Mathematics 0580 · syllabus E2.12
Cambridge IGCSE Mathematics 0580 — Differentiation topical past papers
Differentiating sums of terms axⁿ, finding gradients of curves, and locating turning points and deciding whether each is a maximum or a minimum.
- Questions
- 44
- Years
- 2020–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 31
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What Differentiation covers
Differentiation finds the gradient of a curve at any point. The syllabus point begins with estimating a gradient by drawing a tangent to the curve and finding the tangent's gradient. Then comes the rule: the derivative of axⁿ is naxⁿ⁻¹, so for y = 2x³ − 5x² + 4 the gradient function is dy/dx = 6x² − 10x, and the constant 4 disappears because its derivative is 0. The powers are limited to whole numbers from 0 upwards, in sums of up to three terms, and the dy/dx notation is expected.
Substituting a value of x into dy/dx gives the gradient of the curve at that point. At a turning point, or stationary point, the gradient is zero, so you solve dy/dx = 0 to find the x-coordinates and substitute back into the equation of the curve for the y-coordinates. You then decide whether each point is a maximum or a minimum by any valid method: the sign of the second derivative d²y/dx², the sign of the gradient on either side of the point, or an accurate sketch. Points of inflection are not required.
Differentiation is Extended content only, examined on Papers 2 and 4; on the non-calculator paper, substituting values into dy/dx has to be done by hand. It brings together graphs of functions, quadratic equations, which is what dy/dx = 0 gives for a cubic, and the gradient of a straight line from coordinate geometry.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- E2.12 Differentiation
What you should be able to do
- Differentiate sums of up to three terms of the form axⁿ, where n is 0 or a positive whole number
- Find the gradient of a curve at a given point
- Find the x-value at which a curve has a given gradient
- Find the coordinates of turning points by solving dy/dx = 0
- Decide whether a turning point is a maximum or a minimum, by the second derivative or the gradient on either side
Key ideas
- If y = axⁿ, then dy/dx = naxⁿ⁻¹; the derivative of a constant is 0
- The value of dy/dx at x = a is the gradient of the curve, and of its tangent, at that point
- At a turning point, dy/dx = 0
- d²y/dx² > 0 at a turning point means a minimum; d²y/dx² < 0 means a maximum
- A gradient that changes from negative to positive marks a minimum; from positive to negative, a maximum
How Differentiation has been examined
From 2022 to 2026, Differentiation was on Paper 4 · Calculator (Extended) in 13 of 14 exam series, worth about 6% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 2 · Non-calculator (Extended) | 10 | 3% | 6 of 14 |
| Paper 4 · Calculator (Extended) | 21 | 6% | 13 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 9.
Filed in Momentum under: Differentiation.
A worked example
Find the coordinates of the turning points of y = x³ − 3x² − 9x + 2, and determine whether each is a maximum or a minimum.
- Differentiate: dy/dx = 3x² − 6x − 9.
- At a turning point dy/dx = 0: 3(x² − 2x − 3) = 0, so 3(x − 3)(x + 1) = 0, giving x = 3 or x = −1.
- Substitute into the curve: at x = 3, y = 27 − 27 − 27 + 2 = −25; at x = −1, y = −1 − 3 + 9 + 2 = 7.
- Second derivative: d²y/dx² = 6x − 6, which is 12 at x = 3 and −12 at x = −1.
- A positive second derivative means a minimum and a negative one a maximum.
Answer: (−1, 7) is a maximum and (3, −25) is a minimum
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Forgetting to multiply by the power: the derivative of 4x³ is 12x², not 4x².
- Keeping the constant term: the derivative of x² + 7 is 2x, because a constant does not change.
- Giving only the x-coordinate of a turning point when its coordinates are asked for: substitute back into the equation of the curve for y.
- Substituting into the equation of the curve instead of into dy/dx when a gradient is asked for.
- Stating 'maximum' or 'minimum' without a reason: show the second derivative or the gradients on either side, because the decision carries its own mark.
Exam technique
- Write dy/dx = … on its own line before using it; the derivative is usually the first mark.
- Solve dy/dx = 0 by factorising, and find y for each x-value by substituting into the original equation, not the derivative.
- When asked for the nature of a turning point, write the value of d²y/dx² there and the conclusion in words.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.