Product and Quotient Rule (AA SL)
When two functions of \(x\) are multiplied together, like \(x^2e^x\), or divided, like \(\dfrac{x}{x+1}\), the power rule alone isn't enough - you need the product rule or the quotient rule. Both are given in the formula booklet, but knowing which one to reach for, and keeping the algebra tidy, is what separates full marks from a slip. This page covers both formulas, worked examples, and where marks are usually lost. It's part of the broader Differentiation topic.
15 questions on this sub-topic.
The two rules
Covered under IB syllabus reference SL5.6, alongside the chain rule and the derivatives of \(x^n\), \(\sin x\), \(\cos x\), \(e^x\) and \(\ln x\).
Product rule
\[\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
For a product of two functions - differentiate one at a time and add the results.
✓ In the formula bookletQuotient rule
\[\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
For a fraction of two functions - order matters in the numerator, unlike the product rule.
✓ In the formula bookletNeed the full derivative and integral reference table? See Differentiation.
Worked examples
Differentiate \(y = x^2 e^{x}\).
Worked solution
Apply the product rule \((uv)'=u'v+uv'\) with \(u=x^2,\ v=e^x\). M1
\(u'=2x,\ v'=e^x.\) A1
This simplifies to \(=xe^x(2+x).\) A1
Differentiate \(y = \dfrac{x}{x + 1}\).
Worked solution
Apply the quotient rule with \(u=x,\ v=x+1.\) M1
\(u'=1,\ v'=1.\) A1
Applying the quotient rule formula and simplifying: \(\dfrac{(x+1)-x}{(x+1)^2}=\dfrac{1}{(x+1)^2}.\) A1
\(y = \ln(5x).\)
(a) Find \(\dfrac{dy}{dx}\).
(b) Differentiate \(y = x\ln x.\)
Worked solution
(a) Chain rule on \(\ln(5x)\): derivative of \(\ln u\) is \(\tfrac{1}{u}\) times \(u'\). \(\dfrac{dy}{dx}=\dfrac{1}{5x}\cdot 5\) M1
\(=\dfrac{1}{x}.\) A1
(b) Product rule with \(u=x,\ v=\ln x\): \(\dfrac{dy}{dx}=(1)\ln x+x\cdot\dfrac{1}{x}\) M1
\(=\ln x+1.\) A1
Differentiate \(y=\dfrac{x}{x^2+1}\) and simplify.
Worked solution
\(\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{u'v-uv'}{v^2}.\) Let \(u=x,\ v=x^2+1,\) so \(u'=1,\ v'=2x.\) M1
\(\dfrac{dy}{dx} = \dfrac{(x^2+1)(1) - x(2x)}{(x^2+1)^2}\) A1
\(= \dfrac{x^2+1-2x^2}{(x^2+1)^2}\) M1
\(= \dfrac{1-x^2}{(x^2+1)^2}.\) A1
Common mistakes
- Getting the sign wrong in the quotient rule numerator. It's \(v\frac{du}{dx}-u\frac{dv}{dx}\), not the other way round - swapping \(u\) and \(v\) here flips the sign of the whole answer.
- Using the product rule when a term should just be scaled. \(\dfrac{d}{dx}(5x^3)\) is a constant multiple, not a product of two functions of \(x\) - the product rule is only needed when both factors actually depend on \(x\) in a way that needs differentiating separately.
- Forgetting the chain rule inside a product or quotient rule term. If \(v=\sin(2x)\) inside a product rule calculation, \(v'\) is \(2\cos(2x)\), not \(\cos(2x)\) - the inner derivative from the chain rule still has to be applied before combining terms.
Ready to practise properly?
15 product and quotient rule questions, marked instantly like the real exam.
Quick answers
What is the product rule?
If \(y=uv\), then \(\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\). Differentiate each factor in turn, keeping the other one unchanged, and add the two results.
What is the quotient rule?
If \(y=\dfrac{u}{v}\), then \(\dfrac{dy}{dx}=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\). The order in the numerator matters - it isn't symmetric like the product rule.