Pearson Edexcel International Advanced Level Mathematics: Product, quotient and chain rules (YMA01) – 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 are aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019), Unit P3 section 4 (Differentiation, 4.1 to 4.4). The specification states that P3 is compulsory for IAL Mathematics and Pure Mathematics and has no IAS weighting. Angles are in radians.

Learn the methods first in the study guide and the revision notes. The Edexcel A Level Mathematics hub, the printable checklist and the free 10-minute diagnostics help you plan the rest. Earlier work is in differentiating powers of x.

Questions

1. Differentiate y = 7e^(−x/2) + 3 ln 2x − 2 sin 3x. [3]

2. Given y = x⁴ cos 3x, find the exact gradient of the curve at x = π/3. [4]

3. f(x) = (1 + ln x) / x², x > 0.

(a) Show that f′(x) = −(1 + 2 ln x) / x³. [3] (b) Find the exact coordinates of the stationary point of the curve y = f(x). [3]

4. Differentiate with respect to x:

(a) y = ln(2x² + 5) [2] (b) y = √(1 + cos 2x) [2]

5. (a) By writing sec x as (cos x)⁻¹, show that d/dx(sec x) = sec x tan x. [3] (b) Differentiate y = cot 4x. [2]

6. Find the equation of the tangent to y = x e^(3x) at the point where x = 1/3. Give your answer in the form y = mx + c, with m and c exact. [5]

7. Given x = cos 2y, for 0 < y < π/2:

(a) find dy/dx in terms of y. [2] (b) Show that dy/dx = −1 / (2√(1 − x²)). [2]

8. Find the exact gradient of y = 5^(3x) at x = 1/3. [3]

9. A cup of tea cools so that its temperature T °C after t minutes is T = 20 + 65e^(−0.04t).

(a) State the initial temperature and the temperature the tea approaches. [2] (b) Find the rate of change of temperature when t = 10, to 3 significant figures. [3] (c) Find the time taken to reach 40 °C, to 3 significant figures. [3] (d) Show that dT/dt = −0.04(T − 20). [2]

10. The number of followers of a channel is modelled by N = 250e^(0.12t), where t is the number of days since launch.

(a) Show that dN/dt = 0.12N. [2] (b) Find the value of t, to 3 significant figures, when N is increasing at 60 followers per day. [3] (c) Find the number of followers predicted when t = 60, to 3 significant figures, and comment on whether this is sensible. [2] (d) A second model is N = 4000 − 3750e^(−0.12t). Find, to 3 significant figures, the time at which this model gives N = 3000. [3]

11. f(x) = e^(−x) sin 2x, 0 ≤ x ≤ π.

(a) Find f′(x). [3] (b) Find the values of x, to 3 significant figures, at which f′(x) = 0. [3] (c) Find the y-coordinate of the stationary point with the smaller x-value, to 3 significant figures. [2]

12. A curve has equation x = 3y + e^(2y).

(a) Find dy/dx at the point where y = 0. [3] (b) Find the equation of the normal at that point, in the form y = mx + c. [3]

Answers

1. d/dx(7e^(−x/2)) = −(7/2)e^(−x/2) [1] d/dx(3 ln 2x) = 3/x [1] d/dx(−2 sin 3x) = −6 cos 3x [1] dy/dx = −(7/2)e^(−x/2) + 3/x − 6 cos 3x [3] Examiner insight: Each term earns credit separately, so write each derivative on its own line. The 2 inside ln 2x does not appear in the derivative.

2. Product rule with u = x⁴ and v = cos 3x [1] dy/dx = 4x³ cos 3x − 3x⁴ sin 3x [1] At x = π/3: cos π = −1 and sin π = 0 [1] Gradient = −4π³/27 [1] [4] Examiner insight: “Exact” means the answer stays in terms of π. A decimal answer does not meet the instruction, and showing cos π and sin π makes your method clear.

3. (a) u = 1 + ln x, u′ = 1/x, v = x², v′ = 2x [1] f′(x) = [(1/x)x² − (1 + ln x)(2x)] / x⁴ [1] = (x − 2x − 2x ln x) / x⁴ = −(1 + 2 ln x) / x³ [1] [3] (b) f′(x) = 0 gives 1 + 2 ln x = 0 [1] ln x = −½, so x = e^(−1/2) [1] y = (1 − ½) / e^(−1) = e/2, so the point is (e^(−1/2), e/2) [1] [3] Examiner insight: In a “show that” question, include the line where the numerator is expanded and simplified. In (b), substitute the exact x-value into f, not into f′.

4. (a) Chain rule: (1/(2x² + 5)) × 4x [1] dy/dx = 4x / (2x² + 5) [1] [2] (b) y = (1 + cos 2x)^(1/2): dy/dx = ½(1 + cos 2x)^(−1/2) × (−2 sin 2x) [1] dy/dx = −sin 2x / √(1 + cos 2x) [1] [2] Examiner insight: A chain rule answer is easiest to check when the inner derivative is written separately. A missing factor of 2 from the inner function is a common slip in (b).

5. (a) sec x = (cos x)⁻¹ [1] d/dx = −(cos x)⁻² × (−sin x) [1] = sin x / cos² x = (1/cos x)(sin x/cos x) = sec x tan x [1] [3] (b) d/dx(cot kx) = −k cosec² kx, so with k = 4: [1] dy/dx = −4 cosec² 4x [1] [2] Examiner insight: A “show that” needs every step, including the final rewriting in terms of sec and tan. Writing the result first and working backwards does not demonstrate the derivation.

6. Product rule: dy/dx = e^(3x) + 3x e^(3x) [1] At x = 1/3: gradient = e + e = 2e [1] y = (1/3)e^1 = e/3 [1] Tangent: y − e/3 = 2e(x − 1/3) [1] y = 2e x − e/3 [1] [5] Examiner insight: Both the gradient and the y-coordinate must be found before writing the line. Leave e as e, as the question asks for exact values.

7. (a) dx/dy = −2 sin 2y [1] dy/dx = −1 / (2 sin 2y) [1] [2] (b) sin² 2y = 1 − cos² 2y = 1 − x² [1] For 0 < y < π/2, sin 2y > 0, so sin 2y = √(1 − x²) and dy/dx = −1 / (2√(1 − x²)) [1] [2] Examiner insight: Justify the positive square root using the given interval. Without that statement, the sign of sin 2y is unexplained.

8. 5^(3x) = e^(3x ln 5) [1] dy/dx = 3 ln 5 × 5^(3x) [1] At x = 1/3: 5^1 = 5, so gradient = 15 ln 5 [1] [3] Examiner insight: The chain rule factor 3 is easy to omit when a^x is differentiated. Substitute x = 1/3 before simplifying 5^(3x).

9. (a) At t = 0: T = 20 + 65 = 85 °C [1] As t → ∞, e^(−0.04t) → 0, so T approaches 20 °C [1] [2] (b) dT/dt = −2.6e^(−0.04t) [1] At t = 10: −2.6e^(−0.4) [1] = −1.74 °C per minute [1] [3] (c) 40 = 20 + 65e^(−0.04t), so e^(−0.04t) = 20/65 [1] −0.04t = ln(4/13), so t = ln(13/4) / 0.04 [1] t = 29.5 minutes [1] [3] (d) dT/dt = −0.04 × 65e^(−0.04t) [1] Since T − 20 = 65e^(−0.04t), dT/dt = −0.04(T − 20) [1] [2] Examiner insight: A negative rate needs a sentence of interpretation, such as “the temperature is falling”. In (d), start from your derivative and replace the exponential term, rather than starting from the result.

10. (a) dN/dt = 0.12 × 250e^(0.12t) [1] = 0.12N [1] [2] (b) 0.12N = 60, so N = 500 [1] 250e^(0.12t) = 500, so e^(0.12t) = 2 [1] t = ln 2 / 0.12 = 5.78 days [1] [3] (c) N = 250e^(7.2) = about 335 000 [1] Not sensible: the model has no upper limit, and this is far more followers than a single channel is likely to have. [1] [2] (d) 3000 = 4000 − 3750e^(−0.12t), so e^(−0.12t) = 1000/3750 = 4/15 [1] −0.12t = ln(4/15), so t = ln(15/4) / 0.12 [1] t = 11.0 days [1] [3] Examiner insight: A comment on a model should refer to the context, not just say “too big”. Keep unrounded values in your working until the final line.

11. (a) Product rule, u = e^(−x), v = sin 2x [1] f′(x) = −e^(−x) sin 2x + 2e^(−x) cos 2x [1] f′(x) = e^(−x)(2 cos 2x − sin 2x) [1] [3] (b) e^(−x) ≠ 0, so 2 cos 2x = sin 2x, giving tan 2x = 2 [1] 2x = 1.1071 or 1.1071 + π = 4.2487 (0 ≤ 2x ≤ 2π) [1] x = 0.554 or x = 2.12 [1] [3] (c) f(0.5536) = e^(−0.5536) sin(1.1071) [1] y = 0.514 [1] [2] Examiner insight: List all the solutions in the interval by extending the range for 2x, not x. For (c), use the unrounded value of x to avoid a rounding error in the third significant figure.

12. (a) dx/dy = 3 + 2e^(2y) [1] At y = 0: dx/dy = 5 [1] dy/dx = 1/5 [1] [3] (b) At y = 0: x = 0 + e^0 = 1, so the point is (1, 0) [1] Normal gradient = −5 [1] y = −5(x − 1), that is y = 5 − 5x [1] [3] Examiner insight: The gradient of the normal comes from dy/dx, not dx/dy. Find the point first, then use the reciprocal, then take the negative reciprocal.

Where marks are usually lost

• Writing the derivative of cos kx as k sin kx.
• Omitting the inner derivative in chain rule questions.
• Putting the terms in the wrong order in the quotient rule numerator.
• Dropping a common factor, such as e^(−x), when simplifying a product rule answer.
• Giving k/x for the derivative of ln kx.
• Treating a^x as a power of x.
• Using dx/dy as the gradient of the curve.
• Giving a decimal where an exact answer is required, or the reverse.
• Reporting only the first solution in a trigonometric equation within an interval.
• Giving a rate or time without units, or commenting on a model without referring to the context.

Next steps

• Go back to the revision notes for any rule you missed.
• Read the study guide for the full explanations.
• Browse the Edexcel A Level Mathematics hub and tick off topics on the printable checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, published by Pearson Education Limited. Unit P3 section 4, topics 4.1 to 4.4.