Maths›Calculus — Differentiation
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Unit 2 · AlgebraPearson Edexcel IGCSE · 4MA1 Higher

Calculus — Differentiation

Differentiating integer powers, gradients, stationary points, maxima and minima. Higher tier IGCSE 4MA1 only.

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Corbett MathsDifferentiation — gradients and stationary points
Key facts & methods

Differentiation rules

  • If y = xⁿ, then dy/dx = nxⁿ⁻¹. (Bring the power down, reduce power by 1.)
  • If y = axⁿ, then dy/dx = naxⁿ⁻¹.
  • E.g. y = 3x⁴ → dy/dx = 12x³.
  • E.g. y = 5x² − 3x + 7 → dy/dx = 10x − 3.
  • Constant terms disappear (derivative of a constant = 0).
  • dy/dx gives the gradient of the curve at any point.

Stationary points and applications

  • Stationary points: where dy/dx = 0 (gradient = 0). These are maxima, minima or points of inflection.
  • To find: set dy/dx = 0 and solve for x. Substitute back to find y.
  • Nature of stationary point: find d²y/dx² (differentiate again).
  • If d²y/dx² > 0 → minimum (curve is concave up).
  • If d²y/dx² < 0 → maximum (curve is concave down).
  • Rate of change: dy/dx at a specific point gives the instantaneous rate of change.
  • Speed = ds/dt. Acceleration = dv/dt = d²s/dt².
Exam questions — 5 questions · 12 marks · Edexcel 4MA1 style
⚡ Show ALL working — the AI awards method marks just like a real examiner. Partial credit for correct working even if the final answer is wrong.
3 marksDifferentiate and find gradient4MA1 style ✏️ No calculator

y = 4x³ − 6x² + 5x − 2. (a) Find dy/dx. (b) Find the gradient when x = 2. (3 marks)

Hint: (a) Differentiate each term: bring the power down, reduce power by 1. Constants disappear. (b) Substitute x=2 into dy/dx.
+30 XP
4 marksStationary points4MA1 style ✏️ No calculator

Find the coordinates of the stationary points of y = 2x³ − 9x² + 12x − 3. Determine whether each is a maximum or minimum. (4 marks)

Hint: Differentiate to get dy/dx. Set dy/dx = 0 and solve. Find y values. Differentiate again for d²y/dx²: positive → minimum, negative → maximum.
+40 XP
1 markDifferentiation rule4MA1 style

If y = 5x³ − 4x² + 2x − 7, what is dy/dx?

A 15x² − 8x + 2
B 15x² − 8x
C 5x² − 4x + 2
D 15x³ − 8x² + 2
3 marksApplied rate of change4MA1 style ✏️ No calculator

The height h metres of a ball thrown upward is given by h = 20t − 5t² where t is time in seconds. (a) Find the velocity (dh/dt) at t = 1. (b) Find the time at which the ball is at maximum height. (3 marks)

Hint: (a) Differentiate h with respect to t, then substitute t=1. (b) Maximum height is where dh/dt = 0 (ball momentarily stationary at top).
+30 XP

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