OCR A Level Mathematics A: Differentiation techniques and differential equations (H240) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions test differentiation techniques for OCR A Level Mathematics A (H240), written against specification Version 3.1 (August 2026), statements 1.07q to 1.07t. The specification lists all of them under Stage 2 only, so they are A Level only. Learn the methods first from the study guide and revision notes. The wider course is on the OCR A Level Mathematics hub, the printable checklist lists every skill, and the free 10-minute diagnostics help you choose what to practise. Marks are given in bold brackets. Give exact answers unless told otherwise, and decimals to 3 significant figures.

Questions

1. Differentiate y = (2x + 1)e^(−3x), giving your answer in a fully factorised form. [3]

2. Differentiate y = e^(2x)/(x − 1) and simplify. [4]

3. Find dy/dx when y = sin³ 4x. [3]

4. Given f(x) = x ln(x² + 1), find the exact value of f′(1). [5]

5. A solid metal cylinder is heated so that it expands, with its height always equal to its radius r mm. The radius increases at 0.01 mm per second. Find, in exact form, how fast the volume is increasing at the instant when r = 20. [5]

6. A curve has equation x = y⁵ + 3y.

(a) Find dx/dy. [1] (b) Hence write down dy/dx in terms of y. [1] (c) Evaluate the gradient where y = 1, which is the point (4, 1). [1]

7. A curve has equation 2x² + xy + y³ = 4. The point P(1, 1) lies on the curve.

(a) Find an expression for dy/dx. [3] (b) Find the gradient of the curve at P. [1] (c) Find the equation of the normal at P in the form ax + by + c = 0, where a, b and c are integers. [2]

8. A curve has parametric equations x = t² + 1, y = 2t³ − 3t.

(a) Find dy/dx in terms of t. [3] (b) Find the normal to the curve at t = 1, in the form ax + by + c = 0 with integer a, b and c. [3]

9. A curve has parametric equations x = 2 + 3 sin θ, y = 1 − 2 cos 2θ. Show that dy/dx = (8/3) sin θ, and find the gradient when θ = π/6. [4]

10. A tank is empty at time t = 0 minutes. A tap adds water at 0.5 m³ per minute. Water also leaks out at a rate proportional to the volume V m³ in the tank. When V = 8 the leak rate is 0.4 m³ per minute.

(a) Show that dV/dt = 0.5 − 0.05V, explaining how you find the constant. [3] (b) Show that V = 10 − 10e^(−0.05t) satisfies the equation and the initial condition. [3] (c) State the volume the tank approaches in the long term. [1]

11. A spherical raindrop of radius r mm evaporates so that its volume decreases at a rate proportional to its surface area. For a sphere, V = (4/3)πr³ and S = 4πr².

(a) Show that dr/dt is a negative constant. Call it −λ. [4] (b) The radius is 2 mm at t = 0 and 1.5 mm when t = 10 seconds. Find λ and the time for the drop to vanish. [3] (c) Find the rate of change of volume when r = 1, giving your answer to 3 significant figures. [1]

Answers

1. u = 2x + 1 and v = e^(−3x), so u′ = 2 and v′ = −3e^(−3x) [1] dy/dx = 2e^(−3x) − 3(2x + 1)e^(−3x) [1] = −(6x + 1)e^(−3x) [1] [3] Examiner insight: write the product rule in full before substituting. If the derivative of e^(−3x) is wrong, the working still shows the rule was understood.

2. u = e^(2x), v = x − 1, u′ = 2e^(2x), v′ = 1 [1] dy/dx = ( (x − 1)·2e^(2x) − e^(2x)·1 ) / (x − 1)² [1] = e^(2x)(2x − 2 − 1)/(x − 1)² [1] = e^(2x)(2x − 3)/(x − 1)² [1] [4] Examiner insight: the minus sign and the order of the terms in the numerator are where most slips happen. Bracket v′ and u′ terms so the sign is clear.

3. Write y = (sin 4x)³ and let u = sin 4x [1] dy/dx = 3(sin 4x)² × 4 cos 4x [1] = 12 sin² 4x cos 4x [1] [3] Examiner insight: the chain rule is needed twice here. Say which function is the inside each time.

4. u = x, v = ln(x² + 1), u′ = 1, v′ = 2x/(x² + 1) [1] f′(x) = x · 2x/(x² + 1) + 1 · ln(x² + 1) [1] = 2x²/(x² + 1) + ln(x² + 1) [1] f′(1) = 2/2 + ln 2 [1] = 1 + ln 2 (about 1.69) [1] [5] Examiner insight: “exact value” means leave ln 2 in the answer. A decimal alone does not show the exact form.

5. Height equals radius, so V = πr²·r = πr³ [1] dV/dr = 3πr², which is 1200π at r = 20 [1] dV/dt = (dV/dr)(dr/dt) [1] dV/dt = 1200π × 0.01 [1] = 12π mm³ per second (about 37.7) [1] [5] Examiner insight: write the volume in terms of r alone before differentiating. Differentiating with r and h both present is a common route to a wrong answer.

6. (a) dx/dy = 5y⁴ + 3 [1] (b) dy/dx = 1/(5y⁴ + 3) [1] (c) At y = 1, dy/dx = 1/8, so the gradient is 1/8 [1] [3] Examiner insight: the reciprocal relationship needs dx/dy with y as the variable. Do not substitute x = 4 into an expression in y.

7. (a) Differentiate 2x² + xy: 4x + y + x(dy/dx) [1] Differentiate y³: 3y²(dy/dx), and the right-hand side is 0 [1] dy/dx = −(4x + y)/(x + 3y²) [1] (b) At (1, 1): −5/4 [1] (c) Normal gradient 4/5 [1] y − 1 = (4/5)(x − 1), so 4x − 5y + 1 = 0 [1] [6] Examiner insight: the product rule on xy is the step most often missed. Check that your final line has integer coefficients if the question asks for them.

8. (a) dx/dt = 2t and dy/dt = 6t² − 3 [1] dy/dx = (dy/dt)/(dx/dt) [1] = (6t² − 3)/(2t) [1] (b) At t = 1: x = 2, y = −1, gradient 3/2 [1] Normal gradient −2/3 [1] y + 1 = −(2/3)(x − 2), so 2x + 3y − 1 = 0 [1] [6] Examiner insight: find the point from both parametric equations. The gradient alone does not locate the normal.

9. dx/dθ = 3 cos θ and dy/dθ = 4 sin 2θ [1] dy/dx = 4 sin 2θ/(3 cos θ) [1] = 8 sin θ cos θ/(3 cos θ) = (8/3) sin θ [1] At θ = π/6: (8/3)(1/2) = 4/3 [1] [4] Examiner insight: in a “show that” question, state the identity sin 2θ = 2 sin θ cos θ. Without it the step is unexplained.

10. (a) The leak is kV, so 0.4 = k × 8 and k = 0.05 [1] Net rate = inflow minus outflow [1] dV/dt = 0.5 − 0.05V [1] (b) dV/dt = 10 × 0.05e^(−0.05t) = 0.5e^(−0.05t) [1] Right-hand side: 0.5 − 0.05(10 − 10e^(−0.05t)) = 0.5e^(−0.05t), so both sides agree [1] At t = 0, V = 10 − 10 = 0, which matches the empty tank [1] (c) As t increases, e^(−0.05t) approaches 0, so V approaches 10 m³ [1] [7] Examiner insight: in a verification question, show both sides separately and state that they agree. Do not forget the initial condition.

11. (a) dV/dr = 4πr² [1] dV/dt = (dV/dr)(dr/dt) = 4πr²(dr/dt) [1] Model: dV/dt = −μ · 4πr², with μ > 0 [1] Dividing by 4πr² gives dr/dt = −μ, a constant, so λ = μ [1] (b) The radius changes at a constant rate, so r = 2 − λt [1] 1.5 = 2 − 10λ, so λ = 0.05 [1] The drop vanishes when 2 = 0.05t, so after 40 seconds [1] (c) dV/dt = 4π(1)²(−0.05) = −0.628 mm³ per second [1] [8] Examiner insight: the minus sign should come from “decreases” in the model. Cancelling 4πr² is the step that makes the constant visible, so show it.

Where marks are usually lost

• Omitting the minus sign or reversing the numerator in the quotient rule.
• Leaving out the derivative of the inside function in a chain rule question.
• Dropping the product rule when differentiating xy implicitly.
• Not writing dy/dx beside the derivative of a y term.
• Dividing dx/dt by dy/dt in a parametric question.
• Using the tangent gradient for the normal, or forgetting to take the negative reciprocal.
• Differentiating a volume with two variables instead of eliminating one first.
• Forgetting the sign of a rate when a quantity is decreasing.
• Leaving a modelling answer without units.

Next steps

• Revisit the revision notes for the method boxes.
• Reread the study guide for the full worked examples.
• Check progress on the OCR A Level Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics for wider Pure Mathematics.
• Book a free trial class.

Official syllabus

Source: OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Statements tested: 1.07q to 1.07t.