Product and Quotient Rule (AA HL)
The chain rule handles functions inside functions, but many expressions are two functions multiplied or divided together instead - \(x^2\sin x\), \(\dfrac{2x+1}{x^2+3}\), and similar combinations. That's what the product and quotient rules are for. This page covers both formulas with worked examples and the sign errors that trip people up most. It's part of the broader Differentiation topic.
13 questions on this sub-topic.
The two rules
Covered under IB syllabus reference SL5.6. Both formulas are in the formula booklet, so the skill is recognising which one a question needs and applying it without a sign slip.
Product rule
\[y=uv \Rightarrow y'=u'v+uv'\]
Differentiate each factor in turn, keeping the other factor unchanged, and add the two results.
✓ In the formula bookletQuotient rule
\[y=\dfrac{u}{v} \Rightarrow y'=\dfrac{u'v-uv'}{v^2}\]
"Low d-high minus high d-low, over low squared" - order matters because subtraction isn't commutative.
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table? See Differentiation.
Worked examples
Differentiate \(y = x^2 \sin x\).
Worked solution
\((uv)' = u'v + uv'\) with \(u = x^2,\ v = \sin x,\ u'=2x,\ v'=\cos x.\) M1
\(\dfrac{dy}{dx} = 2x\sin x + x^2\cos x.\) A1
Differentiate \(y = \dfrac{2x+1}{x^2+3}.\)
Worked solution
\(\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}\) with \(u = 2x+1,\ v = x^2+3.\) M1
\(u' = 2,\ v' = 2x,\) so \(\dfrac{dy}{dx} = \dfrac{2(x^2+3) - (2x+1)(2x)}{(x^2+3)^2}\) A1
\(\dfrac{dy}{dx} = \dfrac{-2x^2 - 2x + 6}{(x^2+3)^2}.\) A1
Differentiate \(y = \dfrac{x}{x + 1}\).
Worked solution
\(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\) with \(u=x,\ v=x+1.\) M1
\(u'=1,\ v'=1.\) A1
\(\dfrac{dy}{dx}=\dfrac{(1)(x+1)-x(1)}{(x+1)^2}=\dfrac{1}{(x+1)^2}.\) A1
\(y = \ln(5x^2 + 2).\)
(a) Find \(\dfrac{dy}{dx}\).
(b) Differentiate \(y = e^{3x}\cos x.\)
Worked solution
(a) Chain rule: \(\dfrac{dy}{dx}\) M1
\(= \dfrac{10x}{5x^2+2}.\) A1
(b) Product rule: \(\dfrac{dy}{dx} = 3e^{3x}\cos x - e^{3x}\sin x\) M1
\(= e^{3x}(3\cos x - \sin x).\) A1
Common mistakes
- Getting the quotient rule's order backwards. \(\dfrac{u'v-uv'}{v^2}\) is not symmetric - swapping \(u\) and \(v\) in the numerator gives the wrong sign on the whole answer, not just a small error.
- Forgetting the chain rule inside a product or quotient factor. If \(u\) or \(v\) is itself a composite function - like \(\sqrt{1+x^2}\) - you still need the chain rule to differentiate it before plugging it into the product or quotient formula.
- Not simplifying (or over-simplifying) the final answer. Exam mark schemes often award the last A1 for a specific simplified form, so factorise or combine fractions where the algebra allows it, but don't force an unnecessary expansion that risks a new error.
Ready to practise properly?
13 product-and-quotient-rule questions, marked instantly like the real exam.
Quick answers
What is the product rule?
For \(y = uv\), \(y'=u'v+uv'\). Differentiate each factor in turn, keeping the other factor unchanged, and add the two results.
What is the quotient rule?
For \(y = \dfrac{u}{v}\), \(y'=\dfrac{u'v-uv'}{v^2}\). Order matters because subtraction isn't commutative - a common shortcut is "low d-high minus high d-low, over low squared".
Need a calculator refresher? See Using your GDC on the full Differentiation page.