Differential Calculus is worth ~35 marks in Paper 1 and rewards practice more than any other topic. The same five question types repeat every year.
Practice 1 — First principles
Use first principles to find f'(x) if f(x) = x² − 4x. Apply f'(x) = lim h→0 [f(x+h) − f(x)] / h. The h-terms cancel and you get f'(x) = 2x − 4.
Practice 2 — Derivative using rules
Differentiate y = 3x³ − 5/x² + √x. Rewrite as 3x³ − 5x⁻² + x^½, then dy/dx = 9x² + 10x⁻³ + ½x^(−½).
Practice 3 — Equation of a tangent
Find the equation of the tangent to f(x) = x² − 3x at x = 4. f(4) = 4, f'(x) = 2x − 3, so f'(4) = 5. Tangent: y − 4 = 5(x − 4) → y = 5x − 16.
Practice 4 — Cubic graph stationary points
Given f(x) = x³ − 3x² − 9x + 5. Set f'(x) = 3x² − 6x − 9 = 0. Factor: (x − 3)(x + 1) = 0 → x = 3 (local min) or x = −1 (local max).
Practice 5 — Optimisation
A rectangle has perimeter 40 cm. Maximise its area. Let length = x, then width = 20 − x. A(x) = x(20 − x) = 20x − x². A'(x) = 20 − 2x = 0 → x = 10. Max area = 100 cm².
Practice 6 — Point of inflection
For f(x) = x³ − 3x² − 9x + 5, f″(x) = 6x − 6 = 0 gives x = 1. f(1) = 1 − 3 − 9 + 5 = −6, so the point of inflection is (1; −6).
Practice 7 — Rate of change
The volume of water in a tank is V(t) = 100 + 20t − t² litres after t minutes. The rate of change at t = 4 is V′(4) = 20 − 2(4) = 12 litres per minute. The volume stops increasing when V′(t) = 0, at t = 10 minutes.
Practice 8 — Concavity
For which values of x is f(x) = x³ − 6x² concave up? f″(x) = 6x − 12, which is positive when x > 2.
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