Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 1: Pure Mathematics 1 · official syllabus
DIFFERENTIATION LAB
Learn it by playing it
Four short missions on what a derivative really is. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Shrink the chord onto the tangent
The line through P = (1, 1) and its neighbour Q is a chord of y = x^2, and h is the horizontal gap between them. Its gradient has h on the bottom of the fraction - so what happens as h slides down to nothing? Shrink h below 0.15 and watch the chord settle onto the dashed tangent.
Before you slide: as the neighbour point slides home (h towards 0), the chord's gradient will...
chord gradient = 2 + h = 4.00 at P(1, 1) - the tangent gradient is 2 once h reaches 0 (h = 2.00)
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Derivative as a limiting gradient 1.7
The gradient of a chord from x to x + h is [f(x + h) - f(x)]/h. As h approaches zero, this approaches the gradient of the tangent, when that limit exists. This is the derivative f′(x), also written dy/dx.
For y = x³ at x = 2, the chord gradient is [(2 + h)³ - 8]/h = 12 + 6h + h² for h nonzero. It approaches 12 as h approaches zero. The tangent gradient is 12, not the height y = 8.
The second derivative f″(x) or d²y/dx² is the derivative of f′(x). It measures how the gradient changes. Formal general differentiation from first principles is not required for this syllabus section.
Power rule, rational powers and the chain rule 1.7
For any rational n, d(xⁿ)/dx = nxⁿ⁻¹ where the real function is defined and differentiable. A constant differentiates to zero; differentiate sums and constant multiples term by term.
Write roots and reciprocals as powers first: y = 3√x - 4/x² becomes 3x^(1/2) - 4x^(-2), so dy/dx = (3/2)x^(-1/2) + 8x^(-3), for x > 0.
For a composite power y = [u(x)]ⁿ, use dy/dx = n[u(x)]ⁿ⁻¹u′(x). Example: y = (2x + 5)³ gives dy/dx = 6(2x + 5)². The inner derivative 2 is essential.
For y = √(3x² + 1), dy/dx = 3x/√(3x² + 1). At x = 1 the gradient is 3/2. Check domain restrictions before evaluating fractional or negative powers.
Tangents and normals 1.7
At x = a, first find the point (a,f(a)) and the tangent gradient m = f′(a). The tangent is y - f(a) = m(x - a). Do not substitute the gradient into the original curve as though it were a coordinate.
The normal is perpendicular to the tangent. For a finite nonzero tangent gradient m, its gradient is -1/m. A horizontal tangent has a vertical normal; write x = a instead of dividing by zero.
For y = x² + 1 at x = 2, the point is (2,5) and dy/dx = 2x gives m = 4. The tangent is y = 4x - 3; the normal is y - 5 = -(x - 2)/4.
To find tangents parallel to y = 11x + 4 on y = x³ - x, solve 3x² - 1 = 11. This gives x = ±2, with points (2,6) and (-2,-6). The tangent lines are y = 11x - 16 and y = 11x + 16.
Increasing, decreasing and stationary points 1.7
A differentiable function is increasing where f′(x) > 0 and decreasing where f′(x) < 0. A stationary point has f′(x) = 0; substitute its x-coordinate into f, not f′, to find its height.
At a stationary input a, f″(a) > 0 identifies a local minimum and f″(a) < 0 a local maximum. If f″(a) = 0, this test is inconclusive; inspect the sign of f′ on either side instead.
For y = x³ - 3x² + 2, y′ = 3x(x - 2) gives x = 0 and 2. The points are (0,2) and (2,-2). Since y″ = 6x - 6, the first is a maximum and the second a minimum.
The same derivative is positive for x < 0 and x > 2, and negative for 0 < x < 2. Use these intervals and intercepts to sketch the curve. A stationary point need not be a turning point: y = x³ is increasing on both sides of x = 0.
In a bounded optimisation problem, compare all relevant stationary points and allowed endpoints. A local maximum is not automatically the greatest value on a stated interval.
Rates of change and connected variables 1.7
If variables are linked, differentiate their relationship and multiply rates with the chain rule: dA/dt = (dA/dr)(dr/dt). Include units and use a negative rate when a quantity decreases.
For a circle A = πr² with dr/dt = 0.4 cm/s, dA/dt = 2πr×0.4. At r = 5 cm the area grows at 4π cm²/s, not at the constant radial rate 0.4.
For a sphere V = 4πr³/3, dV/dt = 4πr² dr/dt. If volume increases at 12π cm³/s when r = 3 cm, then dr/dt = 12π/(36π) = 1/3 cm/s.
For a rectangle with perimeter 40 cm and sides x and 20 - x, A = 20x - x². A′ = 20 - 2x is zero at x = 10 and A″ = -2, giving the greatest area 100 cm² for a square (0 < x < 20).
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