Differentiation (Cambridge IGCSE Maths): Revision notes

Section 1

What is differentiation?

Differentiation is a method for finding the gradient (rate of change) of a curve at any point, using algebra rather than drawing a tangent by eye.

The result of differentiating $y$ with respect to $x$ is written $\dfrac{dy}{dx}$ , called the derivative
$\dfrac{dy}{dx}$ is itself a function of $x$ — it tells you the gradient of the curve at any given $x$ -value
On this platform, differentiation questions apply only to functions built from terms of the form $ax^n$ , where $a$ is rational and $n$ is a positive integer or zero, with no more than three such terms added together

Key termsdifferentiationgradientderivative

Section 2

The rule for differentiating $ax^n$

To differentiate a term of the form $ax^n$ , multiply by the power and reduce the power by 1: $\text{if } y = ax^n, \text{ then } \frac{dy}{dx} = anx^{n-1}$

For a sum of up to three such terms, differentiate each term separately.

Example: Differentiate $y = 3x^4 + 2x^2 - 5x$ . $\frac{dy}{dx} = 12x^3 + 4x - 5$

A constant term (e.g. $+7$ ) differentiates to $0$ , since it doesn't change as $x$ changes.

Exam tip

Write each term's power and coefficient multiplication explicitly as your method line — this earns the M1 even if a later term is mis-simplified.

Common mistake

A very common error is forgetting that differentiating a constant term gives 0, not the constant itself.

Section 3

Finding gradients using the derivative

Once you have $\dfrac{dy}{dx}$ , substitute a specific $x$ -value to find the gradient of the curve at that point.

Differentiate $y$ to get $\dfrac{dy}{dx}$
Substitute the given $x$ -value into $\dfrac{dy}{dx}$
The result is the gradient of the tangent to the curve at that point

Example: Find the gradient of $y = x^2 - 4x$ at $x = 3$ . $\frac{dy}{dx} = 2x - 4$ At $x=3$ : gradient $= 2(3) - 4 = 2$ .

Key termstangent

Section 4

Stationary points: maxima and minima

A stationary point (turning point) is a point where the gradient of the curve is zero — the curve is momentarily flat.

Differentiate $y$ to find $\dfrac{dy}{dx}$
Set $\dfrac{dy}{dx} = 0$ and solve for $x$
Substitute back into the original equation to find the corresponding $y$ -value

To decide whether a stationary point is a maximum or a minimum, use any valid method:

Inspect the gradient just before and just after the point (positive→negative = maximum; negative→positive = minimum)
Use the second derivative (not required by name, but checking sign either side is always valid)
Consider an accurate sketch of the curve's shape

Example: For $y = x^2 - 4x + 3$ , $\dfrac{dy}{dx}=2x-4=0 \Rightarrow x=2$ , $y=-1$ . Gradient is negative just before $x=2$ and positive just after, so $(2,-1)$ is a minimum.

Key termsstationary pointmaximumminimum

Exam tip

Points of inflection are not required on this specification — you only need to classify stationary points as maxima or minima.

Must Know

Differentiating $ax^n$ gives $anx^{n-1}$ ; differentiate each term of a sum separately
A constant term differentiates to 0
Substituting an $x$ -value into $\dfrac{dy}{dx}$ gives the gradient of the curve at that point
Stationary points occur where $\dfrac{dy}{dx} = 0$
Classify stationary points as maxima or minima by checking the gradient sign either side of the point
Only functions with terms of the form $ax^n$ (n a positive integer or zero, up to three terms) are examined

That's the notes covered.

Carry on to the next subtopic.