What are stationary points?

A stationary point occurs where dydx = 0. Classify it with the second derivative: d^2ydx^2 > 0 means a minimum, d^2ydx^2 < 0 means a maximum, and if it is zero you must test the sign of the first derivative either side.

What is not checking the nature of an optimisation solution?

Always confirm you have a maximum or minimum as required.

Pearson Edexcel Edexcel 2019-style practice: A closed cylinder has total surface area 600π cm^2. Show that the volume V is given by V = 300π r - π r^3, and hence find the maximum volume, justifying that it is a maximum.

The surface area is 2π r^2 + 2π r h = 600π, so h = 600π - 2π r^22π r = 300 - r^2r (M1). Volume V = π r^2 h = π r^2 · 300 - r^2r = π r(300 - r^2) = 300π r - π r^3 (A1, the printed result). Differentiate: dVdr = 300π - 3π r^2 (M1). Set to zero: 300π = 3π r^2, so r^2 = 100 and r = 10 (M1, A1). Second derivative d^2Vdr^2 = -6π r < 0, confirming a maximum (M1). Maximum volume V = 300π(10) - π(10)^3 = 3000π - 1000π = 2000π cm^3 (A1). Markers reward eliminating h, the printed expression, differentiating, solving for r, the second-derivative justification, and the final volume.

EnglandMathsSyllabus dot point

How do you find the rate at which a quantity changes, and how do you use that to analyse curves and solve optimisation problems?

Differentiation from first principles, the rules for powers, the chain, product and quotient rules, derivatives of standard functions, implicit and parametric differentiation, stationary points and connected rates of change.

A focused answer to the Edexcel A-Level Mathematics differentiation content, covering first principles, the chain, product and quotient rules, derivatives of standard functions, implicit and parametric differentiation, stationary points and their nature, and applications to optimisation.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Edexcel wants you to differentiate from first principles, use the power, chain, product and quotient rules, differentiate standard functions including trigonometric, exponential and logarithmic ones, differentiate implicitly and parametrically, find and classify stationary points, and apply differentiation to tangents, normals, rates of change and optimisation.

The answer

First principles

The derivative is defined as the limit $f'(x) = \lim_{h \to 0}\dfrac{f(x+h) - f(x)}{h}$ .

The rules

Standard derivatives include $\frac{d}{dx}(e^x) = e^x$ , $\frac{d}{dx}(\ln x) = \frac{1}{x}$ , $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$ .

Implicit and parametric differentiation

For an implicit relation such as $x^2 + y^2 = 25$ , differentiate every term with respect to $x$ and treat $y$ as a function of $x$ , giving $2x + 2y\frac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{x}{y}$ . For a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .

Stationary points

A stationary point occurs where $\dfrac{dy}{dx} = 0$ . Classify it with the second derivative: $\dfrac{d^2y}{dx^2} > 0$ means a minimum, $\dfrac{d^2y}{dx^2} < 0$ means a maximum, and if it is zero you must test the sign of the first derivative either side.

Examples in context

Connected rates and optimisation

If two quantities both depend on time, the chain rule links their rates: $\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$ . In optimisation you write the quantity to be maximised or minimised as a function of one variable, differentiate, set the derivative to zero, and check the nature of the stationary point.

Try this

Q1. Differentiate $y = (2x + 1)^4$ . [2 marks]

Cue. Chain rule: $\frac{dy}{dx} = 8(2x+1)^3$ .

Q2. Find the maximum value of $y = 12x - x^3$ for $x > 0$ . [4 marks]

Cue. $\frac{dy}{dx} = 12 - 3x^2 = 0$ gives $x = 2$ ; the value is $y = 16$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20197 marksA closed cylinder has total surface area

600\pi

^2

. Show that the volume

V

is given by

V = 300\pi r - \pi r^3

, and hence find the maximum volume, justifying that it is a maximum.

Show worked answer →

The surface area is $2\pi r^2 + 2\pi r h = 600\pi$ , so $h = \dfrac{600\pi - 2\pi r^2}{2\pi r} = \dfrac{300 - r^2}{r}$ (M1).

Volume $V = \pi r^2 h = \pi r^2 \cdot \dfrac{300 - r^2}{r} = \pi r(300 - r^2) = 300\pi r - \pi r^3$ (A1, the printed result).

Differentiate: $\dfrac{dV}{dr} = 300\pi - 3\pi r^2$ (M1). Set to zero: $300\pi = 3\pi r^2$ , so $r^2 = 100$ and $r = 10$ (M1, A1).

Second derivative $\dfrac{d^2V}{dr^2} = -6\pi r < 0$ , confirming a maximum (M1).

Maximum volume $V = 300\pi(10) - \pi(10)^3 = 3000\pi - 1000\pi = 2000\pi$ cm $^3$ (A1).

Markers reward eliminating $h$ , the printed expression, differentiating, solving for $r$ , the second-derivative justification, and the final volume.

Edexcel 20235 marksDifferentiate

y = x^2 \ln x

with respect to

x

, and hence find the exact coordinates of the stationary point of the curve.

Show worked answer →

Use the product rule with $u = x^2$ and $v = \ln x$ , so $\dfrac{du}{dx} = 2x$ and $\dfrac{dv}{dx} = \dfrac{1}{x}$ (M1).

$\dfrac{dy}{dx} = x^2 \cdot \dfrac{1}{x} + \ln x \cdot 2x = x + 2x\ln x = x(1 + 2\ln x)$ (A1).

At a stationary point $\dfrac{dy}{dx} = 0$ . Since $x > 0$ , set $1 + 2\ln x = 0$ (M1), so $\ln x = -\dfrac{1}{2}$ and $x = e^{-1/2}$ (A1).

Then $y = (e^{-1/2})^2 \ln(e^{-1/2}) = e^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{1}{2e}$ , so the point is $\left(e^{-1/2}, -\dfrac{1}{2e}\right)$ (A1).

Markers reward the product rule, the simplified derivative, solving the logarithm, and the exact coordinates.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)