In Year 1 you could differentiate sums of powers of $x$. That is enough for polynomials, but almost nothing in the real world is a polynomial. Year 2 differentiation (DfE content area G) gives you the tools to differentiate any function you can write down: exponentials, logarithms, trigonometric functions, and every combination of them.
Implicit and parametric curves · concavity and points of inflection · connected rates of change
| Word | Meaning | Example |
|---|---|---|
| Composite function | A function of a function — needs the chain rule. | $\sin(x^2)$, $(3x - 1)^5$ |
| Product / quotient | Two functions multiplied / divided. | $x^2\mathrm{e}^x$, $\dfrac{\ln x}{x}$ |
| Parametric | $x$ and $y$ both given in terms of a parameter $t$. | $x = t^2$, $y = 2t$ |
| Implicit | An equation in $x$ and $y$ not solved for $y$. | $x^2 + xy + y^2 = 7$ |
| Convex | $\mathrm{f}''(x) \geqslant 0$: the curve bends upwards. | $y = \mathrm{e}^x$ |
| Concave | $\mathrm{f}''(x) \leqslant 0$: the curve bends downwards. | $y = \ln x$ |
| Point of inflection | Where the curve changes between concave and convex. | $(0, 0)$ on $y = x^3$ |
| Rate of change | A derivative with respect to time. | $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ |
$y = x^2\ln x$, $x > 0$. Find the exact coordinates of the stationary point, determine its nature, and find where the curve is convex.
Each part of DfE content area G covered in Year 2, and the page that covers it. The Year 1 basics are in A Level Maths Year 1 · Differentiation.
- GExponentials, Logs and TrigDerivatives of $\mathrm{e}^{kx}$, $a^{kx}$, $\ln x$, $\sin kx$, $\cos kx$, $\tan kx$, with first-principles proofs for $\sin x$ and $\cos x$.
- GChain, Product and Quotient RulesDifferentiating composite functions, products and quotients, and $\frac{\mathrm{d}y}{\mathrm{d}x} = 1/\frac{\mathrm{d}x}{\mathrm{d}y}$.
- GImplicit and Parametric DifferentiationGradients, tangents, normals and stationary points of implicit and parametric curves.
- GConcavity and Points of InflectionUsing the second derivative to find where a curve is convex or concave.
- GConnected Rates of ChangeRelated rates with the chain rule, and setting up simple differential equations.
Exponentials
$\mathrm{e}^{kx} \to k\mathrm{e}^{kx}$; $a^x \to a^x\ln a$
Logs
$\ln x \to \tfrac1x$; $\ln\mathrm{f}(x) \to \tfrac{\mathrm{f}'(x)}{\mathrm{f}(x)}$
Trig
$\sin \to \cos$, $\cos \to -\sin$, $\tan \to \sec^2$ (times $k$)
Chain
Outside derivative × inside derivative.
Product
$uv' + vu'$
Quotient
$\dfrac{vu' - uv'}{v^2}$
Parametric
$\dot y / \dot x$
Implicit
Differentiate $y$-terms, then multiply by $\frac{\mathrm{d}y}{\mathrm{d}x}$.
Inflection
$\mathrm{f}'' = 0$ and changes sign.
Ten questions drawing on the whole Differentiation topic. Work each one through before opening the solution.
Differentiate $y = \mathrm{e}^{5x} + 3\ln x - 2\sin x$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 5\mathrm{e}^{5x} + \dfrac{3}{x} - 2\cos x$.
Differentiate $y = (x^2 + 3)^4$.
▶ Show solution
Chain rule: $4(x^2 + 3)^3 \times 2x = 8x(x^2 + 3)^3$.
Differentiate $y = x\cos x$.
▶ Show solution
Product rule: $x(-\sin x) + \cos x \cdot 1 = \cos x - x\sin x$.
Show that the derivative of $y = \dfrac{\mathrm{e}^x}{x + 1}$ is $\dfrac{x\mathrm{e}^x}{(x + 1)^2}$.
▶ Show solution
Quotient rule: $\dfrac{(x + 1)\mathrm{e}^x - \mathrm{e}^x \cdot 1}{(x + 1)^2} = \dfrac{\mathrm{e}^x(x + 1 - 1)}{(x + 1)^2} = \dfrac{x\mathrm{e}^x}{(x + 1)^2}$.
A curve has parametric equations $x = t^2 + 1$, $y = 4t$. Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$ and the gradient at $(5, 8)$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4}{2t} = \dfrac{2}{t}$.
$y = 8 \Rightarrow t = 2$ (check $x = 5$ ✓), so the gradient is $1$.
Find the gradient of the curve $x^2 + xy + y^2 = 7$ at the point $(1, 2)$.
▶ Show solution
$2x + x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2x + y}{x + 2y}$.
At $(1, 2)$: $-\dfrac{4}{5}$.
Find the point of inflection on $y = x^3 - 3x^2 + 2x$, showing that it is a point of inflection.
▶ Show solution
$y'' = 6x - 6 = 0 \Rightarrow x = 1$; $y''$ is negative for $x < 1$ and positive for $x > 1$, so it changes sign.
$y = 1 - 3 + 2 = 0$: point of inflection $(1, 0)$.
The radius of a spherical balloon increases at $0.5$ cm s$^{-1}$. Find the rate of increase of its volume when $r = 2$ cm, in terms of $\pi$.
▶ Show solution
$\dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2 = 16\pi$. $\;\dfrac{\mathrm{d}V}{\mathrm{d}t} = 16\pi \times 0.5 = 8\pi$ cm$^3$ s$^{-1}$.
Prove from first principles, using small angle approximations, that $\dfrac{\mathrm{d}}{\mathrm{d}x}(\cos x) = -\sin x$.
▶ Show solution
$\dfrac{\cos(x + h) - \cos x}{h} = \dfrac{\cos x\cos h - \sin x\sin h - \cos x}{h}$.
With $\cos h \approx 1 - \tfrac12 h^2$ and $\sin h \approx h$: $\approx \dfrac{-\tfrac12 h^2\cos x - h\sin x}{h} = -\tfrac12 h\cos x - \sin x$.
As $h \to 0$ this tends to $-\sin x$. $\blacksquare$
$\mathrm{f}(x) = \mathrm{e}^{-x}\sin x$, $\;0 \leqslant x \leqslant \pi$.
(a) Show that $\mathrm{f}'(x) = \mathrm{e}^{-x}(\cos x - \sin x)$.
(b) Find the exact coordinates of the stationary point.
(c) Show that $\mathrm{f}''(x) = -2\mathrm{e}^{-x}\cos x$, and hence determine the nature of the stationary point.
(d) Find the exact coordinates of the point of inflection.
▶ Show solution
(a) Product rule: $\mathrm{e}^{-x}\cos x + \sin x \cdot (-\mathrm{e}^{-x}) = \mathrm{e}^{-x}(\cos x - \sin x)$.
(b) $\mathrm{e}^{-x} > 0$, so $\cos x = \sin x \Rightarrow \tan x = 1 \Rightarrow x = \dfrac{\pi}{4}$. $\;\mathrm{f}\!\left(\tfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}\mathrm{e}^{-\pi/4}$ ($\approx 0.322$).
(c) $\mathrm{f}''(x) = -\mathrm{e}^{-x}(\cos x - \sin x) + \mathrm{e}^{-x}(-\sin x - \cos x) = -2\mathrm{e}^{-x}\cos x$.
At $x = \tfrac{\pi}{4}$: $-2\mathrm{e}^{-\pi/4} \cdot \tfrac{\sqrt2}{2} < 0$, so a maximum.
(d) $\mathrm{f}''(x) = 0$ when $\cos x = 0$, so $x = \dfrac{\pi}{2}$; $\cos x$ changes sign there, so it is a point of inflection: $\left(\dfrac{\pi}{2}, \mathrm{e}^{-\pi/2}\right)$.