IGCSE Additional Maths · Cambridge 0606 · Edexcel 4PM1

Differentiation: notes and questions

Revision for standard derivatives, the chain, product and quotient rules, tangents and normals, stationary points, connected rates of change, small changes and optimisation. Both boards. Cambridge 0606 also differentiates tan x and ln x.

Cambridge 0606: section 14.1–14.9
Edexcel 4PM1: section 9B, 9D–G
14 questions

Practise differentiation →

Key points

d/dx(xⁿ) = nxⁿ⁻¹; d/dx(eᵃˣ) = aeᵃˣ; d/dx(sin ax) = a cos ax; d/dx(cos ax) = −a sin ax.
Chain rule: dy/dx = dy/du × du/dx. Product rule: (uv)′ = u′v + uv′. Quotient rule: (u/v)′ = (vu′ − uv′)/v².
Tangent gradient = dy/dx at the point; normal gradient = −1 ÷ (tangent gradient).
Stationary points: dy/dx = 0. If d²y/dx² > 0 it is a minimum; if d²y/dx² < 0 it is a maximum.
Connected rates: dA/dt = dA/dr × dr/dt. Small changes: δy ≈ (dy/dx) δx.
In optimisation questions, form the expression in one variable first, then differentiate.

Watch out: Trigonometric calculus works in radians only. Keep exact values (like 10π) until the last line.

Worked example · 2 marks

Find \(\frac{dy}{dx}\) when \(y = 4x^3 - 6x^2 + 2x - 7\).

M1 powers reduced by one
A1 12x² − 12x + 2

Differentiate term by term: d/dx(axⁿ) = naxⁿ⁻¹.
12x² − 12x + 2 (the constant −7 differentiates to 0).

Answer: \(\frac{dy}{dx} = 12x^2 - 12x + 2\)

Differentiation questions

Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).

Question 1 · 2 marks06064PM1no calculator

Differentiate \(y = (2x + 1)^5\) with respect to \(x\).

Show the mark scheme and worked solution

M1 chain rule: 5(2x + 1)⁴ × 2
A1 10(2x + 1)⁴

Chain rule: bring down the power, reduce it by one, multiply by the derivative of the inside.
5(2x + 1)⁴ × 2 = 10(2x + 1)⁴.

Answer: \(\frac{dy}{dx} = 10(2x + 1)^4\)

Question 2 · 3 marks06064PM1no calculator

Find \(\frac{dy}{dx}\) when \(y = \frac{3x - 1}{x + 2}\), simplifying your answer.

Show the mark scheme and worked solution

M1 quotient rule
A1 (3(x + 2) − (3x − 1))/(x + 2)²
A1 7/(x + 2)²

Quotient rule: (v u′ − u v′)/v² with u = 3x − 1, v = x + 2.
((x + 2) × 3 − (3x − 1) × 1)/(x + 2)².
= (3x + 6 − 3x + 1)/(x + 2)² = 7/(x + 2)².

Answer: \(\frac{dy}{dx} = \frac{7}{(x + 2)^2}\)

Question 3 · 5 marks06064PM1no calculator

Find the coordinates of the stationary points of the curve \(y = x^3 - 6x^2 + 9x + 2\) and determine their nature.

Show the mark scheme and worked solution

M1 3x² − 12x + 9 = 0
A1 x = 1 and x = 3
A1 (1, 6) and (3, 2)
M1 second derivative 6x − 12 used
A1 (1, 6) maximum, (3, 2) minimum

dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0, so x = 1 or x = 3.
y(1) = 1 − 6 + 9 + 2 = 6; y(3) = 27 − 54 + 27 + 2 = 2.
d²y/dx² = 6x − 12: at x = 1 it is −6 < 0 (maximum); at x = 3 it is 6 > 0 (minimum).

Answer: \((1, 6)\) maximum, \((3, 2)\) minimum

Question 4 · 2 marks06064PM1no calculator

Differentiate \(y = \sin 3x + \cos 2x\) with respect to \(x\).

Show the mark scheme and worked solution

B1 3cos 3x
B1 −2sin 2x

d/dx(sin ax) = a cos ax, so sin 3x gives 3 cos 3x.
d/dx(cos ax) = −a sin ax, so cos 2x gives −2 sin 2x.

Answer: \(\frac{dy}{dx} = 3\cos 3x - 2\sin 2x\)

9 more differentiation questions

Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.

Next steps

Cambridge 0606: papers, syllabus map and official past papers
Edexcel 4PM1: papers, syllabus map and official past papers
Going on to A Level? Differentiation continues in A Level differentiation (A Level Math Revision). See also from Additional Maths to A Level.