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Learn7357Paper 1 · PureDifferentiationThe product and quotient rules
Differentiation · Subtopic 3 of 10

The Product and Quotient Rules

How to differentiate two functions multiplied together, uvuv, or divided, uv\dfrac uv, and how to tidy the result by factorising, so it is ready for finding stationary points.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use the product rule ddx(uv)=udvdx+vdudx\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}.
  • Use the quotient rule ddx(uv)=vdudx−udvdxv2\dfrac{d}{dx}\left(\dfrac uv\right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}.
  • Combine either rule with the chain rule.
  • Factorise the result to simplify it and find where it is zero.
Pause and think

Is the derivative of x⋅xx\cdot x equal to (derivative of xx) ×\times (derivative of xx)? Check with x⋅x=x2x\cdot x=x^2.

Section 2 of 14

Core idea

Start hereA rectangle that grows both ways

A rectangle's width and height are both growing. How fast is its area growing? Not just (rate of width) × (rate of height). The new area comes from two strips: width-growth × height and height-growth × width. That's the product rule, and it's a picture.

The big idea

ddx(uv)=udvdx+vdudx\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}. ddx(uv)=vdudx−udvdxv2\dfrac{d}{dx}\left(\dfrac uv\right)=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.

1. The product rule

Area of a growing rectangle: why the product rule has two terms
u × vuvv δuu δv
If uu grows a little (δu\delta u) and vv grows a little (δv\delta v), the area uvuv gains two strips: v δuv\,\delta u and u δvu\,\delta v (the tiny corner is negligible). That's udvdx+vdudxu\dfrac{dv}{dx}+v\dfrac{du}{dx}.
Proof
  1. δ(uv)=(u+δu)(v+δv)−uv\delta(uv)=(u+\delta u)(v+\delta v)-uv
    Change in area when xx increases by δx\delta x.
  2. =u δv+v δu+δu δv=u\,\delta v+v\,\delta u+\delta u\,\delta v
    Expand; uvuv cancels.
  3. δ(uv)δx=uδvδx+vδuδx+δuδvδx\frac{\delta(uv)}{\delta x}=u\frac{\delta v}{\delta x}+v\frac{\delta u}{\delta x}+\delta u\frac{\delta v}{\delta x}
    Divide by δx\delta x.
  4. →udvdx+vdudx+0\to u\frac{dv}{dx}+v\frac{du}{dx}+0
    As δx→0\delta x\to0, δu→0\delta u\to0: the corner vanishes.
Worked exampley=x3e2xy=x^3e^{2x}

Differentiate.

Goal
Factorised derivative.
Strategy
u=x3u=x^3, v=e2xv=e^{2x}.
  1. u′=3x2,v′=2e2xu'=3x^2,\quad v'=2e^{2x}
    Derivatives.
  2. dydx=x3⋅2e2x+e2x⋅3x2\frac{dy}{dx}=x^3\cdot2e^{2x}+e^{2x}\cdot3x^2
    uv′+vu′uv'+vu'.
  3. =x2e2x(2x+3)=x^2e^{2x}(2x+3)
    Factorise.
Answerx2e2x(2x+3)x^2e^{2x}(2x+3).
✓ CheckStationary at x=0x=0 and x=−32x=-\frac32 (plausible: a dip for negative xx) ✓.

2. The quotient rule

From the product and chain rules
  1. uv=uv−1\frac uv=uv^{-1}
    Rewrite.
  2. ddx(uv−1)=u⋅(−v−2v′)+v−1u′\frac{d}{dx}(uv^{-1})=u\cdot(-v^{-2}v')+v^{-1}u'
    Product rule; chain rule on v−1v^{-1}.
  3. =vu′−uv′v2=\frac{vu'-uv'}{v^2}
    Common denominator v2v^2.
Worked exampleDerive ddxtan⁡x\frac{d}{dx}\tan x

Show ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x=\sec^2x.

Goal
Proof.
Strategy
tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}.
  1. cos⁡x⋅cos⁡x−sin⁡x⋅(−sin⁡x)cos⁡2x\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^2x}
    Quotient rule.
  2. =cos⁡2x+sin⁡2xcos⁡2x=1cos⁡2x=sec⁡2x=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac1{\cos^2x}=\sec^2x
    Pythagoras.
Answersec⁡2x\sec^2x ∎.
✓ CheckAt x=0x=0: gradient 11, matching tan⁡x≈x\tan x\approx x near 00 ✓.
Worked exampley=exx+1y=\dfrac{e^x}{x+1}

Differentiate and find the stationary point.

Goal
Derivative; point.
Strategy
Quotient rule.
  1. (x+1)ex−ex⋅1(x+1)2=xex(x+1)2\frac{(x+1)e^x-e^x\cdot1}{(x+1)^2}=\frac{xe^x}{(x+1)^2}
    Simplify the top.
  2. xex=0⇒x=0xe^x=0\Rightarrow x=0
    ex≠0e^x\neq0.
Answerxex(x+1)2\frac{xe^x}{(x+1)^2}; stationary point (0,1)(0,1).
✓ Checky(0)=11=1y(0)=\frac11=1 ✓.
Common misconceptionOrder in the quotient rule "Bottom × d(top) − top × d(bottom), over bottom squared." Swapping gives the wrong sign.
Pause and think

Differentiate x2ln⁡xx^2\ln x.

Summary

  • (uv)′=uv′+vu′(uv)'=uv'+vu' (growing rectangle).
  • (u/v)′=(vu′−uv′)/v2(u/v)'=(vu'-uv')/v^2.
  • Factorise answers.
Key words
Product rule
Derivative of a product.
Quotient rule
Derivative of a fraction.
HistoryLeibniz got it wrong first. In his notes from November 1675, Leibniz wondered whether d(uv)d(uv) might simply be du⋅dvdu\cdot dv. He checked it, found it failed, and within days worked out the correct rule d(uv)=u dv+v dud(uv)=u\,dv+v\,du. Even the inventors of calculus made the mistake in the Think box.
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