Launch price £9.99 until 31 Jan, then £14.99

TMUA Differentiation and Stationary Points

The derivative dydx\dfrac{dy}{dx} gives the gradient of a curve and the rate of change of yy. It is used for tangents, stationary points and finding maximum or minimum values.

Paper 1 and Paper 2 topic. Specification codes: MM6.1, MM6.2, MM6.3, MM8.5. Last checked 12 November 2026.

Rules

For any rational nn, ddxxn=nxn−1\dfrac{d}{dx}x^n=nx^{n-1}. Differentiate sums term by term, and simplify first, for example rewriting 1x\dfrac{1}{x} as x−1x^{-1} and x\sqrt x as x1/2x^{1/2}.

Tangents and normals

The tangent at a point has gradient equal to the derivative there. The normal is perpendicular to the tangent.

Stationary points

Solve f′(x)=0f'(x)=0. If f′′(x)>0f''(x)>0 it is a minimum, if f′′(x)<0f''(x)<0 a maximum. If f′′(x)=0f''(x)=0, check the sign of f′f' either side.

Increasing and decreasing functions

A function is increasing where f′(x)>0f'(x)>0 and decreasing where f′(x)<0f'(x)<0. Questions can combine this with algebra: a cubic has two stationary points exactly when f′(x)=0f'(x)=0 has two distinct roots, so the discriminant of f′f' must be positive.

The specification treats points of inflexion qualitatively only: they are where the curve changes from bending one way to the other.

Always check what the question wants: the coordinates, the value of yy, the nature of the point, or an equation of a line.

Worked examples

Example 1

Find the tangent to y=x3−3x2+2y=x^3-3x^2+2 at x=3x=3.

Show the worked solution

At x=3x=3, y=27−27+2=2y=27-27+2=2. y′=3x2−6x=9y'=3x^2-6x=9 there, so y−2=9(x−3)y-2=9(x-3), i.e. y=9x−25y=9x-25.

Answer: y=9x−25y=9x-25

Example 2

Find and classify the stationary points of f(x)=x3−6x2+9x+1f(x)=x^3-6x^2+9x+1.

Show the worked solution

f′(x)=3(x−1)(x−3)f'(x)=3(x-1)(x-3), so x=1x=1 or x=3x=3. f′′(x)=6x−12f''(x)=6x-12 is negative at 1 and positive at 3. So (1,5)(1,5) is a maximum and (3,1)(3,1) a minimum.

Answer: Max at (1,5)(1,5), min at (3,1)(3,1)

Example 3

Find the minimum value of x2+16xx^2+\dfrac{16}{x} for x>0x>0.

Show the worked solution

f′(x)=2x−16x−2=0f'(x)=2x-16x^{-2}=0 gives x3=8x^3=8, so x=2x=2. f′′(x)=2+32x−3>0f''(x)=2+32x^{-3}>0, so it is a minimum, with value 4+8=124+8=12.

Answer: 12

Common mistakes

  • Giving the xx-value of a stationary point when the question asks for the value of yy.
  • Differentiating 16x\dfrac{16}{x} as 161\dfrac{16}{1} instead of −16x−2-16x^{-2}.
  • Using the gradient of the tangent as the gradient of the normal.
  • Assuming every stationary point is a maximum or minimum.

Test yourself on this topic with our free practice questions.

Quick answers

How do you find a stationary point?

Differentiate, set the derivative equal to zero and solve for xx. Then substitute back to get yy.

How do you tell a maximum from a minimum?

Use the second derivative: positive means minimum, negative means maximum.

What is the gradient of a normal?

The negative reciprocal of the tangent's gradient.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

Related topics

Written by Proven Papers, an independent practice site, from the public UAT-UK content specification. Not affiliated with UAT-UK, Pearson VUE or any university.