Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 1 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 1.7 Differentiation of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). The section is part of Pure Mathematics 1 and is examined on Paper 1, compulsory for AS and A Level. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Give non-exact answers to 3 significant figures.

Related: the study guide, the revision notes, the A Level Mathematics hub, the printable 9709 checklist, the free AS Level diagnostic and the 9709 self-check bank.

Questions

1. Find dy/dx when y = 5x⁴ − 6/x³ + 4√x. [3]

2. It is given that f(x) = 3/(4 − x)² for x ≠ 4. Find f′(x) and f″(x). [4]

3. The points P and Q lie on the curve y = x³ − x. P has x-coordinate 1 and Q has x-coordinate 1 + h.

(a) Show that the gradient of the chord PQ is 2 + 3h + h². [3] (b) State the gradient of the curve at P, explaining how it follows from part (a). [1]

4. The function f is defined by f(x) = (2x − 3)³ + 6x. Show that f is an increasing function. [3]

5. Find the x-coordinate of the stationary point of the curve y = 2x^(3/2) − 6x + 3, for x > 0, and determine its nature. [5]

6. Water is poured into an inverted cone at a constant rate of 20 cm³ per second. When the depth of water is h cm, the volume of water is V = (π/12)h³ cm³. Find the rate at which the depth is increasing when h = 8. [4]

7. The curve y = kx² + 12/x, where k is a constant, has a stationary point at x = 2.

(a) Find the value of k. [3] (b) Find the y-coordinate of the stationary point and determine its nature. [2]

8. Given y = 6x^(2/3), find the value of d²y/dx² when x = 8. [3]

9. The point P(2, 2) lies on the curve y = x² − 4/x. The tangent to the curve at P meets the x-axis at A, and the normal at P meets the x-axis at B.

(a) Find the equation of the tangent and the equation of the normal at P. [5] (b) Find the area of triangle PAB. [2]

10. The equation of a curve is y = x³ − 6x².

(a) Find the coordinates of the stationary points and determine their nature. [5] (b) Sketch the curve, showing the coordinates of the stationary points and where the curve meets the axes. [2] (c) State the set of values of x for which y is decreasing. [1]

11. A closed cylindrical tin has radius r cm and height h cm. Its volume is 250π cm³.

(a) Show that the total surface area, S cm², is given by S = 2πr² + 500π/r. [2] (b) Find the value of r for which S is stationary. [3] (c) Show that this value of r gives a minimum, and find the minimum value of S. [3]

12. The equation of a curve is y = 12/(x² + 2).

(a) Find dy/dx. [2] (b) Find the coordinates of the stationary point. By considering the sign of dy/dx on either side, determine its nature. [3] (c) A point moves along the curve so that its x-coordinate increases at 0.5 units per second. Find the rate of change of its y-coordinate when x = 1. [2] (d) Find the equation of the normal to the curve at x = 1, in the form ax + by = c. [2]

Answers

1. y = 5x⁴ − 6x⁻³ + 4x^(1/2) [1]. dy/dx = 20x³ + 18x⁻⁴ [1] + 2x^(−1/2), so dy/dx = 20x³ + 18/x⁴ + 2/√x [1] [3] Examiner insight: the first mark is for rewriting as powers; −6x⁻³ differentiated to −18x⁻⁴ is a sign error that loses the second mark.

2. f(x) = 3(4 − x)⁻² [1]. f′(x) = −6(4 − x)⁻³ × (−1) [1] = 6/(4 − x)³ [1]. f″(x) = −18(4 − x)⁻⁴ × (−1) = 18/(4 − x)⁴ [1] [4] Examiner insight: each derivative needs the ×(−1) from the inside; missing it twice gives two wrong signs and loses both accuracy marks.

3. (a) At P, y = 0; at Q, y = (1 + h)³ − (1 + h) [1]. (1 + h)³ − (1 + h) = 1 + 3h + 3h² + h³ − 1 − h = 2h + 3h² + h³ [1]. Gradient = (2h + 3h² + h³)/h = 2 + 3h + h² [1] (b) As h → 0 the chord gradient tends to 2, so the gradient at P is 2 [1] Examiner insight: “show that” needs the expanded numerator written out before dividing by h; jumping straight to the given answer earns no marks.

4. f′(x) = 3(2x − 3)² × 2 + 6 [1] = 6(2x − 3)² + 6 [1]. (2x − 3)² ≥ 0, so f′(x) ≥ 6 > 0 for all x, so f is increasing [1] [3] Examiner insight: the final mark needs both the reason ((2x − 3)² ≥ 0) and the conclusion f′(x) > 0 for all x.

5. dy/dx = 3x^(1/2) − 6 [1]. 3√x − 6 = 0 [1], so √x = 2 and x = 4 [1]. d²y/dx² = (3/2)x^(−1/2) = 3/4 at x = 4 [1]. 3/4 > 0, so it is a minimum [1] [5] Examiner insight: the nature mark depends on the correct second derivative evaluated at x = 4; “positive, so minimum” with no value shown can lose it.

6. dV/dh = (π/4)h² [1] = 16π when h = 8 [1]. dV/dt = (dV/dh) × (dh/dt), so 20 = 16π × dh/dt [1]. dh/dt = 5/(4π) = 0.398 cm per second (3 s.f.) [1] [4] Examiner insight: the chain-rule statement earns a method mark on its own, so write it out even if you then make an arithmetic slip.

7. (a) dy/dx = 2kx − 12/x² [1]. At x = 2: 4k − 3 = 0 [1], so k = 3/4 [1] (b) y = (3/4)(4) + 12/2 = 9 [1]. d²y/dx² = 2k + 24/x³ = 3/2 + 3 = 9/2 > 0, so minimum [1] Examiner insight: in (b) the second derivative must use the value of k found in (a); follow-through is normally allowed on a wrong k if the method is right.

8. dy/dx = 4x^(−1/3) [1]. d²y/dx² = −(4/3)x^(−4/3) [1]. At x = 8, 8^(4/3) = 16, so d²y/dx² = −1/12 [1] [3] Examiner insight: give the exact fraction; a truncated decimal such as −0.08 is not accurate to 3 s.f. and loses the final mark.

9. (a) dy/dx = 2x + 4x⁻² [1] (both terms correct [1]). At x = 2, m = 4 + 1 = 5 [1]. Tangent: y − 2 = 5(x − 2), y = 5x − 8 [1]. Normal: gradient −1/5, y − 2 = −(1/5)(x − 2), x + 5y = 12 [1] (b) A = (8/5, 0) and B = (12, 0) [1]. Area = ½ × (12 − 8/5) × 2 = 52/5 = 10.4 [1] Examiner insight: −4/x differentiates to +4/x²; a sign error here makes m = 3, and only the method marks for the lines can then be earned.

10. (a) dy/dx = 3x² − 12x = 3x(x − 4) [1]. Stationary at x = 0 and x = 4 [1], giving (0, 0) and (4, −32) [1]. d²y/dx² = 6x − 12: at x = 0 it is −12 < 0, so (0, 0) is a maximum; at x = 4 it is 12 > 0, so (4, −32) is a minimum [1] [1] (b) Positive cubic shape with a maximum at (0, 0) and a minimum at (4, −32) [1]; crosses the x-axis at (6, 0) and touches it at the origin [1] (c) 0 < x < 4 [1] Examiner insight: a sketch mark needs the key points labelled with coordinates; an unlabelled correct shape usually earns only the shape mark.

11. (a) πr²h = 250π, so h = 250/r² [1]. S = 2πr² + 2πrh = 2πr² + 2πr(250/r²) = 2πr² + 500π/r [1] (b) dS/dr = 4πr − 500πr⁻² [1]. 4πr − 500π/r² = 0 gives r³ = 125 [1], so r = 5 [1] (c) d²S/dr² = 4π + 1000π/r³ = 12π at r = 5 [1]. 12π > 0, so minimum [1]. S = 50π + 100π = 150π = 471 cm² (3 s.f.) [1] Examiner insight: in (a), writing h in terms of r and substituting must both be seen, since the answer is given.

12. (a) y = 12(x² + 2)⁻¹, dy/dx = −12(x² + 2)⁻² × 2x [1] = −24x/(x² + 2)² [1] (b) dy/dx = 0 when x = 0, so the stationary point is (0, 6) [1]. At x = −1, dy/dx = 24/9 > 0; at x = 1, dy/dx = −24/9 < 0 [1]. Positive then negative, so maximum [1] (c) dy/dt = (dy/dx) × (dx/dt) = (−8/3) × 0.5 [1] = −4/3, so y decreases at 1.33 units per second (3 s.f.) [1] (d) At x = 1, y = 4 and the normal gradient is 3/8 [1]. y − 4 = (3/8)(x − 1), so 3x − 8y = −29 [1] Examiner insight: in (b) the question fixes the method, so a second-derivative argument is not accepted here, however correct.

Where marks are usually lost

• Roots and reciprocals not rewritten as powers of x before differentiating.
• The inside derivative left out of the chain rule, especially when it is −1 or −2.
• A “show that” with the given result appearing without the algebra that produces it.
• Normal gradient taken as −m instead of −1/m.
• Rates of change found by substituting the value of h or r before differentiating.
• Stationary points given without y-coordinates, or their nature stated without evidence.
• A method different from the one the question specifies (“by considering the sign of dy/dx”).
• Rounded intermediate values carried forward, so the final answer is wrong at 3 s.f.

Next steps

• Paper 1 differentiation revision notes
• Paper 1 differentiation study guide
• Pure Mathematics 1 mixed practice
• A Level Mathematics hub
• Printable 9709 checklist
• AS Level diagnostic, 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 1, Pure Mathematics 1 (for Paper 1): section 1.7 Differentiation.