IGCSE Maths differentiation: gradients, turning points and kinematics

What each board expects

The key ideas

The derivative dy/dx is a formula for the gradient of the curve. For each term ax^n, multiply by the power and reduce the power by one: 3x^4 becomes 12x^3, 5x becomes 5, and a constant like 7 becomes 0. Differentiate a sum term by term. Expand brackets first: y = x(x + 2) = x^2 + 2x, so dy/dx = 2x + 2.

Substitute an x value into dy/dx to get the gradient at that point. At a turning point the gradient is zero, so solve dy/dx = 0 to find the x-coordinates, then substitute into y to find the y-coordinates.

To decide maximum or minimum, Cambridge accepts any valid method: an accurate sketch, the second derivative d^2y/dx^2 (negative means maximum, positive means minimum), or the sign of the gradient either side. Edexcel asks you to use the general shape of the graph, for example a positive cubic has its maximum on the left and its minimum on the right.

For motion (Edexcel), if s is displacement at time t, then velocity v = ds/dt and acceleration a = dv/dt. A particle is momentarily at rest when v = 0.

Worked example 1: turning points of y = x^3 - 3x^2 + 4

1. Differentiate: dy/dx = 3x^2 - 6x.
2. Set equal to zero: 3x^2 - 6x = 0, so 3x(x - 2) = 0, giving x = 0 or x = 2.
3. Find y: x = 0 gives y = 4; x = 2 gives y = 8 - 12 + 4 = 0. Turning points: (0, 4) and (2, 0).
4. Second derivative: d^2y/dx^2 = 6x - 6. At x = 0 it is -6, negative, so (0, 4) is a maximum. At x = 2 it is 6, positive, so (2, 0) is a minimum.
5. Check by gradient signs: at x = 1, dy/dx = 3 - 6 = -3, so the curve falls between the two points, from a maximum down to a minimum. Consistent.

Worked example 2 (Edexcel Higher): kinematics

1. A particle's displacement is s = 24t^2 - t^3 metres after t seconds, for 0 ≤ t ≤ 20. Find v and a, and when the particle is at rest.
2. v = ds/dt = 48t - 3t^2. a = dv/dt = 48 - 6t.
3. At t = 2: v = 96 - 12 = 84 m/s and a = 48 - 12 = 36 m/s^2.
4. At rest when v = 0: 3t(16 - t) = 0, so t = 0 or t = 16 seconds.
5. Maximum velocity when a = 0: t = 8, v = 384 - 192 = 192 m/s.

Worked example 3: an optimisation problem

1. A rectangle has perimeter 40 cm. Find the maximum possible area.
2. If one side is x, the other is 20 - x, so A = x(20 - x) = 20x - x^2.
3. dA/dx = 20 - 2x. Set to zero: x = 10.
4. d^2A/dx^2 = -2, which is negative, so this is a maximum. A = 10 × 10 = 100 cm^2: the best rectangle is a square.
5. Edexcel extra with a negative power: y = 2x^2 + 3x + 1/x = 2x^2 + 3x + x^-1, so dy/dx = 4x + 3 - x^-2 = 4x + 3 - 1/x^2. At x = 1 the gradient is 4 + 3 - 1 = 6.

Common mistakes that cost marks

• Differentiating a constant to itself instead of 0.
• Forgetting to expand brackets before differentiating, or differentiating a product term by term.
• Giving only the x-coordinate of a turning point when the question asks for coordinates.
• Substituting x into y instead of dy/dx when asked for a gradient.
• Getting the second derivative test the wrong way round: negative means maximum.
• Differentiating 1/x as 1 or as ln x. Rewrite it as x^-1 first, giving -x^-2.

Exam technique and how a tutor helps

Differentiation questions are worth several marks and follow a predictable sequence: differentiate, set to zero, solve, find y, decide the nature. Write each step on its own line with the dy/dx notation, which Cambridge says is expected. Where the question says 'show that', give every algebraic step.

Link calculus to the graph. A turning point you find should match the shape of the curve you would sketch. If you find a maximum to the right of a minimum on a positive cubic, something is wrong.

Differentiation is often the first calculus a student meets, and the rule is easy while the applications are not. In one-to-one lessons a tutor connects each step to the picture on the shared whiteboard, so that 'dy/dx = 0' visibly means a flat point on the curve. Students who go on to Additional Maths or A Level benefit from building that understanding early.

Self-check: can you do these?

• Differentiate y = 5x^4 - 2x + 7. (Answer: 20x^3 - 2)
• Find the gradient of y = x^3 at x = 2. (Answer: 12)
• Find the turning point of y = 4x^2 - 8x + 1 and say whether it is a maximum or minimum. (Answer: (1, -3), minimum)
• Differentiate y = (x + 3)(x - 1). (Answer: 2x + 2)
• Edexcel Higher: differentiate y = 3/x. (Answer: -3/x^2)