OCR A Level Mathematics A: Differentiation fundamentals: gradients, second derivatives, first principles and standard derivatives (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 cover differentiation fundamentals in OCR A Level Mathematics A (H240), content references 1.07a to 1.07l, specification Version 3.1 (August 2026). Questions marked A Level only use Stage 2 content; the rest are Stage 1, common to AS and A Level. The marks are our own guide to the size of each step. Start with the study guide if a topic is unfamiliar, and use the revision notes for a quick recap. The course hub and printable checklist show the wider course, and the free diagnostics can find your weak spots.

Questions

1. Use differentiation from first principles to show that the derivative of f(x) = 2x³ + x is 6x² + 1. [4]

2. The curve y = 2x² − x passes through A(1, 1). The point B on the curve has x-coordinate 1 + h.

(a) Find the gradient of the chord AB, in terms of h, in its simplest form. [2] (b) Use your answer to find the gradient of the curve at A, and say what happens to B in the limit. [1]

3. Work this out by hand.

(a) Find dy/dx when y = 6x^(5/2) − 4/x³ + 9/√x. [3] (b) Find g’(x) when g(x) = (2x − 1)(x + 4)/x. [3]

4. A Level only. Differentiate f(x) = 5e^(−3x) + 2 × 7^(2x) + 4 tan(x/2) + 6 ln x. [4]

5. A Level only. Given y = 3 sin 2x − 8 cos(x/2), with x in radians:

(a) find dy/dx [2] (b) find d²y/dx². [2]

6. A cubic curve y = f(x) has a positive coefficient of x³, a local maximum at x = −2 and a local minimum at x = 4.

(a) Describe the graph of y = f’(x), giving its shape and where it meets the x-axis. [2] (b) State the x-coordinate of the lowest point of y = f’(x), with a reason. [1] (c) State the value of f’’(x) at that point. [1]

7. A Level only. f(x) = x⁴ − 6x² + x.

(a) Find f’’(x). [2] (b) Find the intervals on which f is convex and on which it is concave. [2] (c) Show that f has points of inflection at x = 1 and x = −1, and give their coordinates. [2]

8. A Level only. A cup of tea cools so that its temperature after t minutes is T = 20 + 30e^(−t/4) degrees Celsius.

(a) Find dT/dt. [2] (b) Find the rate of change of temperature when t = 2, to 3 significant figures, with units. [1] (c) Find d²T/dt². [1] (d) Explain what the sign of d²T/dt² tells you about the cooling. [1]

9. A Level only. Prove from first principles that the derivative of cos x is −sin x. You may use that, in radians, sin h / h → 1 and (cos h − 1)/h → 0 as h → 0. [5]

10. A Level only. g(x) = e^(2x) − 4e^x.

(a) Find g’(x) and g’‘(x). [2] (b) Show that g’’(x) = 0 only when x = 0. [2] (c) Determine where g is concave and where it is convex, and give the coordinates of the point of inflection. [4] (d) Find the gradient of the curve at the point of inflection. [1]

11. A Level only. The height of the tide above a marker is h = 6 + 2.5 sin(πt/6) metres, where t is in hours and the angle is in radians.

(a) Find dh/dt. [2] (b) Find the exact rate of change at t = 2, then give it to 3 significant figures. [2] (c) Find d²h/dt². [2] (d) At t = 3, state dh/dt, and use d²h/dt² to say what the height is doing, giving the height. [2]

Answers

1. f(x + h) − f(x) = 2(3x²h + 3xh² + h³) + h [1]. Divide by h: 6x² + 6xh + 2h² + 1 [1]. Write the limit as h → 0 [1]. The terms in h vanish, so f’(x) = 6x² + 1 [1] [4] Examiner insight: A “show that” answer needs the expansion and the limit visible. Writing only the final 6x² + 1 shows none of the reasoning.

2. (a) y at B = 2(1 + h)² − (1 + h) = 1 + 3h + 2h² [1]. Gradient AB = (3h + 2h²)/h = 3 + 2h [1] (b) As h → 0, B moves along the curve towards A and the gradient is 3 [1] [3] Examiner insight: Cancel the h before taking the limit. Putting h = 0 into the unsimplified fraction gives 0/0, which tells you nothing about the gradient.

3. (a) 15x^(3/2) [1]; +12x⁻⁴ [1]; −(9/2)x^(−3/2) [1] (b) Expand the numerator: 2x² + 7x − 4 [1]. Divide by x: g(x) = 2x + 7 − 4x⁻¹ [1]. g’(x) = 2 + 4/x² [1] [6] Examiner insight: Rewrite 4/x³ as 4x⁻³ before differentiating. Applying the power rule to the fraction as printed is a common source of lost marks.

4. −15e^(−3x) [1]; 4 ln 7 × 7^(2x) [1]; 2 sec²(x/2) [1]; 6/x [1]. So f’(x) = −15e^(−3x) + 4 ln 7 · 7^(2x) + 2 sec²(x/2) + 6/x [4] Examiner insight: The factor ln 7 and the factor 2 are separate pieces of working. Show both so a slip on one does not hide the other.

5. (a) dy/dx = 6 cos 2x [1] + 4 sin(x/2) [1] (b) d²y/dx² = −12 sin 2x [1] + 2 cos(x/2) [1] [4] Examiner insight: The minus in front of the cosine term combines with the minus from differentiating cos, giving a plus. Write the sign of each term explicitly.

6. (a) The graph is a parabola (a quadratic) [1] that is U-shaped and meets the x-axis at x = −2 and x = 4 [1] (b) The lowest point is at x = 1, midway between the roots, since a parabola is symmetrical [1] (c) f’‘(1) = 0, as the gradient of f’ is zero at its lowest point [1] [4] Examiner insight: Mention the shape and both intercepts. A sketch with no scale or labels cannot show where the roots lie.

7. (a) f’(x) = 4x³ − 12x + 1 [1]; f’‘(x) = 12x² − 12 [1] (b) f’’ = 12(x − 1)(x + 1) > 0 for x < −1 and x > 1, so f is convex there [1]. f’’ < 0 for −1 < x < 1, so f is concave there [1] (c) f’’ = 0 at x = ±1 and its sign changes at each, so both are points of inflection [1]. f(1) = −4 and f(−1) = −6, so the points are (1, −4) and (−1, −6) [1] [6] Examiner insight: Say that f’’ changes sign. Showing only f’’ = 0 does not distinguish an inflection from, say, y = x⁴ at the origin.

8. (a) dT/dt = 30 × (−1/4)e^(−t/4) [1] = −7.5e^(−t/4) [1] (b) At t = 2, dT/dt = −7.5e^(−0.5) = −4.55 °C per minute [1] (c) d²T/dt² = 1.875e^(−t/4) [1] (d) It is positive, so the rate of change is increasing from a negative value towards 0: the tea cools more slowly as time passes [1] [5] Examiner insight: Give units in (b), and say in (d) what the sign means for the situation. A bare “it is positive” earns less than a sentence in context.

9. The quotient is [cos(x + h) − cos x]/h, and cos(x + h) = cos x cos h − sin x sin h [1]. The numerator becomes cos x (cos h − 1) − sin x sin h [1]. Divide by h: cos x · (cos h − 1)/h − sin x · (sin h / h) [1]. As h → 0, the first fraction tends to 0 and the second to 1 [1]. So the derivative is −sin x [1] [5] Examiner insight: Name the limits you use and keep the lim notation until the final line. A jump straight from the expansion to the answer is unlikely to show the reasoning a proof needs.

10. (a) g’(x) = 2e^(2x) − 4e^x [1]; g’‘(x) = 4e^(2x) − 4e^x [1] (b) g’’ = 4e^x(e^x − 1) [1]. Since e^x > 0, g’’ = 0 only when e^x = 1, so x = 0 [1] (c) For x < 0, e^x < 1 so g’’ < 0: concave [1]. For x > 0, g’’ > 0: convex [1]. The sign changes at x = 0, so it is a point of inflection [1]. g(0) = 1 − 4 = −3, so the point is (0, −3) [1] (d) g’(0) = 2 − 4 = −2 [1] [9] Examiner insight: Factorise g’’ so that the sign on each side of x = 0 is easy to state. Remember e^x is never zero, which is why the single root is enough.

11. (a) dh/dt = 2.5 × (π/6) cos(πt/6) [1] = (5π/12) cos(πt/6) [1] (b) At t = 2: (5π/12) cos(π/3) = 5π/24 m/h [1] = 0.654 m/h [1] (c) d²h/dt² = (5π/12)(−π/6) sin(πt/6) [1] = −(5π²/72) sin(πt/6) [1] (d) At t = 3, cos(π/2) = 0, so dh/dt = 0 [1]. d²h/dt² = −5π²/72 < 0, so the gradient is decreasing: the height is at a maximum, 6 + 2.5 = 8.5 m [1] [8] Examiner insight: Keep the exact value alongside the decimal, and keep the angle in radians throughout. The factor π/6 appears once in the first derivative and again in the second.

Where marks are usually lost

• Skipping the expansion in a first-principles answer, or dropping the limit notation.
• Putting h = 0 before cancelling the h.
• Differentiating a fraction or root without first writing it as a power.
• Missing ln a in the derivative of a^kx, or the factor k in an exponential or trigonometric derivative.
• Sign errors on cos kx and on the second derivative of sin kx.
• Calling a point an inflection from f’’ = 0 without testing the sign either side.
• Describing f’ as having the shape of f, rather than one degree lower.
• Giving a rate of change with no units, or an interpretation with no context.
• Using degrees when the standard results need radians.

Next steps

• Recap with the revision notes.
• Re-read the study guide for any topic you missed.
• Open the OCR A Level Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR: section 1.07, Differentiation, references 1.07a to 1.07l.