Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 2: Pure Mathematics 2 · official syllabus
DIFFERENTIATION LAB
Learn it by playing it
Four short missions on the new derivatives, the chain rule's multiplier, parametric gradients, and laying a tangent exactly onto a curve. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
The self-gradient curve
y = e^x is the curve whose gradient everywhere equals its own height: d/dx e^x = e^x. The dashed bar sits at gradient 5. Your solid bar shows the gradient of e^x at your x - which is simply e^x. Slide x until the bars coincide.
Before you slide: the gradient of y = e^x at any point equals:
your gradient at x = 1: e^1 = 2.7183 - target is 5. e^x grows with x - slide up.
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 five new derivatives: e^x, ln x, sin x, cos x, tan x 2.4
The core table: d/dx e^x = e^x, d/dx ln x = 1/x, d/dx sin x = cos x, d/dx cos x = -sin x, and d/dx tan x = sec^2 x. Every Paper 2 differentiation problem is built from these five plus the power rule from Paper 1.
e^x is the only function that equals its own derivative: at every point its gradient is its height, so the gradient of e^x is 5 exactly where x = ln 5. Any other base carries a factor: d/dx a^x = (ln a) a^x.
Trig derivatives work in radians. The formulas above assume x is measured in radians; with degrees an extra factor of pi/180 would appear, so switch to radians before differentiating.
Constant multiples, sums and differences behave as in Paper 1: d/dx (3e^x - 2 sin x) = 3e^x - 2 cos x, differentiating term by term and pulling constants out.
The chain rule for composites 2.4
d/dx f(g(x)) = f'(g(x)) g'(x): differentiate the outer function, keep the inner one inside it, then multiply by the derivative of the inner function.
The multiplier comes out in front: d/dx sin(2x) = 2 cos(2x), d/dx e^(3x) = 3e^(3x), and d/dx cos(5x - 1) = -5 sin(5x - 1). Forgetting the inner derivative is the most common slip in this topic.
Some composites simplify away: d/dx ln(5x) = 5/(5x) = 1/x, because ln(5x) = ln 5 + ln x and the constant ln 5 vanishes. The chain rule and the log law agree, as they must.
Powers inside a function chain the same way: d/dx e^(1 - x^2) = -2x e^(1 - x^2), and d/dx (3x + 2)^5 = 15(3x + 2)^4. Read the outer and inner functions before touching anything.
The product rule 2.4
d/dx (uv) = u'v + uv': differentiate the first and keep the second, then keep the first and differentiate the second. The derivative of a product is not the product of the derivatives.
x ln x differentiates to 1 . ln x + x . (1/x) = ln x + 1, so its gradient at x = e is exactly 2.
Mixing the new functions: d/dx (x e^(1 - x^2)) = e^(1 - x^2) + x . (-2x) e^(1 - x^2) = (1 - 2x^2) e^(1 - x^2), and d/dx (x^2 sin x) = 2x sin x + x^2 cos x.
Factor out common pieces at the end; the factored form is what later questions (tangents, stationary reasoning, integration recognition) expect.
The quotient rule 2.4
d/dx (u/v) = (u'v - uv') / v^2: bottom times derivative of top, minus top times derivative of bottom, over the bottom squared. The minus sign makes the order matter.
Worked example: d/dx [(2x - 4)/(3x + 2)] = [2(3x + 2) - 3(2x - 4)] / (3x + 2)^2 = 16/(3x + 2)^2 - the x terms cancel completely, leaving a constant numerator.
The quotient rule explains a table entry: writing tan x = sin x / cos x gives (cos x . cos x - sin x . (-sin x)) / cos^2 x = (cos^2 x + sin^2 x) / cos^2 x = sec^2 x.
When the denominator is a plain constant, skip the rule and treat it as a constant multiple: (sin x)/2 differentiates to (cos x)/2 with no quotient work.
Tangents and normals 2.4
The tangent at x = a has gradient f'(a) and passes through (a, f(a)): y - f(a) = f'(a)(x - a). Evaluate the height and the derivative separately, then assemble the line.
Worked example: for y = ln x at x = 2, the point is (2, ln 2) and the gradient is 1/2, so the tangent is y = 0.5x + ln 2 - 1 (intercept -0.307 to three decimal places; keep ln 2 - 1 exact unless a decimal is requested).
The normal is perpendicular to the tangent, so its gradient is -1/f'(a) when f'(a) is not zero. For ln x at x = 2 the normal has gradient -2: y = -2x + 4 + ln 2.
If a tangent is stated to be parallel to a given line, set f'(x) equal to that line's gradient and solve for x first - the point follows from the curve.
Parametric differentiation 2.4
When x and y are both given in terms of a parameter t, differentiate each with respect to t and divide: dy/dx = (dy/dt) / (dx/dt), valid where dx/dt is not zero.
Syllabus-style example: x = t - e^(2t), y = t + e^(2t). Then dx/dt = 1 - 2e^(2t) and dy/dt = 1 + 2e^(2t), so dy/dx = (1 + 2e^(2t)) / (1 - 2e^(2t)). At t = 0 the rates are -1 and 3, giving dy/dx = -3 at the point (-1, 1).
A negative dx/dt only means x decreases as t increases; the quotient carries the sign automatically, so there is nothing extra to memorise.
Tangents and normals work exactly as before once dy/dx is known: at t = 0 above, the tangent is y - 1 = -3(x + 1).
Implicit differentiation 2.4
When y is tangled up with x in one equation, differentiate every term with respect to x, multiplying each y-term by dy/dx (the chain rule: d/dx y^2 = 2y dy/dx, and d/dx xy = y + x dy/dx by the product rule).
Syllabus-style example: x^2 + y^2 = xy + 7 gives 2x + 2y dy/dx = y + x dy/dx, so (2y - x) dy/dx = y - 2x and dy/dx = (y - 2x)/(2y - x).
Evaluate at a given point by substituting both coordinates. The curve above passes through (1, 3) - check: 1 + 9 = 3 + 7 - and there dy/dx = (3 - 2)/(6 - 1) = 1/5.
The answer normally contains both x and y; that is expected, because the gradient of an implicitly defined curve depends on which point of the curve you stand on.
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