Differentiation from first principles, differentiating powers of $x$, the gradient of a curve, tangents and normals, increasing and decreasing functions, and locating and classifying stationary points.

What this dot point is asking

CCEA wants you to differentiate from first principles, differentiate powers of $x$ using the standard rule, interpret the derivative as the gradient of a curve, find equations of tangents and normals, identify where a function is increasing or decreasing, and locate and classify stationary points using the first and second derivatives. Differentiation is the core of AS calculus and underpins optimisation and rates of change.

The answer

Differentiation from first principles

Differentiating powers of x

Tangents, normals, increasing and decreasing

The derivative gives the gradient of the tangent at a point. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal. A function is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$.

Stationary points

Worked example: optimisation

Examples in context

Example 1. Velocity from displacement. If displacement is $s = 5t^2 - t^3$, the velocity is $\frac{ds}{dt} = 10t - 3t^2$ and the acceleration is the second derivative $10 - 6t$. Differentiation links the kinematics quantities, which is the bridge into the mechanics content.

Example 2. Minimising cost. A manufacturer with cost $C = 2x + \frac{800}{x}$ minimises it by setting $\frac{dC}{dx} = 2 - \frac{800}{x^2} = 0$, giving $x = 20$. Differentiation finds the most economical production level, the everyday meaning of an optimisation problem.

Try this

Q1. Differentiate $y = 3x^4 - 2x + 7$. [2 marks]

• Cue. $\frac{dy}{dx} = 12x^3 - 2$.

Q2. Find the gradient of $y = x^2 - 5x$ at $x = 4$. [2 marks]

• Cue. $\frac{dy}{dx} = 2x - 5 = 3$ at $x = 4$.

Q3. A curve has $\frac{dy}{dx} = 3x^2 - 12$. Find the $x$-coordinates of its stationary points. [2 marks]

• Cue. $3x^2 - 12 = 0$, so $x = \pm 2$.