Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
DIFFERENTIATION LAB
Learn it by playing it
Four short missions on the chain, product and quotient rules and on implicit differentiation. In each one you set the slope of a tangent line until it kisses the curve - when a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
The chain multiplier
e^(2x) grows faster than e^x because the inside function doubles every step. The chain rule says: differentiate the outside, keep the inside, then multiply by the derivative of the inside. The white curve is e^(2x); the purple line is your tangent at x = 0.5 with the slope you set. Slide the gradient until the line kisses the curve.
Before you slide: d/dx e^(2x) equals...
Your slope 4.00 against the true 5.44 - the tangent only kisses when the chain multiplier 2 is in.
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.
The chain rule 3.4
Composite function y = f(g(x)): dy/dx = f'(g(x)) * g'(x), or in Leibniz form dy/dx = (dy/du)(du/dx) with u = g(x).
Differentiate the outside and keep the inside, then multiply by the derivative of the inside: d/dx (5x + 1)^4 = 4(5x + 1)^3 * 5 = 20(5x + 1)^3.
The multiplier pattern runs through the standard results: d/dx e^(kx) = k e^(kx), d/dx sin(kx) = k cos(kx), d/dx ln(kx) = 1/x for k > 0.
Worked example: d/dx e^(2x) = 2 e^(2x), so at x = 0.5 the gradient is 2e = 5.4366 (rounded to 4 d.p.) - exactly double the curve's height there.
Product: d/dx (u v) = u dv/dx + v du/dx. Each factor is differentiated once, multiplied by the other factor untouched, then the terms add.
Worked example: d/dx (x ln x) = 1 * ln x + x * (1/x) = ln x + 1, so at x = e the gradient is exactly 2 - the x * (1/x) term collapses to 1.
Quotient: d/dx (u/v) = (v du/dx - u dv/dx) / v^2. The numerator's minus order matters, and the denominator is v squared, never just v.
Worked example: d/dx (x/(x + 1)) = ((x + 1) - x)/(x + 1)^2 = 1/(x + 1)^2, so at x = 1 the gradient is exactly 1/4 = 0.25.
Multiplying the separate derivatives, (du/dx)(dv/dx), is the classic wrong answer - check a numeric point if you are unsure.
Derivatives of e^x, ln x and the trigonometric functions 3.4
d/dx e^x = e^x (the only function equal to its own derivative); d/dx ln x = 1/x for x > 0.
d/dx sin x = cos x, d/dx cos x = -sin x, d/dx tan x = sec^2 x - all in radians; in degrees every one of these picks up an extra factor of pi/180.
The reciprocal trig derivatives follow by the quotient or chain rule: d/dx sec x = sec x tan x, d/dx cot x = -cosec^2 x.
Exam questions combine these with the rules above: d/dx (e^x sin x) = e^x sin x + e^x cos x = e^x (sin x + cos x) by the product rule.
For y = a^x, rewrite as e^(x ln a) first: d/dx a^x = a^x ln a.
Implicit differentiation 3.4
When y is mixed into the equation (circles, folium curves), differentiate every term with respect to x, treating y as a function of x: d/dx y^2 = 2y dy/dx and d/dx (x y) = y + x dy/dx.
Then collect the dy/dx terms on one side and divide - no need to solve for y first.
Worked example: x^2 + y^2 = 25 gives 2x + 2y dy/dx = 0, so dy/dx = -x/y; at (3, 4) the tangent slope is -3/4 = -0.75 exactly, downhill at the top-right of the circle.
The answer normally contains both x and y - that is expected, because the slope depends on which point of the curve you stand on.
Check the point actually lies on the curve before quoting a slope: (3, 4) works because 9 + 16 = 25.
Parametric differentiation 3.4
For x = x(t) and y = y(t), the chain rule gives dy/dx = (dy/dt) / (dx/dt), valid where dx/dt is not zero.
Worked example: x = t^2, y = 2t gives dx/dt = 2t and dy/dt = 2, so dy/dx = 1/t, which equals 1/2 at t = 2 (the point (4, 4)).
Second derivatives need the chain rule again: d^2y/dx^2 = d/dt (dy/dx) divided by dx/dt - differentiating dy/dx with respect to t alone is the classic slip.
Tangents and normals in parametric form: find the point and dy/dx at the given t, then use the straight-line equation as usual.
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