0606 Syllabus Topic 14 of 14

Calculus

Teacher Rig, IGCSE Add Math tutor

Written by Teacher Rig

8 years teaching IGCSE Add Math · Updated 12 June 2026

Calculus is the summit of the 0606 syllabus, the largest topic, the most marks, and the content that carries furthest into A Level. It splits into differentiation (with four applied question families) and integration (with two). Each family has a fixed exam routine; the two dedicated technique guides, differentiation and integration & area, drill them. These notes are the map.

Differentiation: the rules

dydx\frac{dy}{dx} is the gradient function. The core kit (none of it given in the exam, memorise list):

  • Power rule: xnnxn1x^n \to nx^{n-1}
  • Product rule: (uv)=uv+uv(uv)' = u'v + uv'
  • Quotient rule: (u/v)=uvuvv2(u/v)' = \frac{u'v - uv'}{v^2}
  • Chain rule: dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}, for composites like (3x25)4(3x^2 - 5)^4, e2xe^{2x}, ln(x2+1)\ln(x^2 + 1)
  • Standard derivatives: sinxcosx\sin x \to \cos x, cosxsinx\cos x \to -\sin x, tanxsec2x\tan x \to \sec^2 x, exexe^x \to e^x, lnx1x\ln x \to \frac{1}{x} (radians assumed for trig)

The universal working habit: write the setup (uu, uu', vv, vv' or the chain decomposition) before differentiating. That line carries the method mark and prevents the sign slips that dominate examiner reports.

The four differentiation question families

  1. Tangents and normals. Differentiate \to substitute xx \to tangent gradient mm (normal: 1m-\frac{1}{m}) \to line equation yy1=m(xx1)y - y_1 = m(x - x_1). Read the question twice: tangent or normal?
  2. Stationary points. Set dydx=0\frac{dy}{dx} = 0 (write the line, it’s a mark), solve, classify with d2ydx2\frac{d^2y}{dx^2}: negative \to maximum, positive \to minimum, stating the conclusion in words. Applied max/min (“find the dimensions minimising the area”) add a modelling step: build the function, then run the routine.
  3. Rates of change. Connected rates via the chain: dAdt=dAdrdrdt\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}. State the chain before substituting numbers, the stated chain is the method mark.
  4. Small changes/approximations where included: δydydxδx\delta y \approx \frac{dy}{dx} \cdot \delta x.

Integration: the reverse

xndx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} +c+ c (n1n \ne -1), plus the bracket forms (ax+b)ndx=(ax+b)n+1a(n+1)+c\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c and the standard integrals of eax+be^{ax+b}, sin(ax+b)\sin(ax+b), cos(ax+b)\cos(ax+b), each “divide by the inside coefficient”. The dropped +c+c is the syllabus’s most predictable lost mark. Finding cc from a given point (“the curve passes through (1,3)(1, 3)”) is its own question family: integrate, substitute, solve.

Definite integrals: write the bracketed antiderivative, then the substituted line F(b)F(a)F(b) - F(a) before evaluating, visible substitution is where the method credit sits, especially on Paper 1 where answers land as exact fractions.

Area under a curve

The four-step routine: sketch \to identify the region \to set up the integral(s) \to evaluate with visible substitution. Below-axis regions integrate negative (split and take magnitudes); curve-and-line regions use (topbottom)\int (\text{top} - \text{bottom}) between intersection points. The full set of traps and variants is in the integration technique guide.

Kinematics: calculus in motion

For a particle on a line with displacement s(t)s(t):

v=dsdtv = \frac{ds}{dt}, a=dvdta = \frac{dv}{dt}, differentiate forward; integrate to go back, constants from initial conditions

The translation dictionary does the marking work: at rest v=0\to v = 0; at the origin/O s=0\to s = 0; velocity is minimum dvdt=0\to \frac{dv}{dt} = 0; total distance (vs displacement) \to check for direction changes (v=0v = 0) inside the interval and integrate v|v| piecewise. Each translated statement, written down, is typically a mark.

Worked exam-style question

A curve has equation y=x36x2+9x+2y = x^3 - 6x^2 + 9x + 2. Find the stationary points and determine their nature.

dydx=3x212x+9\frac{dy}{dx} = 3x^2 - 12x + 9 (M, A) dydx=0\frac{dy}{dx} = 0: 3(x24x+3)=03(x1)(x3)=0x=1,33(x^2 - 4x + 3) = 0 \to 3(x - 1)(x - 3) = 0 \to x = 1, 3 (M, A) Points: (1,6)(1, 6) and (3,2)(3, 2) (A) d2ydx2=6x12\frac{d^2y}{dx^2} = 6x - 12. At x=1x = 1: 6<0-6 < 0 \to (1,6)(1, 6) is a maximum. At x=3x = 3: +6>0+6 > 0 \to (3,2)(3, 2) is a minimum (M, A, with conclusions in words)

Seven marks; every one attached to a written line. This question, in some costume, appears in essentially every session.

Common mistakes in this topic

  • Quotient-rule numerator order flipped; chain rule’s inner derivative forgotten
  • "dydx=0\frac{dy}{dx} = 0" used but never written, the silent mark donation
  • +c+c dropped; initial conditions never applied in kinematics
  • Tangent given where the normal was asked
  • Degrees used in trig calculus (radians always, see circular measure)
  • Distance/displacement conflated when velocity changes sign

Calculus gets weeks 1–2 of the revision plan because mastery here moves more marks than anywhere else, and it’s where an A* is genuinely built.

Calculus is also where 1-to-1 marking feedback compounds fastest. Teacher Rig has marked eight years of these scripts, free 1-hour trial class, booked on WhatsApp.

Common questions

How much of the 0606 exam is calculus?
It's the largest single topic and appears on both papers every session, directly and inside applied questions. Across a typical session, calculus-dependent marks commonly reach 25–35 of the 160 total. It is the highest-return topic to master.
What's the difference between dy/dx and d²y/dx²?
dy/dx is the gradient function, the rate of change of y. d²y/dx² differentiates again: the rate of change of the gradient, used to classify stationary points (negative → maximum, positive → minimum).
Do I need radians for calculus with trig functions?
Yes. The derivatives sin x → cos x and cos x → −sin x are only true in radians. Any calculus question involving trig assumes radian measure throughout.

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