OxfordAQA International AS and A-level Mathematics: Product, quotient and chain rules; e^x, ln x, trig (9660) – 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 the first part of specification section P2.6 (Differentiation) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification states that this content is part of International A-level (A2) Unit P2 (Pure Maths) and is not examined at International AS. Learn the methods first from the study guide or the revision notes. See also the course hub, the printable checklist and the free diagnostics. Angles are in radians throughout.

Questions

1. Differentiate with respect to x. (a) y = 6e^(x/2) + 2 sin 3x [2] (b) y = 5 ln x − tan 2x [2]

2. Differentiate with respect to x. (a) y = (5x² − 2)⁴ [2] (b) y = e^(sin x) [2]

3. A curve has equation y = x⁴ cos 2x. (a) Find dy/dx. [3] (b) Find the exact gradient of the curve where x = π/4. [2]

4. Show that the gradient of y = (3x + 1)/(2x − 5) is negative at every point where x ≠ 5/2. [4]

5. Given y = (1 − cos x)/sin x, show that dy/dx = 1/(1 + cos x). [5]

6. A curve has equation x = y³ + 4y. Find dy/dx at the point (5, 1). [3]

7. A curve has equation x = 3y² − 12y + 5. (a) Find dy/dx in terms of y. [2] (b) Find the coordinates of the point on the curve where dy/dx = 1/6. [2]

8. A curve has equation y = x³ sin 2x. (a) Find dy/dx. [3] (b) Find the exact value of dy/dx when x = π/4. [2]

9. The function f is defined by f(x) = e^(2x)/x for x > 0. (a) Show that f′(x) = e^(2x)(2x − 1)/x². [4] (b) Find the exact coordinates of the point where the gradient of the curve y = f(x) is zero. [2] (c) Write down the exact value of f′(1). [1]

10. A flywheel has speed v metres per second t seconds after its power is cut, where v = 12e^(−0.25t). (a) Find dv/dt. [2] (b) Calculate the value of dv/dt when t = 3. Give your answer to 3 significant figures and include its units. [2] (c) The speed is falling at a rate of 0.5 metres per second per second at time T. Find T to 3 significant figures. [3]

11. A curve has equation y = e^x cos 2x. (a) Show that dy/dx = e^x(cos 2x − 2 sin 2x). [3] (b) Find the smallest positive value of x at which the gradient is zero. Give your answer to 3 significant figures. [3]

Answers

1. (a) The derivative of 6e^(x/2) is 6 × (1/2)e^(x/2) = 3e^(x/2) [1]. The derivative of 2 sin 3x is 6 cos 3x [1]. dy/dx = 3e^(x/2) + 6 cos 3x. (b) The derivative of 5 ln x is 5/x [1]. The derivative of tan 2x is 2 sec²2x [1]. dy/dx = 5/x − 2 sec²(2x). [4] Examiner insight: the factor from the inside of each function, 1/2 and 2 here, is the usual place to lose credit, so write it in as you differentiate.

2. (a) Let u = 5x² − 2, so y = u⁴ and du/dx = 10x [1]. dy/dx = 4u³ × 10x = 40x(5x² − 2)³ [1]. (b) Let u = sin x, so y = e^u and du/dx = cos x [1]. dy/dx = cos x · e^(sin x) [1]. [4] Examiner insight: naming the inner function u makes the chain rule visible, and it keeps your working easy to follow if a later slip occurs.

3. (a) Let u = x⁴ and v = cos 2x, so u′ = 4x³ [1] and v′ = −2 sin 2x [1]. dy/dx = x⁴(−2 sin 2x) + cos 2x (4x³) = 4x³ cos 2x − 2x⁴ sin 2x [1]. (b) At x = π/4, 2x = π/2, so cos 2x = 0 and sin 2x = 1 [1]. The gradient is −2(π/4)⁴ = −2π⁴/256 = −π⁴/128 [1]. [5] Examiner insight: when the question says exact, evaluate cos(π/2) and sin(π/2) as 0 and 1 and leave π in the answer. A decimal approximation would not answer it.

4. Let u = 3x + 1 and v = 2x − 5, so u′ = 3 and v′ = 2 [1]. dy/dx = (3(2x − 5) − 2(3x + 1))/(2x − 5)² [1]. The numerator is 6x − 15 − 6x − 2 = −17, so dy/dx = −17/(2x − 5)² [1]. A square is positive for x ≠ 5/2, so dividing −17 by it always gives a negative value, which shows the gradient is negative everywhere [1]. [4] Examiner insight: in a “show that” question, give the reason as well as the algebra. A sentence about the sign of a square is part of the argument.

5. Let u = 1 − cos x and v = sin x, so u′ = sin x and v′ = cos x [1]. dy/dx = (sin x · sin x − (1 − cos x) cos x)/sin²x [1]. The numerator is sin²x − cos x + cos²x = 1 − cos x [1]. Using sin²x = 1 − cos²x = (1 − cos x)(1 + cos x) [1], dy/dx = (1 − cos x)/((1 − cos x)(1 + cos x)) = 1/(1 + cos x) [1]. [5] Examiner insight: the given answer means every line must lead to it. State which identity you use, and do not skip the cancelling step.

6. dx/dy = 3y² + 4 [1]. At y = 1, dx/dy = 7 [1]. dy/dx = 1/(dx/dy) = 1/7 [1]. [3] Examiner insight: it is easy to stop at dx/dy = 7. The question asks for dy/dx, so show the reciprocal as a separate final line.

7. (a) dx/dy = 6y − 12 [1]. dy/dx = 1/(6y − 12) [1]. (b) Setting 1/(6y − 12) = 1/6 gives 6y − 12 = 6, so y = 3 [1]. Then x = 3(9) − 12(3) + 5 = −4, giving (−4, 3) [1]. [4] Examiner insight: give coordinates when coordinates are asked for. The y-value found from the derivative is only half of the answer.

8. (a) Let u = x³ and v = sin 2x, so u′ = 3x² [1] and v′ = 2 cos 2x [1]. dy/dx = x³(2 cos 2x) + sin 2x (3x²) = 2x³ cos 2x + 3x² sin 2x [1]. (b) At x = π/4, cos 2x = 0 and sin 2x = 1 [1]. The gradient is 3(π/4)² = 3π²/16 [1]. [5] Examiner insight: this question combines the product rule with a chain-rule step in v′. Differentiate sin 2x on its own line first, so the factor 2 is not lost.

9. (a) Let u = e^(2x) and v = x, so u′ = 2e^(2x) and v′ = 1 [1]. f′(x) = (x · 2e^(2x) − e^(2x) · 1)/x² [1]. The numerator is 2x e^(2x) − e^(2x) [1] = e^(2x)(2x − 1), so f′(x) = e^(2x)(2x − 1)/x² [1]. (b) e^(2x) is never zero, so 2x − 1 = 0 and x = 1/2 [1]. f(1/2) = e/(1/2) = 2e, so the point is (1/2, 2e) [1]. (c) f′(1) = e²(1)/1 = e² [1]. [7] Examiner insight: once the derivative is factorised, part (b) takes one line. If you leave it as two separate terms, finding where it equals zero is much harder.

10. (a) The chain rule gives a factor −0.25 for the exponent [1]. dv/dt = 12(−0.25)e^(−0.25t) = −3e^(−0.25t) [1]. (b) At t = 3, dv/dt = −3e^(−0.75) [1] = −1.42 m s⁻² (3 s.f.) [1]. (c) A fall of 0.5 means dv/dt = −0.5, so −3e^(−0.25T) = −0.5 [1]. Then e^(−0.25T) = 1/6 and −0.25T = −ln 6 [1]. T = 4 ln 6 = 7.17 seconds (3 s.f.) [1]. [7] Examiner insight: the negative sign carries the meaning that the speed falls. State the unit in part (b), and in part (c) turn “falling at 0.5” into dv/dt = −0.5 before solving.

11. (a) Let u = e^x and v = cos 2x, so u′ = e^x and v′ = −2 sin 2x [1]. dy/dx = e^x(−2 sin 2x) + cos 2x · e^x [1] = e^x(cos 2x − 2 sin 2x) [1]. (b) The gradient is zero when cos 2x − 2 sin 2x = 0, since e^x is never zero [1]. So tan 2x = 1/2 [1]. The smallest positive solution has 2x = arctan(0.5) = 0.4636, so x = 0.232 (3 s.f.) [1]. [6] Examiner insight: divide by e^x by reasoning that it is never zero, not by cancelling silently. Radians matter here, since the answer depends on the arctangent in radians.

Where marks are usually lost

• Omitting the factor from the inner function in e^(kx), sin(kx), cos(kx) or a bracket raised to a power.
• Writing sin x instead of −sin x as the derivative of cos x.
• Reversing the quotient-rule numerator, so that it reads u v′ − v u′.
• Stopping at dx/dy and forgetting to take the reciprocal.
• Approximating when the question asks for an exact value such as π⁴/128 or e².
• Expanding a product-rule answer instead of factorising, which makes later steps harder.
• Failing to explain the sign of a squared denominator in a “show that” question.
• Working in degrees when the derivative results require radians.

Next steps

• Revise the method with the revision notes.
• Re-read the explanations in the study guide.
• Return to the OxfordAQA Mathematics hub and tick off the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: section 3.3, International A-level Unit P2 (Pure Maths), content P2.6 Differentiation (differentiation of e^x, ln x, sin x, cos x and tan x, and the product, quotient and chain rules, and dy/dx = 1/(dx/dy)).