DifferentiationEdexcel IGCSE Maths: Revision notes

Differentiation finds the gradient function of a curve — a formula that tells you the gradient (rate of change) at any point on the curve, rather than just at one point.

For a curve $y = f(x)$, the gradient function is written $\dfrac{dy}{dx}$ or $f'(x)$.

The basic rule for a term $ax^n$: $\dfrac{d}{dx}(ax^n) = nax^{n-1}$

Multiply by the power, then reduce the power by 1. Differentiate each term of a polynomial separately.

Example: if $y = 4x^3 - 5x^2 + 7x - 3$, then $\dfrac{dy}{dx} = 12x^2 - 10x + 7$

Note that a constant term (like $-3$) always differentiates to $0$ — a constant has zero gradient because it never changes.

Once you have $\dfrac{dy}{dx}$, substitute the $x$-value of the point into it — this gives the actual numerical gradient (i.e. the gradient of the tangent to the curve at that point).

Worked example: for $y = x^3 - 4x$, find the gradient at $x = 2$.

1. Differentiate: $\dfrac{dy}{dx} = 3x^2 - 4$
2. Substitute $x = 2$: $3(2)^2 - 4 = 12 - 4 = 8$

So the gradient of the curve at $x = 2$ is $8$.

This value can then be used with the point $(2, y)$ to find the equation of the tangent line at that point using $y - y_1 = m(x - x_1)$.

At a turning point (also called a stationary point), the gradient is momentarily zero — the tangent is horizontal. So: $\dfrac{dy}{dx} = 0$

Steps to find and classify a turning point:

1. Differentiate $y$ to get $\dfrac{dy}{dx}$.
2. Set $\dfrac{dy}{dx} = 0$ and solve for $x$.
3. Substitute each $x$-value back into the original equation for $y$ to get the coordinates.
4. Classify the point using the second derivative $\dfrac{d^2y}{dx^2}$ (differentiate $\dfrac{dy}{dx}$ again):
   • If $\dfrac{d^2y}{dx^2} > 0$, the point is a minimum (curve bends upwards, like a smile).
   • If $\dfrac{d^2y}{dx^2} < 0$, the point is a maximum (curve bends downwards, like a frown).

Alternatively, check the sign of $\dfrac{dy}{dx}$ just before and just after the point (positive-to-negative = maximum; negative-to-positive = minimum).

In kinematics, displacement, velocity and acceleration are linked by differentiation with respect to time, $t$:

$\text{displacement } s \xrightarrow{\text{differentiate}} \text{velocity } v = \dfrac{ds}{dt} \xrightarrow{\text{differentiate}} \text{acceleration } a = \dfrac{dv}{dt}$

Worked example: a particle has displacement $s = t^3 - 6t^2 + 9t$ (metres, $t$ in seconds).

• Velocity: $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9$
• Acceleration: $a = \dfrac{dv}{dt} = 6t - 12$

To find when the particle is momentarily at rest (stationary), set $v = 0$ and solve for $t$. To find when acceleration is zero (constant velocity), set $a = 0$.