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

If $y=uv$, where $u$ and $v$ are functions of $x$: $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.$ Use it when two functions containing $x$ are multiplied. Example: $y=2x^{4}\sin x$ with $u=2x^4$ and $v=\sin x$ gives $\frac{dy}{dx}=2x^{4}\cos x+8x^{3}\sin x$. For $y=x^{3}e^{2x}$: $\frac{dy}{dx}=3x^{2}e^{2x}+2x^{3}e^{2x}=x^{2}e^{2x}(3+2x)$. Factorising the answer makes stationary points easy: $e^{2x}$ is never zero.

If $y=\dfrac{u}{v}$: $\frac{dy}{dx}=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}.$ The order in the numerator matters: $v$ times $u'$ minus $u$ times $v'$. Example: $y=\dfrac{e^{3x}}{x}$ gives $\frac{dy}{dx}=\dfrac{3xe^{3x}-e^{3x}}{x^{2}}=\dfrac{e^{3x}(3x-1)}{x^{2}}$. A stationary point needs the numerator to be zero: $3x-1=0$, so $x=\frac13$.

For a function of a function, $y=f(g(x))$, let $u=g(x)$. Then $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.$ Examples: $y=\cos(x^{2})$ with $u=x^2$ gives $\frac{dy}{dx}=-\sin u\times2x=-2x\sin(x^{2})$. For $y=\tan^{2}2x=(\tan2x)^2$, let $u=\tan2x$: $\frac{dy}{dx}=2u\times2\sec^{2}2x=4\tan2x\sec^{2}2x$. Chain rule can be combined with the product or quotient rule in a single question.

These results are required (for $x$ in radians): $\frac{d}{dx}(\sec x)=\sec x\tan x,\quad\frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x,\quad\frac{d}{dx}(\cot x)=-\operatorname{cosec}^{2}x.$ They are proved from $\sec x=\frac{1}{\cos x}$, $\operatorname{cosec}x=\frac{1}{\sin x}$ and $\cot x=\frac{\cos x}{\sin x}$ using the chain or quotient rule. For example, $\frac{d}{dx}\left(\frac{1}{\cos x}\right)=\frac{\sin x}{\cos^{2}x}=\sec x\tan x$. Remember also $\frac{d}{dx}(\tan x)=\sec^2x$. Functions of $kx$ pick up a factor $k$ from the chain rule.

Decide the structure first: a product (two factors containing $x$), a quotient, or a composite (a function inside another). Often more than one applies. After differentiating, factorise before solving $\frac{dy}{dx}=0$ and discard factors that can never be zero (such as $e^{kx}$). Worked example: $y=x^2\ln x$. $\frac{dy}{dx}=2x\ln x+x$. Stationary when $\ln x=-\frac12$, i.e. $x=e^{-1/2}$, $y=-\frac{1}{2e}$. The tangent at $x=e$ has gradient $3e$, so $y=3ex-2e^2$.