Differentiation from first principles, differentiating powers of $x$, gradients, tangents and normals, increasing and decreasing functions, and stationary points.

What this dot point is asking

WJEC wants you to understand differentiation as the gradient of a curve, to derive the result from first principles for a simple power, to differentiate powers of $x$ fluently, and to apply the derivative to find tangents and normals, decide where a function is increasing or decreasing, and locate and classify stationary points. This is the gateway to the whole calculus strand and recurs in mechanics through rates of change.

The answer

Differentiation from first principles

The gradient of a curve at a point is the limit of the gradient of a chord as the chord shrinks to zero.

For $f(x) = x^2$: $\dfrac{(x+h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h$, and as $h \to 0$ this gives $f'(x) = 2x$.

The power rule

Rewrite roots and reciprocals as powers first: $\sqrt{x} = x^{1/2}$ differentiates to $\tfrac{1}{2}x^{-1/2}$, and $\dfrac{1}{x^2} = x^{-2}$ differentiates to $-2x^{-3}$. Expand any products or quotients into a sum of powers before differentiating, because the AS power rule applies term by term and there is no product or quotient rule until A2. For example $y = x^2(x + 3)$ should be written as $x^3 + 3x^2$ first, giving $\dfrac{dy}{dx} = 3x^2 + 6x$. The notation $\dfrac{dy}{dx}$ and $f'(x)$ mean the same thing; the second derivative is $\dfrac{d^2y}{dx^2}$ or $f''(x)$.

Tangents and normals

Increasing, decreasing and stationary points

A function is increasing where its derivative is positive and decreasing where it is negative. Stationary points (maxima, minima or points of inflection) occur where $\dfrac{dy}{dx} = 0$.

Examples in context

Example 1. Optimisation. A farmer encloses a rectangular field of area $200\,\text{m}^2$ against a wall, using fencing on three sides. With width $x$, the length is $\dfrac{200}{x}$ and the fence length is $L = 2x + \dfrac{200}{x}$. Then $\dfrac{dL}{dx} = 2 - \dfrac{200}{x^2} = 0$ gives $x = 10\,\text{m}$, the width that minimises the fencing. Differentiation finds the most economical design.

Example 2. A rate of change. The volume of a balloon is $V = \tfrac{4}{3}\pi r^3$. Then $\dfrac{dV}{dr} = 4\pi r^2$, which is exactly the surface area, so a small increase in radius adds a shell of volume equal to the surface area times the thickness. The derivative gives the instantaneous rate of growth.

Try this

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

• Cue. $\dfrac{dy}{dx} = 12x^3 - 2$.

Q2. Find the gradient of $y = \dfrac{1}{x}$ at $x = 2$. [3 marks]

• Cue. $y = x^{-1}$, so $\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{4}$ at $x = 2$.

Q3. Find the range of values of $x$ for which $y = x^2 - 8x$ is decreasing. [3 marks]

• Cue. $\dfrac{dy}{dx} = 2x - 8 < 0$, so $x < 4$.