

dy/dx = x²e²ˣ(3 + 2x).

IB-style question — differentiate a quotient

Differentiate y = (3x − 1)/(x² + 2).

Step by step:

u = 3x−1 (u′ = 3), v = x²+2 (v′ = 2x). Apply (u′v − uv′)/v².
$\frac{3(x^{2}+2) - (3x-1)(2x)}{(x^{2}+2)^{2}}$
Expand the numerator carefully (mind the minus).
$= \frac{3x^{2}+6 - 6x^{2}+2x}{(x^{2}+2)^{2}} = \frac{-3x^{2}+2x+6}{(x^{2}+2)^{2}}$

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.

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.

IB Math AA SL — Chain, product & quotient

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?

🔗 The three rules

\frac{d}{dx}\,f\big(g(x)\big) = f'\big(g(x)\big)\cdot g'(x)

$f$: the outer function (the wrapper)
$g(x)$: the inner function — its derivative g'(x) is the extra factor

\frac{d}{dx}(uv) = u'v + uv'

$u,\ v$: the two factors being multiplied
$u',\ v'$: their derivatives — label all four before you substitute

\frac{d}{dx}\!\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^{2}}

$u$: the top (numerator)
$v$: 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:

Outer is ( )³, inner is 2x² − 5. Differentiate the outer.
$3(2x^{2}-5)^{2}$
Multiply by the inner's derivative (4x).
$\frac{dy}{dx} = 3(2x^{2}-5)^{2}\cdot 4x = 12x(2x^{2}-5)^{2}$

Final answer:

dy/dx = 12x(2x² − 5)².

IB-style question — differentiate a product

Differentiate y = x³·e²ˣ.

Step by step:

u = x³ (u′ = 3x²), v = e²ˣ (v′ = 2e²ˣ by the chain rule).
$u'v + uv' = 3x^{2}e^{2x} + x^{3}\cdot 2e^{2x}$
Factor out the common e²ˣ and x².
$= x^{2}e^{2x}(3 + 2x)$

Final answer:

dy/dx = x²e²ˣ(3 + 2x).

IB-style question — differentiate a quotient

Differentiate y = (3x − 1)/(x² + 2).

Step by step:

u = 3x−1 (u′ = 3), v = x²+2 (v′ = 2x). Apply (u′v − uv′)/v².
$\frac{3(x^{2}+2) - (3x-1)(2x)}{(x^{2}+2)^{2}}$
Expand the numerator carefully (mind the minus).
$= \frac{3x^{2}+6 - 6x^{2}+2x}{(x^{2}+2)^{2}} = \frac{-3x^{2}+2x+6}{(x^{2}+2)^{2}}$

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.

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.

IB Math AA SL — Chain, product & quotient

Key concepts in Chain, product & quotient

🧭 Which rule, when?

🔗 The three rules

✏️ IB-style worked examples

IB-style question — differentiate using the chain rule

IB-style question — differentiate a product

IB-style question — differentiate a quotient

Exam Tips

What you'll learn in Topic 5.6

Study resources — 5.6 Chain, product & quotient

Chain rule

Product rule

Quotient rule

Ready to study Chain, product & quotient?

Ready to practice?

IB Math AA SL — Chain, product & quotient

Key concepts in Chain, product & quotient

🧭 Which rule, when?

🔗 The three rules

✏️ IB-style worked examples

IB-style question — differentiate using the chain rule

IB-style question — differentiate a product

IB-style question — differentiate a quotient

Exam Tips

What you'll learn in Topic 5.6

Study resources — 5.6 Chain, product & quotient

Chain rule

Product rule

Quotient rule

Ready to study Chain, product & quotient?

Ready to practice?