A curve has equation $ y = 2x^2 - \dfrac{12}{x} $. Find the gradient of the curve at the point where $ x = 2 $.

B1 writes 12/x as 12x⁻¹ M1 dy/dx = 4x + 12x⁻² (at least one term correct) M1 substitutes x = 2: 8 + 12/4 A1 11 cao Answer: 11

Edexcel IGCSE Maths (4MA1) · Unit 8: Calculus

Edexcel IGCSE 4MA1 Differentiation: Exam Questions and Answers

Differentiation questions with full solutions for Pearson Edexcel International GCSE Mathematics A (4MA1), Higher tier. Read the 5 worked examples first, then try the 8 exam-style questions yourself and check your working against the mark scheme.

5 worked examples
8 questions with mark schemes
Higher tier · calculator allowed
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Practise differentiation in Unit 8 →Differentiation revision notes

What this topic covers

Differentiation 1: Power Rule: Differentiating polynomial terms with the power rule.
Differentiation 2: Gradients of Curves: Finding gradient at a specific x-value.
Differentiation 3: Tangents & Normals: Equations of tangents and normals at a point: gradient m from dy/dx, normal gradient −1/m, then y − y₁ = m(x − x₁).
Differentiation 4: Stationary Points: Finding stationary points from dy/dx = 0.
Differentiation 5: Turning Points Optimization: Classifying max/min and optimisation problems.

Differentiation worked examples

Worked example 1: Power rule

Differentiate $y = 5 x^{4} - 3 x^{2} + 2 x - 7$ .

Solution

Multiply by the power, then reduce the power by one.
$\frac{d y}{d x} = 20 x^{3} - 6 x + 2$ .

3 steps

Tip: A constant differentiates to 0.

Worked example 2: Gradient of a curve

Find the gradient of $y = x^{2} - 5 x + 2$ at $x = 4$ .

Solution

$\frac{d y}{d x} = 2 x - 5$ .
At $x = 4$ : $8 - 5 = 3$ .

3 steps · 1 to fill in

Tip: Substitute into dy/dx, not into y.

Worked example 3: The tangent and the normal at a point

The curve $y = x^{3} - 2 x$ passes through the point $(2, 4)$ . Find (a) the equation of the tangent to the curve at this point, (b) the equation of the normal to the curve at this point.

Solution

$\frac{d y}{d x} = 3 x^{2} - 2$ ; at $x = 2$ the gradient of the tangent is 10.
(a) Tangent: $y - 4 = 10 (x - 2)$ , so $y = 10 x - 16$ .
(b) The normal is perpendicular to the tangent, so its gradient is $- \frac{1}{10}$ .
Normal: $y - 4 = - \frac{1}{10} (x - 2)$ ; multiply by 10: $10 y - 40 = - x + 2$ , so $x + 10 y = 42$ .

5 steps

Tip: The tangent has the same gradient as the curve at that point; the normal is perpendicular to it, so its gradient is

- \frac{1}{m}

. In 4MA1 the gradient may come from

\frac{d y}{d x}

or from a tangent drawn on a graph.

Worked example 4: Stationary points

Find the stationary points of $y = 2 x^{3} - 3 x^{2} - 12 x$ .

Solution

$\frac{d y}{d x} = 6 x^{2} - 6 x - 12 = 0$ , so $x^{2} - x - 2 = 0$ .
$(x - 2) (x + 1) = 0$ : $x = 2$ or $x = - 1$ .
$(2, - 20)$ and $(- 1, 7)$ : for a positive cubic, $(- 1, 7)$ is the maximum and $(2, - 20)$ the minimum.

4 steps · 1 to fill in

Tip: Find y by substituting into the original equation.

Worked example 5: Optimisation

A farmer has 60 m of fencing to make three sides of a rectangular pen against a wall. Find the maximum area.

Solution

Two sides of $x$ and one of $60 - 2 x$ : $A = 60 x - 2 x^{2}$ .
$\frac{d A}{d x} = 60 - 4 x = 0$ , so $x = 15$ .
$A = 15 \times 30 = 450$ m².

4 steps · 1 to fill in

Tip: Write the area in terms of one variable before differentiating.

Where students lose marks on differentiation

Examiner's noteDifferentiation: forgetting the negative root and the context

Where marks are lost: When dy/dx = 0 leads to x² = k, only the positive root is used; some do not realise that a turning-point or tangent question needs calculus at all.

Examiners flagged this in 4 exam series (November 2023, June 2024, November 2024, June 2025). Summarised in our own words from Pearson Edexcel examiner reports. Read the full note →

Differentiation exam-style questions

Try each question before you open the answer (one has its full worked solution open). The mark schemes use Edexcel-style marks: M for method, A for accuracy (after the method mark), B for an independent result.

Question 1 · easy · 2 marks

$y = 4 x^{3} - 5 x^{2} + 7 x - 2$ . Find $\frac{d y}{d x}$ .

Show the answer and mark scheme

M1 at least two of the three non-zero terms correct
A1 12x² − 10x + 7 cao

Answer: $12 x^{2} - 10 x + 7$

Question 2 · medium · 3 marks

$y = (2 x - 3) (x^{2} + 4)$ . Find $\frac{d y}{d x}$ .

Show the answer and mark scheme

M1 expands to 2x³ − 3x² + 8x − 12 (at least three terms correct)
M1 differentiates their expansion (at least two terms correct)
A1 6x² − 6x + 8 cao

Answer: $6 x^{2} - 6 x + 8$

Question 3 · medium · 4 marks

A curve has equation $y = 2 x^{2} - \frac{12}{x}$ . Find the gradient of the curve at the point where $x = 2$ .

Full worked solution Free, no sign-in

B1 writes 12/x as 12x⁻¹
M1 dy/dx = 4x + 12x⁻² (at least one term correct)
M1 substitutes x = 2: 8 + 12/4
A1 11 cao

Answer: 11

Every question here has a full worked solution. See them all →

Question 4 · medium · 4 marks

A curve has equation $y = x^{3} - 4 x + 1$ . Find the equation of the tangent to the curve at the point where $x = 2$ . Give your answer in the form $y = m x + c$ .

Show the answer and mark scheme

B1 y = 1 when x = 2
M1 dy/dx = 3x² − 4
M1 gradient 8 at x = 2 and y − 1 = 8(x − 2) oe
A1 y = 8x − 15 cao

Answer: $y = 8 x - 15$

Question 5 · medium · 4 marks

Find the coordinates of the two turning points of the curve $y = x^{3} - 3 x^{2} - 9 x + 5$ .

Show the answer and mark scheme

M1 dy/dx = 3x² − 6x − 9
M1 3x² − 6x − 9 = 0, so 3(x − 3)(x + 1) = 0
A1 x = 3 and x = −1
A1 (−1, 10) and (3, −22)

Answer: $(- 1, 10)$ and $(3, - 22)$

Question 6 · hard · 5 marks

The point $P (4, 2)$ lies on the curve $y = x^{2} - 6 x + 10$ . (a) Find the equation of the normal to the curve at P. Give your answer in the form $y = m x + c$ . (b) The normal meets the curve again at the point Q. Find the coordinates of Q.

Show the answer and mark scheme

M1 dy/dx = 2x − 6, so the gradient of the tangent at P is 2
M1 gradient of the normal −1/2 and y − 2 = −½(x − 4) oe
A1 y = −½x + 4 oe
M1 x² − 6x + 10 = −½x + 4, giving 2x² − 11x + 12 = 0 oe
A1 Q(3/2, 13/4) oe (1.5, 3.25)

Answer: (a) $y = - \frac{1}{2} x + 4$ (b) $(\frac{3}{2}, \frac{13}{4})$

Question 7 · hard · 5 marks

A curve has equation $y = 2 x + \frac{8}{x}$ . (a) Find the coordinates of the turning point of the curve for which $x > 0$ . (b) Is this turning point a maximum or a minimum? Give a reason for your answer.

Show the answer and mark scheme

M1 dy/dx = 2 − 8x⁻²
M1 2 − 8/x² = 0, so x² = 4
A1 (2, 8)
M1 gradient either side, e.g. −6 at x = 1 and 10/9 at x = 3 (or d²y/dx² = 16/x³ = 2 at x = 2)
A1 minimum, because the gradient goes from negative to positive (or d²y/dx² > 0)

Answer: (a) $(2, 8)$ (b) minimum

Question 8 · very hard · 6 marks

A closed box is a cuboid. Its base is $2 x$ cm long and $x$ cm wide, and its volume is 72 cm³. (a) Show that the total surface area, $S$ cm², of the box is given by $S = 4 x^{2} + \frac{216}{x}$ . (b) Find the minimum value of $S$ . Show that your value is a minimum.

Show the answer and mark scheme

M1 height h = 72 ÷ 2x² = 36/x²
A1 S = 2(2x·x) + 2(2x·h) + 2(x·h) = 4x² + 6x × 36/x² = 4x² + 216/x shown
M1 dS/dx = 8x − 216x⁻²
M1 8x − 216/x² = 0, so x³ = 27 and x = 3
A1 S = 36 + 72 = 108
B1 minimum shown, e.g. d²S/dx² = 8 + 432/x³ > 0, or dS/dx negative at x = 2 and positive at x = 4

Answer: (a) shown (b) 108

Every question and worked example here is original, written for IGCSE Math Revision rather than copied from Pearson papers, and each answer was re-solved independently before publishing.

Revise differentiation for the exam

Unit 8 practice: Calculus in the student hub (more questions with mark schemes)
IGCSE Maths revision notes for this topic and the IGCSE Maths formula sheet
Edexcel IGCSE Maths 4MA1 past papers to practise under exam conditions
IGCSE Maths 4MA1 examiner report summary: where students lose marks

Questions students ask

Which Edexcel IGCSE Maths papers test differentiation?

Either Higher tier paper, 1H or 2H, can test differentiation: both cover the whole Pearson Edexcel International GCSE Mathematics A (4MA1) specification and both allow a calculator. Show your method, because most marks are for working.

Where can I practise more differentiation questions?

In the IGCSE student hub: Unit 8 (Calculus) has more exam-style questions on this topic, each with a mark scheme. Practice starts with a 7-day free trial; the worked examples and questions on this page are free.

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Edexcel IGCSE 4MA1 Differentiation: Exam Questions and Answers

Try it first

What this topic covers

Differentiation worked examples

Worked example 1: Power rule

Solution

Worked example 2: Gradient of a curve

Solution

Worked example 3: The tangent and the normal at a point

Solution

Worked example 4: Stationary points

Solution

Worked example 5: Optimisation

Solution

Where students lose marks on differentiation

Differentiation exam-style questions

Question 1 · easy · 2 marks

Question 2 · medium · 3 marks

Question 3 · medium · 4 marks

Full worked solution Free, no sign-in

Question 4 · medium · 4 marks

Question 5 · medium · 4 marks

Question 6 · hard · 5 marks

Question 7 · hard · 5 marks

Question 8 · very hard · 6 marks

Revise differentiation for the exam

Questions students ask

Practise differentiation with your progress saved

More Unit 8 topics: Calculus