Learn9709Paper 2 · Pure Mathematics 2DifferentiationStationary points and tangents

Differentiation · Subtopic 5 of 5

Stationary Points and Tangents

Putting the Pure 2/3 rules to work: finding exact stationary points of curves like $y=xe^{2x}$ and $y=\dfrac{\ln x}{x}$ , deciding their nature, and finding tangents and normals.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find stationary points of curves involving $e^x$ , $\ln x$ and trigonometric functions.
Use the fact that $e^x>0$ to simplify $\dfrac{dy}{dx}=0$ .
Give exact coordinates in terms of $e$ , $\ln$ and $\pi$ .
Find equations of tangents and normals to these curves.

Pause and think

$\dfrac{dy}{dx}=(1+2x)e^{2x}$ . For which value of $x$ is this zero? Why don't you need to worry about $e^{2x}$ ?

Section 2 of 14

Core idea

Start hereDrug in the bloodstream

After an injection, the drug concentration in a patient's blood rises, peaks, then decays, often modelled by $C=te^{-t}$ . When is it highest? Set the derivative to zero. All the new derivatives feed into the same tools as before: stationary points, tangents, normals.

The big idea

Use product, quotient and chain rules to differentiate, factorise the derivative, set it to zero, and solve exponential, log or trig equations. $e^{(\ldots)}$ is never zero.

1. Stationary points

y = x e⁻ˣ: a maximum at x = 1

Product rule:

\dfrac{dy}{dx}=e^{-x}-xe^{-x}=e^{-x}(1-x)

. Since

e^{-x}

is never zero, the gradient is zero only when

x=1

Worked example $y=xe^{-x}$

Find and classify the stationary point.

Goal: Point and nature.
Strategy: Product rule; factorise.

$y'=e^{-x}+x(-e^{-x})=e^{-x}(1-x)$
Product rule.
$e^{-x}\neq0\Rightarrow x=1,\ y=e^{-1}$
Solve.
$y''=-e^{-x}(1-x)-e^{-x}=e^{-x}(x-2);\ y''(1)=-e^{-1}<0$
Second derivative.

AnswerMaximum

(1,\frac1e)

✓ Check

y(0.9)=0.366

y(1.1)=0.366

y(1)=0.368

✓.

Worked example $y=\dfrac{\ln x}x$

Find the stationary point.

Goal: Point.
Strategy: Quotient rule.

$y'=\frac{x\cdot\frac1x-\ln x}{x^2}=\frac{1-\ln x}{x^2}$
Quotient rule.
$\ln x=1\Rightarrow x=e$
Numerator zero.

Answer

(e,\frac1e)

, a maximum.

✓ CheckThis proves

\frac{\ln x}x\le\frac1e

; e.g.

\frac{\ln3}3=0.366<0.368

✓.

Worked example $y=\sin x+\cos2x$ , $0<x<\pi$

Find the $x$ -coordinates of the stationary points.

Goal: All $x$ .
Strategy: Differentiate; double angle; factorise.

$y'=\cos x-2\sin2x=\cos x-4\sin x\cos x$
$\sin2x=2\sin x\cos x$ .
$\cos x(1-4\sin x)=0$
Factorise.
$x=\tfrac\pi2;\ \sin x=\tfrac14\Rightarrow x=0.253,\ 2.889$
Solve each.

Answer

x=0.253

\frac\pi2

2.889

✓ CheckThree turning points in

(0,\pi)

; values alternate max/min/max ✓.

2. Tangents and normals

Worked exampleNormal to $y=\ln(2x-1)$ at $x=1$

Find it.

Goal: Line.
Strategy: Point, gradient, $-1/m$ .

$y=\ln1=0$
Point $(1,0)$ .
$y'=\frac2{2x-1}=2$
Chain rule.
$y=-\tfrac12(x-1)$
Normal.

Answer

x+2y-1=0

✓ Check

(1,0)

1+0-1=0

✓.

Pause and think

Find the minimum of $y=e^x-2x$ .

Summary

Factorise derivatives.
Solve exactly with ln and inverse trig.

Stationary point: Gradient zero.

HistoryGalileo's best angle. Galileo found experimentally that a projectile on level ground travels furthest when launched at

45^\circ

. With calculus it is one line: the range is proportional to

\sin2\theta

, whose derivative

2\cos2\theta

is zero when

\theta=45^\circ

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