Every differentiation question from 8 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.12. 8 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/21/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.
0580 Differentiation past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q24
(a) Find the coordinates of the turning points of the graph of y = x³ − 6x² + 7.
(b) By drawing a suitable sketch, determine whether each turning point is a maximum or a minimum.
Show solution
- Differentiate to find the gradient: dy⁄dx = 3x² − 12x.
- At a turning point the gradient is 0: 3x² − 12x = 0.
- Factorise: 3x(x − 4) = 0, so x = 0 or x = 4.
- Substitute into y = x³ − 6x² + 7. x = 0: y = 7. x = 4: y = 64 − 96 + 7 = −25.
- (0, 7) and (4, −25)
- The x³ term is positive, so the graph comes up from the bottom left, turns down, then turns up again to the top right.
- So the first turning point is a peak and the second is a dip.
- (0, 7) is a maximum and (4, −25) is a minimum
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q23
The equation of a curve is y = 2x² − 11x + 12.
(a) Find the gradient of the tangent to the curve at x = 2.
(b) Line L intersects the curve at x = 2. Line L is perpendicular to the tangent to the curve at x = 2. Find the equation of line L.
Show solution
- Differentiate: dy⁄dx = 4x − 11.
- At x = 2: gradient = 4 × 2 − 11 = 8 − 11 = −3
- Find the point on the curve at x = 2: y = 2(4) − 11(2) + 12 = 8 − 22 + 12 = −2. So L passes through (2, −2).
- L is perpendicular to the tangent, so its gradient is the negative reciprocal of −3: 1⁄3.
- Substitute (2, −2) into y = 1⁄3x + c: −2 = 2⁄3 + c, so c = −2 − 2⁄3 = −8⁄3.
- y = 1⁄3x − 8⁄3
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q20
(a) Find the coordinates of the turning points of the graph y = 8x³ − 6x + 5. You must show all your working.
(b) Determine whether each turning point is a maximum or a minimum. Show how you decide.
Show solution
- Turning points are where the gradient is 0. Differentiate: dy⁄dx = 24x² − 6.
- Set the gradient to 0: 24x² − 6 = 0, so 24x² = 6 and x² = 1⁄4.
- x = 1⁄2 or x = −1⁄2.
- Find each y-value. x = 1⁄2: y = 8 × 1⁄8 − 6 × 1⁄2 + 5 = 1 − 3 + 5 = 3.
- x = −1⁄2: y = 8 × (−1⁄8) − 6 × (−1⁄2) + 5 = −1 + 3 + 5 = 7.
- (1⁄2, 3) and (−1⁄2, 7)
- Differentiate again: d²y⁄dx² = 48x. A negative value means a maximum; a positive value means a minimum.
- At x = −1⁄2: 48 × (−1⁄2) = −24 < 0, so (−1⁄2, 7) is a maximum
- At x = 1⁄2: 48 × 1⁄2 = 24 > 0, so (1⁄2, 3) is a minimum
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q17
A curve has equation y = x³ − 5⁄2x² + 3. Find the gradient of the curve at x = 1.5.
Show solution
- Differentiate each term: bring the power down and reduce it by 1. The constant 3 becomes 0.
- dy⁄dx = 3x² − 5⁄2 × 2x = 3x² − 5x.
- Substitute x = 1.5: 3 × 1.5² − 5 × 1.5 = 3 × 2.25 − 7.5 = 6.75 − 7.5.
- Gradient = −0.75
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q23
Find the coordinates of the turning point on the graph of y = 7 − 2x − x².
Show solution
- Differentiate: dy⁄dx = −2 − 2x.
- At a turning point the gradient is 0: −2 − 2x = 0, so x = −1.
- Substitute x = −1 into y = 7 − 2x − x²: y = 7 + 2 − 1 = 8.
- Turning point = (−1, 8)
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q20
y = x³ + 3x² − 13x. (a) Find dy⁄dx. (b) Find the gradient of the curve at the point where x = 3.
Show solution
- Differentiate each term: bring the power down and reduce it by 1.
- x³ → 3x², 3x² → 6x, −13x → −13.
- dy⁄dx = 3x² + 6x − 13
- Substitute x = 3 into the gradient: 3 × 3² + 6 × 3 − 13 = 27 + 18 − 13 = 32
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q22
A curve has equation y = xn + qx² + 9x, and dy⁄dx = 3x² − 12x + 9.
(a) Find the value of n, and the value of q.
(b) Work out the coordinates of the turning points of the curve.
Show solution
- Differentiate y = xⁿ + qx² + 9x term by term: dy⁄dx = nxⁿ⁻¹ + 2qx + 9.
- Compare with 3x² − 12x + 9.
- nxⁿ⁻¹ = 3x² means n = 3. 2qx = −12x means q = −6.
- n = 3, q = −6
- At a turning point the gradient is 0: 3x² − 12x + 9 = 0.
- Divide by 3: x² − 4x + 3 = 0. Factorise: (x − 1)(x − 3) = 0, so x = 1 or x = 3.
- Find y using y = x³ − 6x² + 9x.
- x = 1: y = 1 − 6 + 9 = 4. x = 3: y = 27 − 54 + 27 = 0.
- (1, 4) and (3, 0)
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q21
The diagram shows the graph of y = 3x − x³. The graph crosses the x-axis at A, at O and at B. The turning points of the graph are at P and at Q.
(a) Find the x-coordinate of A and the x-coordinate of B. Give your answers as exact values.
(b)(i) Differentiate 3x − x³.
(b)(ii) Find the coordinates of P and Q.
Show solution
- The graph crosses the x-axis where y = 0: 3x − x³ = 0.
- Factorise: x(3 − x²) = 0, so x = 0 or x² = 3.
- x² = 3 gives x = √3 or x = −√3. O is at x = 0, A is on the left and B is on the right.
- A: x = −√3, B: x = √3
- Differentiate each term: 3x → 3 and x³ → 3x².
- dy⁄dx = 3 − 3x²
- At a turning point the gradient is 0: 3 − 3x² = 0, so x² = 1 and x = 1 or x = −1.
- x = −1: y = 3(−1) − (−1)³ = −3 + 1 = −2.
- x = 1: y = 3(1) − 1³ = 3 − 1 = 2.
- P is on the left (below the axis) and Q is on the right: P(−1, −2) and Q(1, 2)
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