State the quotient rule.

\frac{d}{dx}\left(\frac uv\right)=\frac{vu'-uv'}{v^{2}}.

State the chain rule.

\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.

Product, quotient and chain rulesEdexcel International A Level Maths: Flashcards

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State the product rule.

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State the product rule.: $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ .
State the quotient rule.: $\frac{d}{dx}\left(\frac uv\right)=\frac{vu'-uv'}{v^{2}}$ .
State the chain rule.: $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$ .
$\frac{d}{dx}(\sec x)$ ?: $\sec x\tan x$ .
$\frac{d}{dx}(\operatorname{cosec}x)$ ?: $-\operatorname{cosec}x\cot x$ .
$\frac{d}{dx}(\cot x)$ ?: $-\operatorname{cosec}^{2}x$ .
$\frac{d}{dx}(\tan x)$ ?: $\sec^{2}x$ .
Differentiate $x^{3}e^{2x}$ .: $x^{2}e^{2x}(3+2x)$ .
Differentiate $\frac{e^{3x}}{x}$ .: $\frac{e^{3x}(3x-1)}{x^{2}}$ .
Differentiate $\cos(x^{2})$ .: $-2x\sin(x^{2})$ .
Differentiate $\tan^{2}2x$ .: $4\tan2x\sec^{2}2x$ .
Differentiate $x^{2}\ln x$ .: $2x\ln x+x$ .
Why can you ignore $e^{2x}$ when solving $x^2e^{2x}(3+2x)=0$ ?: $e^{2x}$ is never zero.

Exam questions on Product, quotient and chain rules

The curve $C_1$ has equation $y=x^{3}e^{2x}$ .
Find the exact gradient of $C_1$ at $x=1$ .2 marks
The curve $C_2$ has equation $y=\dfrac{e^{3x}}{x}$ for $x>0$ .
Find the exact $y$ -coordinate of the stationary point of $C_2$ .2 marks
Two curves have equations $C_3:\ y=\cos\left(x^{2}\right)$ and $C_4:\ y=\tan^{2}2x$ , where $x$ is in radians.
Find $\dfrac{dy}{dx}$ on $C_3$ .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).