Product and Quotient Rule (AI HL)
Some functions can't be differentiated term by term - when \(x\) appears inside two factors multiplied together, or as a fraction of two expressions, you need a dedicated rule. This page sets out both, with worked examples and the pitfalls that trip students up under exam pressure. It's part of the broader Differentiation topic.
11 questions on this sub-topic.
The two rules
Covered under IB syllabus reference AHL5.9. Both are given in the formula booklet - the skill is spotting which structure you're looking at, product or quotient, before you start writing.
Product rule
\[(uv)' = u'v+uv'\]
For two functions multiplied together - differentiate each in turn, keeping the other unchanged, and add the results.
✓ In the formula bookletQuotient rule
\[\left(\dfrac{u}{v}\right)' = \dfrac{u'v-uv'}{v^2}\]
For one function divided by another - the order in the numerator matters, and always divide by \(v^2\).
✓ In the formula bookletNeed 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 \(y = \dfrac{2x + 1}{x - 3}.\)
Worked solution
Quotient rule: M1
\(\dfrac{dy}{dx} = \dfrac{2(x-3) - (2x+1)}{(x-3)^2}=\dfrac{2x - 6 - 2x - 1}{(x-3)^2}\) A1
\(= \dfrac{-7}{(x-3)^2}.\) A1
Differentiate.
(a) \(y=\sin(3x)\)
(b) \(y=x\cos x\)
Worked solution
(a) \(\dfrac{dy}{dx} = 3\cos(3x).\) M1 A1
(b) Product rule: \(\dfrac{dy}{dx}\) M1
\(= \cos x - x\sin x.\) A1
The temperature of a cooling object is \(T(t)=20+60e^{-0.1t}\) °C.
Find the rate of change of temperature at \(t=5.\)
Worked solution
\(\dfrac{dT}{dt} = 60(-0.1)e^{-0.1t}\) M1
\(= -6e^{-0.1t}.\) A1
At \(t=5: -6e^{-0.5}\) M1 A1
\(\approx -3.64\) °C/min. A1
\(y = \sin(3x).\)
(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
- Multiplying the derivatives together instead of using the product rule. \(y=x\cos x\) is not differentiated as \(1 \times (-\sin x)\) - you must form \(u'v+uv'\) and add the two pieces.
- Getting the quotient rule's numerator order backwards. It's \(u'v-uv'\), not \(uv'-u'v\) - swapping the order flips the sign of the whole answer.
- Forgetting the chain rule inside a product or quotient. If \(u\) or \(v\) is itself a composite function, such as \(\sin(3x)\), differentiate it correctly with the chain rule before slotting it into the product or quotient rule.
Ready to practise properly?
10 product-and-quotient-rule questions, marked instantly like the real exam.
Quick answers
When do I use the product rule instead of the quotient rule?
Use the product rule when two functions of \(x\) are multiplied together, e.g. \(y = x\cos x\). Use the quotient rule when one function of \(x\) is divided by another, e.g. \(y = \frac{2x+1}{x-3}\).
What is the quotient rule formula?
For \(y=\dfrac{u}{v}\), the derivative is \(\dfrac{u'v-uv'}{v^2}\). The order of the numerator matters, and you always divide by \(v^2\).