Chain Rule (AI HL)
The chain rule is what lets you differentiate a function sitting inside another function - a bracket raised to a power, a sine of something other than \(x\), an exponential with a messy exponent. Almost every calculus question at HL leans on it somewhere. This page covers the rule itself, worked examples, and where marks are typically lost. It's part of the broader Differentiation topic.
11 questions on this sub-topic.
The rule
Covered under IB syllabus reference AHL5.9. It's given in the formula booklet, but the real skill is recognising a composite function - one wrapped around another - before you differentiate.
Chain rule
\[\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\]
For a function nested inside another - differentiate the outside, keep the inside, then multiply by the derivative of the inside.
✓ In the formula bookletPower rule (extended)
\[\dfrac{d}{dx}(x^n)=nx^{n-1}\]
Holds for any rational \(n\) - it's the special case of the chain rule you'll use most often, on brackets raised to a power.
Need the full syllabus wording and formula-booklet reference table, or a refresher on using your GDC to check a derivative? See Differentiation.
Worked examples
Differentiate \(y = (2x - 1)^3.\)
Worked solution
Outer \(u^3\), inner \(u=2x-1\) with \(u'=2.\) M1
\(\dfrac{dy}{dx}=3(2x-1)^2\cdot 2=6(2x-1)^2.\) A1
Differentiate \(y = \sqrt{4x^3 + 1}.\)
Worked solution
\(y=(4x^3+1)^{1/2}.\) Outer \(u^{1/2}\), inner \(u=4x^3+1\) with \(u'=12x^2.\) M1
\(\dfrac{dy}{dx}=\tfrac12(4x^3+1)^{-1/2}\cdot 12x^2.\) A1
\(\dfrac{dy}{dx}=6x^2(4x^3+1)^{-1/2}=\dfrac{6x^2}{\sqrt{4x^3+1}}.\) A1
Differentiate \(y = (3x^2 + 1)^4\).
Worked solution
The chain rule applies when one function sits inside another. Outer \(u^4\), inner \(u=3x^2+1\). M1
\(\dfrac{dy}{du}=4u^3\), \(\dfrac{du}{dx}=6x.\) A1
\(\dfrac{dy}{dx}=4(3x^2+1)^3\cdot 6x=24x(3x^2+1)^3.\) A1
Differentiate \(y = \cos^3(2x).\)
Worked solution
\(y=(\cos(2x))^3\) has three layers: cube of, cosine of, \(2x.\) M1
\(\dfrac{dy}{dx}=3\cos^2(2x)\cdot\dfrac{d}{dx}[\cos(2x)].\) A1
\(\dfrac{d}{dx}[\cos(2x)]=-2\sin(2x).\) A1
\(\dfrac{dy}{dx}=3\cos^2(2x)\cdot(-2\sin(2x))=-6\sin(2x)\cos^2(2x).\) A1
The height of a wave, in metres, is modelled by \(h(t) = 3\sin(2t^2 + 1)\), where \(t\) is the time in seconds.
Find \(\dfrac{dh}{dt}.\)
Worked solution
Outer \(3\sin(u)\), inner \(u=2t^2+1\) with \(u'=4t.\) M1
\(\dfrac{d}{du}[3\sin u]=3\cos u.\) A1
\(\dfrac{du}{dt}=4t.\) A1
\(\dfrac{dh}{dt}=3\cos(2t^2+1)\cdot 4t=12t\cos(2t^2+1).\) A1
Common mistakes
- Forgetting the chain rule on composite functions. Differentiating \(\sin(3x)\) as \(\cos(3x)\) instead of \(3\cos(3x)\) drops the "multiply by the derivative of the inside" step - always check whether the argument of the function is anything other than plain \(x\).
- Not converting roots to fractional powers first. \(\sqrt{4x^3+1}\) needs to become \((4x^3+1)^{1/2}\) before the chain rule can be applied - trying to differentiate a square-root symbol directly usually goes wrong.
- Leaving the inner derivative out of the final answer. \(\dfrac{dy}{dx}=3(2x-1)^2\) is only half the answer for \(y=(2x-1)^3\) - it still needs multiplying by \(u'=2\) to give \(6(2x-1)^2\).
Ready to practise properly?
11 chain-rule questions, marked instantly like the real exam.
Quick answers
What is the chain rule?
The chain rule differentiates a composite function: \(\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\). Differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function.
How do I know when a question needs the chain rule?
Whenever the argument of a function is anything other than plain \(x\) - such as \((2x-1)^3\), \(\sin(3x)\), or \(e^{2x}\) - you need the chain rule, because you're differentiating a function of a function.