IGCSE Additional Maths calculus essentials: differentiation and integration for 0606

Derivatives and integrals to know

Based on 0606 syllabus items 14.3, 14.11 and 14.12. Angles in radians. None of these is printed on the exam paper.

The key ideas

The chain rule differentiates a function of a function: if y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. In practice, differentiate the outside, keep the inside, then multiply by the derivative of the inside. The product rule is d(uv)/dx = u dv/dx + v du/dx, and the quotient rule is d(u/v)/dx = (v du/dx - u dv/dx)/v^2.

Applications use the derivative as a gradient or rate. A tangent at a point has gradient dy/dx there; a normal is perpendicular to it, with gradient -1 ÷ (dy/dx). Stationary points have dy/dx = 0, and the second derivative tells you their nature, with full justification expected. Connected rates of change use the chain rule: dV/dt = dV/dr × dr/dt. Small increments use δy ≈ (dy/dx) δx.

Integration reverses differentiation, so every indefinite integral needs + c. A definite integral gives a number, and the area between a curve and the x-axis is the definite integral between the limits; for the area between a line and a curve, integrate the difference or subtract areas.

Kinematics: displacement s, velocity v = ds/dt, acceleration a = dv/dt. Going backwards, integrate acceleration to get velocity and velocity to get displacement, using given conditions to find the constant.

Worked example 1: chain, product and quotient rules

1. Chain rule: y = (3x^2 + 4)^(1/3). dy/dx = (1/3)(3x^2 + 4)^(-2/3) × 6x = 2x(3x^2 + 4)^(-2/3).
2. Product rule: y = x^2 e^(3x). u = x^2, v = e^(3x), so dy/dx = x^2 × 3e^(3x) + e^(3x) × 2x = xe^(3x)(3x + 2).
3. Quotient rule: y = (ln x)/x. u = ln x, v = x, so dy/dx = (x × 1/x - ln x × 1)/x^2 = (1 - ln x)/x^2.
4. Stationary point of y = (ln x)/x: 1 - ln x = 0, so x = e and y = 1/e. Check with a calculator: the gradient at x = 2.7 and x = 2.75 is very close to zero, changing from positive to negative, so it is a maximum.

Worked example 2: tangent, normal and connected rates

1. Curve y = x^3 - 2x at x = 2: y = 8 - 4 = 4 and dy/dx = 3x^2 - 2 = 10.
2. Tangent: y - 4 = 10(x - 2), so y = 10x - 16.
3. Normal: gradient -1/10, so y - 4 = -0.1(x - 2), which simplifies to x + 10y = 42. Check (2, 4): 2 + 40 = 42.
4. Connected rates: a sphere's radius increases at 0.2 cm/s. V = (4/3)πr^3, so dV/dr = 4πr^2. When r = 5, dV/dt = 4π(25) × 0.2 = 20π, about 62.8 cm^3/s.

Worked example 3: integration, area and kinematics

1. Area between y = 6x - x^2 and the x-axis: the curve meets the axis at x = 0 and x = 6. The integral of (6x - x^2) is 3x^2 - x^3/3. From 0 to 6: (108 - 72) - 0 = 36 square units.
2. Indefinite integrals: the integral of sin(2x + 1) is -(1/2) cos(2x + 1) + c; of 1/(2x + 3) is (1/2) ln(2x + 3) + c; of e^(4x) is (1/4)e^(4x) + c.
3. Kinematics (syllabus example): v = 3t^2 - 30t + 72. Then a = dv/dt = 6t - 30, so at t = 2, a = -18 m/s^2.
4. The particle is at rest when v = 0: t^2 - 10t + 24 = 0, so t = 4 or t = 6. Displacement from t = 0 to t = 4 (starting at s = 0): s = t^3 - 15t^2 + 72t gives 64 - 240 + 288 = 112 m.

Common mistakes that cost marks

• Calculator in degrees for trig calculus. 0606 calculus always uses radians.
• Forgetting to multiply by the derivative of the inside in the chain rule.
• Getting the quotient rule numerator in the wrong order: it is v du/dx minus u dv/dx.
• Leaving out + c on an indefinite integral, which the syllabus explicitly expects.
• Treating area below the x-axis as positive without taking the modulus or splitting the integral.
• Asserting a maximum or minimum without justification; the syllabus expects full justification.

Exam technique and how a tutor helps

Because no calculus formulas are provided, the table above has to be memorised, along with the rules. On Paper 1, which has no calculator, calculus answers are often left exact, in terms of e, ln or π. Write the rule you are using before applying it, for example 'product rule: u = x^2, v = e^(3x)', so that the method is clear even if a sign slips.

Mark schemes for 0606 were updated for the 2025 to 2027 syllabus to award more marks for working, in line with Cambridge's other maths qualifications, so clear working matters even more than before.

Additional Maths students are usually strong at IGCSE Maths, but calculus with e, ln and trig is new territory and the pace in school can be fast. In one-to-one lessons a tutor can slow down on the rule that is not yet secure, build fluency with short sets of mixed derivatives and integrals, and then work through full past-paper questions where calculus is combined with the rest of the syllabus.

Self-check: can you do these?

• Differentiate sin 3x. (Answer: 3 cos 3x)
• Differentiate e^(2x). (Answer: 2e^(2x))
• Integrate 4x^3 + 2. (Answer: x^4 + 2x + c)
• Evaluate the integral of 3x^2 from x = 1 to x = 2. (Answer: 7)
• For y = x^3, estimate the change in y when x increases from 2 to 2.01. (Answer: about 0.12)