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Product, quotient and chain rulesEdexcel A-Level Maths: Revision notes

Section 1

The chain rule

For a composite function y=f(g(x))y=f(g(x)), let u=g(x)u=g(x). Then dydx=dydu×dudx.\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}. Differentiate the outer function, keeping the inner one, then multiply by the derivative of the inner function. Examples: ddxcos⁡2x=2cos⁡x×(−sin⁡x)=−2sin⁡xcos⁡x\frac{d}{dx}\cos^{2}x=2\cos x\times(-\sin x)=-2\sin x\cos x and ddxtan⁡22x=2tan⁡2x×2sec⁡22x=4tan⁡2xsec⁡22x\frac{d}{dx}\tan^{2}2x=2\tan2x\times2\sec^{2}2x=4\tan2x\sec^{2}2x. Chain rule results, such as ddx(2x+1)5=10(2x+1)4\frac{d}{dx}(2x+1)^{5}=10(2x+1)^{4}, are how the standard derivatives ddxekx=kekx\frac{d}{dx}e^{kx}=ke^{kx} arise.

Key termschain rulecomposite function
Common mistake

Forgetting to multiply by the derivative of the inner function, for example differentiating tan⁡22x\tan^{2}2x without the factor 2.

Exam tip

Write cos⁡2x\cos^{2}x as (cos⁡x)2(\cos x)^{2} so the outer and inner functions are obvious.

Section 2

The product rule

If y=uvy=uv where uu and vv are functions of xx, dydx=udvdx+vdudx.\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}. Example: y=2x4sin⁡xy=2x^{4}\sin x with u=2x4u=2x^{4} and v=sin⁡xv=\sin x gives dydx=2x4cos⁡x+8x3sin⁡x\frac{dy}{dx}=2x^{4}\cos x+8x^{3}\sin x. Example: f(x)=x3e2xf(x)=x^{3}e^{2x} gives f′(x)=x3×2e2x+3x2e2x=x2e2x(2x+3)f'(x)=x^{3}\times2e^{2x}+3x^{2}e^{2x}=x^{2}e^{2x}(2x+3), so the non-zero stationary point is at x=−32x=-\frac32 because e2xe^{2x} is never 00.

Key termsproduct rule
Common mistake

Multiplying the two derivatives together. The derivative of a product is not the product of the derivatives.

Exam tip

Factorise the result: it makes solving f′(x)=0f'(x)=0 much easier.

Section 3

The quotient rule

If y=uvy=\frac uv, dydx=vdudx−udvdxv2.\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}. Example: y=e3xxy=\frac{e^{3x}}{x} gives dydx=x×3e3x−e3xx2=e3x(3x−1)x2\frac{dy}{dx}=\frac{x\times3e^{3x}-e^{3x}}{x^{2}}=\frac{e^{3x}(3x-1)}{x^{2}}. Example: h(x)=ln⁡xx2h(x)=\frac{\ln x}{x^{2}} gives h′(x)=x2×1x−2xln⁡xx4=1−2ln⁡xx3h'(x)=\frac{x^{2}\times\frac1x-2x\ln x}{x^{4}}=\frac{1-2\ln x}{x^{3}}. Sometimes it is quicker to write uv=uv−1\frac uv=uv^{-1} and use the product and chain rules.

Key termsquotient rule
Common mistake

Putting the terms in the numerator in the wrong order: it is v u′−u v′v\,u'-u\,v', and the order matters.

Section 4

Derivatives of sec⁡x\sec x, cosec⁡x\operatorname{cosec}x and cot⁡x\cot x

These come from the quotient or chain rule, writing sec⁡x=1cos⁡x\sec x=\frac{1}{\cos x} and so on: ddx(sec⁡x)=sec⁡xtan⁡x,ddx(cosec⁡x)=−cosec⁡xcot⁡x,ddx(cot⁡x)=−cosec⁡2x.\frac{d}{dx}(\sec x)=\sec x\tan x,\quad\frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x,\quad\frac{d}{dx}(\cot x)=-\operatorname{cosec}^{2}x. The derivatives of cosec and cot each carry a minus sign. Example: y=cos⁡x1+sin⁡xy=\frac{\cos x}{1+\sin x} has dydx=−sin⁡x(1+sin⁡x)−cos⁡2x(1+sin⁡x)2=−11+sin⁡x\frac{dy}{dx}=\frac{-\sin x(1+\sin x)-\cos^{2}x}{(1+\sin x)^{2}}=-\frac{1}{1+\sin x}, using sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1.

Key termssecantcosecantcotangent
Exam tip

Use sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1 to simplify numerators after a quotient rule.

Section 5

Connected rates of change

When quantities are linked, use the chain rule to connect their rates: dVdt=dVdr×drdt.\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}. Example: a sphere's volume V=43πr3V=\frac43\pi r^{3} grows at 200200 cm3^3 s−1^{-1}. Then dVdr=4πr2\frac{dV}{dr}=4\pi r^{2} and drdt=2004πr2\frac{dr}{dt}=\frac{200}{4\pi r^{2}}, which is 2π\frac2\pi cm s−1^{-1} when r=5r=5. Surface area S=4πr2S=4\pi r^{2} has dSdt=8πr×drdt=40π×2π=80\frac{dS}{dt}=8\pi r\times\frac{dr}{dt}=40\pi\times\frac2\pi=80 cm2^2 s−1^{-1} at r=5r=5.

Key termsrate of changeconnected rates
Exam tip

Write down the rate you want, the rate you are given, and the link, such as drdt=drdV×dVdt\frac{dr}{dt}=\frac{dr}{dV}\times\frac{dV}{dt}, before substituting.

Common mistake

Substituting the value of rr too early, before differentiating with respect to rr.

Section 6

Inverse functions and dydx=1dx/dy\frac{dy}{dx}=\frac{1}{dx/dy}

When xx is given as a function of yy, find dxdy\frac{dx}{dy} and then invert: dydx=1 dx/dy .\frac{dy}{dx}=\frac{1}{\,dx/dy\,}. Example: x=3tan⁡2yx=3\tan2y gives dxdy=6sec⁡22y\frac{dx}{dy}=6\sec^{2}2y, so dydx=16sec⁡22y\frac{dy}{dx}=\frac{1}{6\sec^{2}2y}. Using sec⁡22y=1+tan⁡22y=1+x29\sec^{2}2y=1+\tan^{2}2y=1+\frac{x^{2}}{9}, this is 32(9+x2)\frac{3}{2(9+x^{2})}, which is 16\frac16 at the origin. Example: x=sin⁡yx=\sin y gives dxdy=cos⁡y\frac{dx}{dy}=\cos y, so dydx=1cos⁡y\frac{dy}{dx}=\frac{1}{\cos y}.

Key termsinverse function
Common mistake

Inverting only part of the expression. The whole dxdy\frac{dx}{dy} goes in the denominator.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Product, quotient and chain rules

  1. The function ff is defined by f(x)=x3e2xf(x)=x^{3}e^{2x} for all real xx.
    Find the equation of the tangent to the curve y=f(x)y=f(x) at the point where x=1x=1.2 marks
  2. The function hh is defined by h(x)=ln⁡xx2h(x)=\frac{\ln x}{x^{2}} for x>0x>0.
    Find the exact value of h(x)h(x) at the stationary point.2 marks
  3. A spherical balloon is inflated so that its volume, VV cm3^3, increases at a constant rate of 200200 cm3^3 s−1^{-1}. The radius of the balloon is rr cm. The volume of a sphere is V=43πr3V=\frac43\pi r^{3} and its surface area is S=4πr2S=4\pi r^{2}.
    Find the exact rate of increase of the radius when r=5r=5.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).