Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Differentiation – 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 cover section 2.4, Differentiation, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.4 is part of Pure Mathematics 2 and is examined on Paper 2, the AS Level Pure Mathematics route; the same outcomes appear as section 3.4 for Paper 3. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Give exact answers where asked, otherwise 3 significant figures. Angles are in radians.

Related: the study guide, the revision notes, the whole-paper Pure Mathematics 2 practice set, the Paper 1 differentiation practice, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

Questions

1. Differentiate each of the following with respect to x.

(a) 5e^(2 − 3x) [1] (b) ln(1 + cos 2x) [2] (c) tan²x [2]

2. Find the exact gradient of the curve y = (2x + 1) ln(2x + 1) at the point where x = 1. [3]

3. The equation of a curve is y = e^(2x)/(1 + x²).

(a) Show that dy/dx = 2e^(2x)(x² − x + 1)/(1 + x²)². [3] (b) Hence show that the curve has no stationary points. [2]

4. Find the exact x-coordinates of the stationary points of the curve y = tan x − 2x, for −½π < x < ½π. [3]

5. A curve is defined by the parametric equations x = e^(2t) − 1, y = t e^(2t).

(a) Show that dy/dx = t + ½. [3] (b) Find the equation of the normal to the curve at the point where t = ½. Give your answer in the form y = −x + c, where c is an exact constant. [3]

6. The equation of a curve is 3x² − 2xy + y² = 24.

(a) Show that dy/dx = (y − 3x)/(y − x). [4] (b) Find the coordinates of the points on the curve at which the tangent is parallel to the x-axis. [3]

7. Find the exact coordinates of the stationary point of the curve y = e^(−x) sin x in the interval 0 < x < π. [4]

8. The normal to the curve y = e^(x/2) + 3 at the point where x = 2 meets the x-axis at P. Calculate the x-coordinate of P, giving your answer correct to 3 significant figures. [4]

9. The equation of a curve is y = x²/ln x, for x > 1.

(a) Find dy/dx. [2] (b) Find the exact coordinates of the stationary point. [2] (c) By considering the sign of dy/dx either side of the stationary point, determine its nature. [2]

10. A curve has parametric equations x = 1 + ln t, y = t + 4/t, for t > 0.

(a) Show that dy/dx = (t² − 4)/t. [3] (b) Find the exact coordinates of the stationary point. [2] (c) Find the equation of the normal to the curve at the point where t = 1, giving your answer in the form ax + by + c = 0. [3]

11. The equation of a curve is x ln y + y² = 4, for y > 0.

(a) Show that dy/dx = −(y ln y)/(x + 2y²). [4] (b) The curve crosses the y-axis at the point Q. Find the equation of the tangent to the curve at Q, giving the gradient in exact form. [3] (c) The tangent at Q meets the x-axis at R. Find the x-coordinate of R, giving your answer correct to 3 significant figures. [2] (d) Show that there is no point on the curve at which the tangent is parallel to the x-axis. [2]

Answers

1. (a) 5 × (−3)e^(2 − 3x) = −15e^(2 − 3x) [1] (b) (1/(1 + cos 2x)) × d/dx(1 + cos 2x), with d/dx(1 + cos 2x) = −2 sin 2x [1]. dy/dx = −2 sin 2x/(1 + cos 2x) [1] (this simplifies to −2 tan x, which is also accepted) (c) tan²x = (tan x)², so dy/dx = 2 tan x × d/dx(tan x) [1] = 2 tan x sec²x [1] Examiner insight: in (b) the mark for the answer needs the factor 2 from the inside of cos 2x; −sin 2x/(1 + cos 2x) earns only the method mark.

2. u = 2x + 1, v = ln(2x + 1), dv/dx = 2/(2x + 1) [1]. dy/dx = 2 ln(2x + 1) + (2x + 1) × 2/(2x + 1) = 2 ln(2x + 1) + 2 [1]. At x = 1, gradient = 2 + 2 ln 3 [1] Examiner insight: “exact” means 2 + 2 ln 3 (or 2 + ln 9); a bare 4.20 from a calculator loses the final mark.

3. (a) Quotient rule with u = e^(2x), v = 1 + x², du/dx = 2e^(2x), dv/dx = 2x [1]. dy/dx = (2e^(2x)(1 + x²) − e^(2x) × 2x)/(1 + x²)² [1]. Taking out 2e^(2x): dy/dx = 2e^(2x)(x² − x + 1)/(1 + x²)², as given [1] (b) e^(2x) > 0 and (1 + x²)² > 0 for all x, so dy/dx = 0 only if x² − x + 1 = 0 [1]. Discriminant = (−1)² − 4(1)(1) = −3 < 0, so there are no real roots and no stationary points [1] Examiner insight: in a “show that”, the unsimplified quotient-rule line must appear; copying the given answer earns nothing in (a).

4. dy/dx = sec²x − 2 [1]. sec²x = 2, so cos²x = ½ and cos x = ±1/√2 [1]. In −½π < x < ½π, cos x > 0, so x = π/4 or x = −π/4 [1] Examiner insight: the last mark needs both values and the reason cos x > 0 for rejecting the others; the domain is in radians, so ±45° does not answer the question.

5. (a) dx/dt = 2e^(2t) [1]. dy/dt = e^(2t) + 2t e^(2t) by the product rule [1]. dy/dx = e^(2t)(1 + 2t)/(2e^(2t)) = (1 + 2t)/2 = t + ½ [1] (b) At t = ½: x = e − 1, y = ½e, and dy/dx = 1 [1]. Normal gradient = −1 [1]. y − ½e = −(x − (e − 1)), so y = −x + (3/2)e − 1 [1] Examiner insight: in (b) the coordinates must come from t = ½; substituting x = ½ anywhere is a wrong method and loses all three marks.

6. (a) d/dx(3x²) = 6x and d/dx(y²) = 2y dy/dx [1]. d/dx(2xy) = 2y + 2x dy/dx [1]. So 6x − 2y − 2x dy/dx + 2y dy/dx = 0 [1]. dy/dx(2y − 2x) = 2y − 6x, so dy/dx = (y − 3x)/(y − x), as given [1] (b) Tangent parallel to the x-axis when y − 3x = 0, so y = 3x [1]. Substituting: 3x² − 6x² + 9x² = 24, so 6x² = 24 and x = ±2 [1]. Points (2, 6) and (−2, −6) [1] Examiner insight: the product-rule mark in (a) needs the minus sign applied to both 2y and 2x dy/dx; −2y + 2x dy/dx loses it.

7. Product rule: dy/dx = e^(−x) cos x + sin x × (−e^(−x)) [1] = e^(−x)(cos x − sin x) [1]. e^(−x) ≠ 0, so cos x = sin x, tan x = 1, and in 0 < x < π, x = π/4 [1]. y = e^(−π/4) sin(π/4), so the point is (π/4, (1/√2)e^(−π/4)) [1] Examiner insight: state e^(−x) ≠ 0 before dividing it out; the answer mark also needs the exact y-coordinate, not 0.322.

8. At x = 2, y = e + 3. dy/dx = ½e^(x/2), which is ½e at x = 2 [1]. Normal gradient = −2/e [1]. Normal: y − (e + 3) = −(2/e)(x − 2) [1]. At P, y = 0: x = 2 + e(e + 3)/2, so x = 9.77 (3 s.f.) [1] Examiner insight: keep e exact until the last line; an early rounded gradient can lose the accuracy mark.

9. (a) Quotient rule with u = x², v = ln x: dy/dx = (2x ln x − x² × (1/x))/(ln x)² [1] = (2x ln x − x)/(ln x)², or x(2 ln x − 1)/(ln x)² [1] (b) x > 1, so x ≠ 0 and 2 ln x − 1 = 0, giving ln x = ½ and x = e^(½) = √e [1]. y = e/(½) = 2e, so the point is (√e, 2e) [1] (c) √e ≈ 1.649. At x = 1.5, dy/dx = −1.73 < 0; at x = 2, dy/dx = 1.61 > 0 [1]. The gradient goes from negative to positive, so it is a minimum [1] Examiner insight: the question fixes the method, so a second-derivative test in (c) earns no marks; test values must lie either side of √e and inside x > 1.

10. (a) dx/dt = 1/t and dy/dt = 1 − 4/t² [1]. dy/dx = (1 − 4/t²) ÷ (1/t) [1] = t − 4/t = (t² − 4)/t [1] (b) t² − 4 = 0 and t > 0, so t = 2 [1]. x = 1 + ln 2, y = 2 + 2 = 4, so the point is (1 + ln 2, 4) [1] (c) At t = 1: x = 1, y = 5, and dy/dx = −3 [1]. Normal gradient = 1/3 [1]. y − 5 = (1/3)(x − 1), so 3y − 15 = x − 1 and x − 3y + 14 = 0 [1] Examiner insight: in (b), t = −2 must be rejected because t > 0 (ln t is undefined for t ≤ 0); keeping it loses the final mark.

11. (a) Product rule on x ln y: ln y + x × (1/y) dy/dx [1], with (1/y) dy/dx as the derivative of ln y [1]. d/dx(y²) = 2y dy/dx and the right-hand side gives 0 [1]. Multiply by y: y ln y + x dy/dx + 2y² dy/dx = 0, so dy/dx = −(y ln y)/(x + 2y²), as given [1] (b) At Q, x = 0, so y² = 4 and y = 2 (since y > 0) [1]. Gradient = −(2 ln 2)/(0 + 8) = −(ln 2)/4 [1]. Tangent: y = 2 − ((ln 2)/4)x [1] (c) 0 = 2 − ((ln 2)/4)x, so x = 8/ln 2 [1] = 11.5 (3 s.f.) [1] (d) dy/dx = 0 requires y ln y = 0; y > 0, so ln y = 0 and y = 1 [1]. Substituting y = 1: x ln 1 + 1 = 4 gives 1 = 4, which is impossible, so no such point exists [1] Examiner insight: in (d) show the contradiction by substituting y = 1 into the curve; “y = 1 is not on the curve” alone earns only the first mark.

Where marks are usually lost

• The inside derivative is dropped, most often the factor from ax + b or from 2x inside sin 2x or cos 2x.
• The quotient-rule numerator is reversed, so a “show that” result appears only after a sign is quietly fixed.
• Implicit terms such as 2xy are differentiated as 2x dy/dx alone, with no product rule.
• In parametric questions, x-values are substituted into a gradient that is written in terms of t.
• The normal’s gradient is written as −m or 1/m instead of −1/m.
• Exact answers are converted to decimals, or decimal gradients are rounded before the last line.
• A factor such as e^(−x) or e^(2x) is cancelled without saying it can never be zero.
• The nature of a stationary point is found by a different method from the one the question names.

Next steps

• Differentiation revision notes
• Differentiation study guide
• Pure Mathematics 2 guide
• A Level Mathematics hub
• Printable 9709 checklist
• AS diagnostic, the 9709 self-check bank, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.4 Differentiation.