What is the purpose of differentiation in Cambridge IGCSE Additional Mathematics?

In Cambridge IGCSE Additional Mathematics, differentiation is used to determine the rate at which a function is changing at any given point. It is a fundamental concept that helps in finding the slope of a curve, solving problems involving motion, and optimizing functions to find maximum or minimum values.

How do you find the derivative of a polynomial function?

To find the derivative of a polynomial function in the context of Cambridge IGCSE Additional Mathematics, apply the power rule. For a term ax^n, the derivative is n*ax^(n-1). This means you multiply the coefficient by the exponent and then decrease the exponent by one. Repeat this process for each term in the polynomial.

What is integration and how is it applied in Additional Mathematics?

Integration is the reverse process of differentiation and is used to find the area under a curve or the total accumulation of quantities. In Cambridge IGCSE Additional Mathematics, integration is applied to solve problems involving areas under curves, volumes of revolution, and finding functions from their derivatives.

How do you perform definite integration for a given function?

To perform definite integration in Cambridge IGCSE Additional Mathematics, first find the indefinite integral (antiderivative) of the function. Then, evaluate this antiderivative at the upper and lower limits of the integral and subtract the lower limit value from the upper limit value. This gives the area under the curve between the two limits.

What are some common applications of differentiation and integration in real-world scenarios?

Differentiation and integration have numerous real-world applications. Differentiation is used in physics to calculate velocity and acceleration, while integration is used to determine distances and areas. In economics, these concepts help in finding cost functions and optimizing profit. In engineering, they are crucial for analyzing forces and designing systems.

Why is "Forgetting that constants differentiate to zero" a common mistake in Differentiation and Integration?

Why it happens: Constants seem like 'stationary' values. How to avoid it: d/dx(c) = 0 for any constant c. Drop constants when differentiating. Source: 0606 Examiner Reports 2022-2024

Why is "Forgetting $+C$ in indefinite integrals" a common mistake in Differentiation and Integration?

Why it happens: Mechanical integration. How to avoid it: ALWAYS write +C for indefinite integrals. Only definite integrals (with bounds) skip it.

Why is "Reporting negative area" a common mistake in Differentiation and Integration?

Why it happens: Definite integral can be negative. How to avoid it: Area is always POSITIVE. If the integral is negative (curve below x-axis), take absolute value or split the integral at x-axis crossings.

Why is "Not handling the case $\frac{d^2 y}{dx^2} = 0$" a common mistake in Differentiation and Integration?

Why it happens: Standard tests assume non-zero. How to avoid it: If second derivative is also 0, use the FIRST DERIVATIVE TEST: check sign of dy/dx on either side of the stationary point.

Differentiation and IntegrationCambridge IGCSE Additional Mathematics (0606)

Differentiation and Integration

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Differentiation and Integration Study Notes — Cambridge IGCSE Additional Mathematics 0606 (2025-2027 syllabus)

The two pillars of calculus. Differentiation: rate of change, slope of curve. Power rule, chain, product, quotient. Trig and exp derivatives. Stationary points and their classification. Integration: reverse of differentiation, area under a curve, kinematics applications.

What you’ll learn

Mapped to the Cambridge IGCSE 0606 syllabus (2025-2027).

1.14.1 — Differentiate using the power, chain, product, quotient rules.
1.14.2 — Differentiate trig and exponential functions.
1.14.3 — Find and classify stationary points.
1.14.4 — Integrate using the power rule.
1.14.5 — Compute definite integrals and areas.
1.14.6 — Apply differentiation and integration to kinematics.

Differentiation rules

Power, chain, product, quotient — four big rules.

Power rule. $\dfrac{d}{dx}(x^n) = nx^{n-1}$ . Bring power down, reduce by 1. Works for any real $n$ (including negative and fractional).

Chain rule. For composite functions $y = f(g(x))$ : $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ where $u = g(x)$ .

Example: $y = (3x + 1)^4$ . Let $u = 3x + 1$ , so $y = u^4$ . Then $\frac{dy}{dx} = 4u^3 \cdot 3 = 12(3x + 1)^3$ .

Product rule. $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$ .

Quotient rule. $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v u' - u v'}{v^2}$ .

Trig derivatives (x in radians).

$(\sin x)' = \cos x$ .
$(\cos x)' = -\sin x$ .
$(\tan x)' = \sec^2 x$ .

Exponential and log.

$(e^x)' = e^x$ .
$(\ln x)' = \frac{1}{x}$ .

Cambridge tip. Memorise the four rules and the four 'special' derivatives ( $\sin, \cos, e^x, \ln$ ). Combine them via chain rule for everything else.

Power, chain, product, quotient — four rules.
Trig and exp special derivatives.
Combine via chain rule for composite functions.

See the full worked example for differentiation and integration →

Stationary points

Where $\frac{dy}{dx} = 0$ . Classify with second derivative.

Stationary point. Where the GRADIENT is ZERO. Could be a maximum, minimum, or stationary point of inflection.

Procedure.

Differentiate to get $\frac{dy}{dx}$ .
Set $\frac{dy}{dx} = 0$ and solve for $x$ .
Find the corresponding $y$ values by substituting back.
Classify using the second derivative test.

Second derivative test.

$\dfrac{d^2 y}{dx^2} > 0$ : MINIMUM (curve concave up at this point).
$\dfrac{d^2 y}{dx^2} < 0$ : MAXIMUM (curve concave down).
$\dfrac{d^2 y}{dx^2} = 0$ : INCONCLUSIVE — use first-derivative test (sign of slope on either side).

Worked walkthrough. $y = x^3 - 6x^2 + 9x + 1$ .

$\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$ . Zeros at $x = 1, 3$ .
$y(1) = 5$ , $y(3) = 1$ .
$\frac{d^2 y}{dx^2} = 6x - 12$ . At $x = 1$ : $-6 < 0$ → MAX. At $x = 3$ : $6 > 0$ → MIN.

Optimisation problems. Real-world max/min: model the quantity to be optimised, differentiate, set to zero, classify.

Cambridge tip. Always classify with the second derivative test (or first-derivative test if needed). Mark scheme awards a separate mark for classification.

$\frac{dy}{dx} = 0$ — stationary point.
Classify with second derivative.
$> 0$ MIN; $< 0$ MAX.

See the full worked example for differentiation and integration →

Integration — the reverse of differentiation

Integrate by adding 1 to the power and dividing by the new power.

Power rule for integration. $\int x^n \, dx = \frac{x^{n+1}}{n + 1} + C, \quad n \ne -1.$

Special case. $\int \frac{1}{x} \, dx = \ln|x| + C$ .

Indefinite integral. Always include $+C$ (constant of integration).

Definite integral. $\int_a^b f(x) \, dx = F(b) - F(a)$ where $F$ is any antiderivative of $f$ . No $+C$ needed (it cancels).

Area under a curve.

If $y > 0$ over $[a, b]$ : area = $\int_a^b y \, dx$ .
If $y < 0$ over $[a, b]$ : area = $\left|\int_a^b y \, dx\right|$ (take absolute value).
If $y$ changes sign: split the integral at zeros, take absolute values, sum.

Trig and exp integrals (reverse the derivatives).

$\int \cos x \, dx = \sin x + C$ .
$\int \sin x \, dx = -\cos x + C$ .
$\int e^x \, dx = e^x + C$ .

Cambridge tip. Mark scheme always penalises missing $+C$ on indefinite integrals. Make it a habit.

Power rule: add 1, divide by new power.
Definite: $F(b) - F(a)$ .
Area: take absolute value when curve below x-axis.

See the full worked example for differentiation and integration →

Kinematics applications

Differentiate position to velocity, then to acceleration. Integrate the reverse.

For a particle moving in 1D:

Position $s(t)$ .
Velocity $v = \frac{ds}{dt}$ (derivative of position).
Acceleration $a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}$ .

Going backwards.

$v = \int a \, dt$ .
$s = \int v \, dt$ .

Each integration introduces a constant of integration — usually determined by initial conditions ( $v(0)$ or $s(0)$ ).

Worked walkthrough. A particle has $v = 6t^2 + 2t$ . Find acceleration and total displacement during $0 \le t \le 2$ .

$a = \frac{dv}{dt} = 12t + 2$ . At $t = 2$ : $a = 26$ .
$s = \int_0^2 (6t^2 + 2t) \, dt = [2t^3 + t^2]_0^2 = 16 + 4 - 0 = 20$ .

Stationary in motion. Velocity = 0 → particle momentarily at rest (e.g. at the highest point of a vertical throw). Often asked: 'Find when the particle is instantaneously at rest.'

Cambridge tip. Always note the initial conditions when integrating to avoid losing constants.

$v = ds/dt$ , $a = dv/dt$ .
Integrate to go backwards; use initial conditions.
$v = 0$ marks instantaneously-at-rest moments.

How it’s examined

Calculus dominates Paper 2. Most-tested: differentiate (5-7 marks), find and classify stationary points (8 marks), integrate to find area (6-8 marks), kinematics (5-7 marks). Examiner reports flag missing $+C$ and chain-rule errors as the most common mistakes.

Sources: Cambridge IGCSE Additional Mathematics 0606 syllabus (2025-2027); 0606 Examiner Reports 2022-2024; 0606/12 Oct/Nov 2024 question paper and mark scheme. Last reviewed 2026-05-09.

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Step-by-step worked examples — Differentiation and Integration

Step-by-step solutions to past-paper-style questions on differentiation and integration, written exactly the way a tutor would explain them at the board.

1Differentiate using power rule (5 marks)
Extended• Adapted from 0606/12 Oct/Nov 2024 Q11• differentiation
Question
Find $\frac{dy}{dx}$ for $y = 3x^4 - 5x^2 + 7x - 2$ . (5 marks)
Step-by-step solution
1. Step 1
  Differentiate term by term (4 marks). Power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$ .
  
  $3x^4 \to 12x^3$ .
  
  $-5x^2 \to -10x$ .
  
  $7x \to 7$ .
  
  $-2 \to 0$ .
2. Step 2
  Combine (1 mark). $\frac{dy}{dx} = 12x^3 - 10x + 7$ .
Answer
$\dfrac{dy}{dx} = 12x^3 - 10x + 7$ .
2Find and classify stationary points (8 marks)
Extended• stationary
Question
Find the stationary points of $y = x^3 - 6x^2 + 9x + 1$ and determine their nature. (8 marks)
Step-by-step solution
1. Step 1
  Differentiate (2 marks). $\frac{dy}{dx} = 3x^2 - 12x + 9$ .
2. Step 2
  Set $\frac{dy}{dx} = 0$ (2 marks). $3x^2 - 12x + 9 = 0$ , divide by 3: $x^2 - 4x + 3 = 0$ . Factorise: $(x - 1)(x - 3) = 0$ , so $x = 1$ or $x = 3$ .
3. Step 3
  Find $y$ values (2 marks). At $x = 1$ : $y = 1 - 6 + 9 + 1 = 5$ . At $x = 3$ : $y = 27 - 54 + 27 + 1 = 1$ .
4. Step 4
  Second derivative test (2 marks). $\frac{d^2 y}{dx^2} = 6x - 12$ . At $x = 1$ : $6 - 12 = -6 < 0$ , so MAXIMUM. At $x = 3$ : $18 - 12 = 6 > 0$ , so MINIMUM.
Answer
Maximum at $(1, 5)$ . Minimum at $(3, 1)$ .
3Find the area under a curve (6 marks)
Extended• integration
Question
Find the area enclosed by the curve $y = x^2 - 4$ and the x-axis. (6 marks)
Step-by-step solution
1. Step 1
  Find x-intercepts (1 mark). $x^2 - 4 = 0$ , so $x = \pm 2$ . Curve crosses x-axis at $x = -2$ and $x = 2$ .
2. Step 2
  Curve is below x-axis between $-2$ and $2$ (1 mark). Area = $\left|\int_{-2}^{2}(x^2 - 4) \, dx\right|$ .
3. Step 3
  Integrate (2 marks). $\int (x^2 - 4) \, dx = \frac{x^3}{3} - 4x$ .
4. Step 4
  Evaluate (2 marks). $\left[\frac{x^3}{3} - 4x\right]_{-2}^{2} = \left(\frac{8}{3} - 8\right) - \left(-\frac{8}{3} + 8\right) = \frac{16}{3} - 16 = -\frac{32}{3}$ . Area $= \frac{32}{3}$ .
Answer
Area = $\dfrac{32}{3}$ square units.

Key Formulae — Differentiation and Integration

The formulae you need to memorise for differentiation and integration on the Cambridge IGCSE 0606 paper, with every variable defined in plain English and a note on when to use it.

Power rule for differentiation
$\frac{d}{dx}(x^n) = nx^{n-1}$
$n$
Any real number (constant)
When to use
Differentiating polynomial terms.
Chain rule
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
$y, u, x$
y depends on u, u depends on x
When to use
Differentiating composite functions $y = f(g(x))$ . Set $u = g(x)$ .
Product rule
$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
$u, v$
Two functions of x
When to use
Differentiating a product of two functions.
Quotient rule
$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$u, v$
Two functions of x, v ≠ 0
When to use
Differentiating a quotient.
Trig derivatives
$\frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x$
$x$
Angle in radians
When to use
Differentiating trig functions. Note: x must be in radians.
Exponential derivative
$\frac{d}{dx}e^x = e^x, \quad \frac{d}{dx}\ln x = \frac{1}{x}$
When to use
Differentiating $e^x$ and $\ln x$ .
Power rule for integration
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \ne -1$
$n$
Real number, ≠ −1
$C$
Constant of integration
When to use
Integrating polynomial terms.
Second derivative test
$\text{If } \frac{dy}{dx} = 0 \text{ at } x = a:\quad \frac{d^2 y}{dx^2} > 0 \Rightarrow \text{MIN}, \quad < 0 \Rightarrow \text{MAX}$
When to use
Classifying stationary points as max, min, or stationary inflection.

Key Definitions and Keywords — Differentiation and Integration

Definitions to memorise and the exact keywords mark schemes credit for differentiation and integration answers — sharpened from recent examiner reports for the 2026 0606 sitting.

Derivative $\frac{dy}{dx}$
Examiner keyword
Rate of change of $y$ with respect to $x$ . Geometric meaning: slope of the curve $y = f(x)$ at a point.
Stationary point
Examiner keyword
A point where $\frac{dy}{dx} = 0$ . Could be a maximum, minimum, or stationary point of inflection.
Maximum
Examiner keyword
A peak of the function. $\frac{dy}{dx} = 0$ AND $\frac{d^2 y}{dx^2} < 0$ .
Minimum
Examiner keyword
A trough of the function. $\frac{dy}{dx} = 0$ AND $\frac{d^2 y}{dx^2} > 0$ .
Integral
Examiner keyword
The reverse of differentiation. Indefinite integrals include $+C$ . Definite integrals give a numerical value.
Definite integral
Examiner keyword
$\int_a^b f(x) \, dx = F(b) - F(a)$ where $F$ is an antiderivative of $f$ . Geometrically: signed area under $y = f(x)$ from $a$ to $b$ .

Common Mistakes and Misconceptions — Differentiation and Integration

The traps other students keep falling into on differentiation and integration questions — taken from recent Cambridge IGCSE 0606 examiner reports and mark schemes — and how to avoid them.

✕Forgetting that constants differentiate to zero
0606 Examiner Reports 2022-2024
Why it happens
Constants seem like 'stationary' values.
How to avoid it
$\frac{d}{dx}(c) = 0$ for any constant $c$ . Drop constants when differentiating.
✕Forgetting $+C$ in indefinite integrals
Why it happens
Mechanical integration.
How to avoid it
ALWAYS write $+C$ for indefinite integrals. Only definite integrals (with bounds) skip it.
✕Reporting negative area
Why it happens
Definite integral can be negative.
How to avoid it
Area is always POSITIVE. If the integral is negative (curve below x-axis), take absolute value or split the integral at x-axis crossings.
✕Not handling the case $\frac{d^2 y}{dx^2} = 0$
Why it happens
Standard tests assume non-zero.
How to avoid it
If second derivative is also 0, use the FIRST DERIVATIVE TEST: check sign of $\frac{dy}{dx}$ on either side of the stationary point.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Differentiation and Integration — frequently asked questions

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Differentiation and Integration

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Differentiate using power rule (5 marks)

2Find and classify stationary points (8 marks)

3Find the area under a curve (6 marks)

Power rule for differentiation

Chain rule

Product rule

Quotient rule

Trig derivatives

Exponential derivative

Power rule for integration

Second derivative test

Derivative dydx\frac{dy}{dx}dxdy​

Stationary point

Maximum

Minimum

Integral

Definite integral

✕Forgetting that constants differentiate to zero

✕Forgetting +C+C+C in indefinite integrals

✕Reporting negative area

✕Not handling the case d2ydx2=0\frac{d^2 y}{dx^2} = 0dx2d2y​=0

Practice questions

Past paper style quiz

Assess your understanding

Differentiation and Integration — frequently asked questions

Derivative $\frac{dy}{dx}$

✕Forgetting $+C$ in indefinite integrals

✕Not handling the case $\frac{d^2 y}{dx^2} = 0$