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

For a function of a function, $y=f(u)$ with $u=g(x)$: $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.$ In practice: differentiate the outer function, leaving the inner function alone, then multiply by the derivative of the inner function. Example: $y=\sqrt{4x^2+9}=(4x^2+9)^{\frac12}$ gives $\frac{dy}{dx}=\frac12(4x^2+9)^{-\frac12}\times8x=\frac{4x}{\sqrt{4x^2+9}}$. Chains can have more than two links: $y=\sin^3(2x)$ gives $3\sin^2(2x)\times\cos(2x)\times2$.

For $y=uv$ where $u$ and $v$ are both functions of $x$: $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.$ Example: $y=x^2e^{3x}$, with $u=x^2$ and $v=e^{3x}$, gives $\frac{dy}{dx}=x^2(3e^{3x})+e^{3x}(2x)=(2x+3x^2)e^{3x}=xe^{3x}(2+3x)$. Factorise the result to find stationary points: $e^{3x}$ is never zero, so $x=0$ or $x=-\frac23$. To classify, differentiate again, using the product rule a second time.

For $y=\frac{u}{v}$: $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.$ Example: $y=\frac{3x+1}{x^2+1}$ gives $\frac{dy}{dx}=\frac{3(x^2+1)-(3x+1)(2x)}{(x^2+1)^2}=\frac{-3x^2-2x+3}{(x^2+1)^2}$. Stationary points need only the numerator to be zero: $3x^2+2x-3=0$. The order in the numerator matters: it is $v\,u'-u\,v'$.

If quantities are linked, chain rule lets you connect their rates: $\frac{dr}{dt}=\frac{dr}{dV}\times\frac{dV}{dt}=\frac{dV}{dt}\div\frac{dV}{dr}.$ Example: a balloon with $V=\frac43\pi r^3$ is inflated at $50$ cm$^3$ s$^{-1}$. Then $\frac{dV}{dr}=4\pi r^2$ and at $r=5$, $\frac{dr}{dt}=\frac{50}{100\pi}=\frac{1}{2\pi}$ cm s$^{-1}$. For the surface area $S=4\pi r^2$: $\frac{dS}{dt}=\frac{dS}{dr}\times\frac{dr}{dt}=8\pi r\times\frac{1}{2\pi}=20$ cm$^2$ s$^{-1}$ when $r=5$.

For a function whose inverse is also differentiable: $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}.$ Example: if $t=\frac{2\pi r^3}{75}$ then $\frac{dt}{dr}=\frac{2\pi r^2}{25}$, so $\frac{dr}{dt}=\frac{25}{2\pi r^2}$. Example: $x=y^3+y$ gives $\frac{dx}{dy}=3y^2+1$, so $\frac{dy}{dx}=\frac{1}{3y^2+1}$. At $y=1$ (so $x=2$), the gradient is $\frac14$. This is useful when $x$ is easier to write in terms of $y$, and when a rate is easier to find in the reverse direction.

Look at the outermost operation. A product, such as $x\sin2x$, uses the product rule. A fraction uses the quotient rule. A function of a function uses the chain rule. Many questions use two: $y=x^2\sqrt{1+x}$ needs the product rule, with the chain rule for the root. Simplify before differentiating when possible, take out common factors afterwards, and use the result for tangents, normals and stationary points.