Chain Rule (AA HL)
Whenever you're differentiating a function nested inside another function - \(\sin(3x)\), \((3x^2+1)^5\), \(e^{3x^2-1}\) - the chain rule is the tool. This page covers the formula, how to spot when it applies (including when you need it twice), and the mistake that costs the most marks: forgetting to multiply by the inner derivative. It's part of the broader Differentiation topic.
12 questions on this sub-topic.
The formula
Covered under IB syllabus reference SL5.6. The chain rule itself is in the formula booklet - the skill is spotting the inner function and remembering to differentiate it too.
Chain rule
\[\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\]
Differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
✓ In the formula bookletSpotting composite functions
If you could write the expression as \(f(u)\) with \(u\) a simpler expression in \(x\) - like \(u = 3x\) in \(\sin(3x)\), or \(u = 3x^2+1\) in \((3x^2+1)^5\) - you're looking at a composite function and need the chain rule. Some expressions, like \(\sin^2(3x)\), need it applied twice.
Not in the formula booklet - recognition strategyNeed the full syllabus wording and formula-booklet reference table? See Differentiation.
Worked examples
Differentiate \(y = (3x^2 + 1)^5.\)
Worked solution
\(\dfrac{dy}{dx} = 5(3x^2+1)^4 \cdot \dfrac{d}{dx}(3x^2+1) = 5(3x^2+1)^4 \cdot 6x.\) M1
\(\dfrac{dy}{dx} = 30x(3x^2+1)^4.\) A1
Differentiate \(y=\sin^2(3x).\)
Worked solution
Chain rule (twice): let \(u = \sin 3x.\) M1
\(\dfrac{dy}{dx} = 2\sin 3x \cdot \cos 3x \cdot 3 = 6\sin 3x\cos 3x\) A1
\(= 3\sin 6x.\) A1
Show that \(y = e^{2x}\) is a solution of \(\dfrac{dy}{dx} = 2y\).
Worked solution
\(\tfrac{dy}{dx} = 2e^{2x}.\) M1
\(2y = 2e^{2x}\), so it satisfies the equation. A1 AG
\(y = \sin(3x).\)
(a) Find \(\dfrac{dy}{dx}\).
(b) Differentiate \(y = \cos(x^2).\)
Worked solution
(a) \(\sin(3x)\) is a composite, so chain rule: differentiate the sine and multiply by the derivative of the inner \(3x\). \(\dfrac{dy}{dx}=\cos(3x)\cdot 3\) M1
\(=3\cos(3x).\) A1
(b) Inner \(x^2\), derivative \(2x\); the derivative of \(\cos\) carries a minus sign. \(\dfrac{dy}{dx}=-\sin(x^2)\cdot 2x\) M1
\(=-2x\sin(x^2).\) A1
Common mistakes
- Forgetting to multiply by the inner derivative. Differentiating \(\sin(3x)\) as \(\cos(3x)\) instead of \(3\cos(3x)\) drops the "multiply by the derivative of the inner function" step - always check for a function inside a function.
- Stopping after one application when two are needed. An expression like \(\sin^2(3x)\) is a function of a function of a function - differentiate the outer square, then the sine, then the \(3x\) inside it, multiplying by each inner derivative in turn.
- Confusing the chain rule with the product rule. \((3x^2+1)^5\) is one function raised to a power (chain rule), not two different functions multiplied together (product rule) - check whether you have one composite expression or two separate factors before choosing a method.
Ready to practise properly?
12 chain-rule questions, marked instantly like the real exam.
Quick answers
What is the chain rule?
\(\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\). Differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
How do I know when to use the chain rule?
Whenever you're differentiating a composite function - a function inside another function, like \(\sin(3x)\), \((3x^2+1)^5\), or \(e^{3x^2-1}\). If substituting \(u\) for the inner part would simplify the expression, you need the chain rule.
Need a calculator refresher? See Using your GDC on the full Differentiation page.