Derivatives

What you will learn

• See the derivative as the gradient of a tangent.
• Use the power rule.
• Find gradients and turning points.

Before you start: you should know gradient and limits. See Understanding Limits.

The idea

For a straight line, the gradient is the same everywhere. A curve is different: it is steep in some places and flat in others. The derivative tells you the gradient of a curve at any particular point. In other words, it is the instantaneous rate of change.

From chords to tangents

Pick a point on a curve and a second point a small distance $h$ further along. The straight line through them (a chord) has gradient

$\frac{f(x + h) - f(x)}{h}$

Now slide the second point closer, making $h$ smaller and smaller. The chord swings round until it just touches the curve at one point, forming the tangent. Its gradient is the derivative:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

This is differentiation from first principles.

Notation

If $y = f(x)$, the derivative can be written $f'(x)$, $\dfrac{dy}{dx}$ or $\dfrac{d}{dx}f(x)$. The $\frac{dy}{dx}$ form reminds us it is a tiny change in $y$ divided by a tiny change in $x$.

The power rule

Working from first principles every time would be slow. Fortunately, patterns emerge. For any power of $x$:

$\frac{d}{dx}\, x^n = n x^{n-1}$

Bring the power down in front, then reduce the power by one. Constants disappear (a flat line has gradient zero), and derivatives of sums are sums of derivatives.

Worked examples

Example 1: First principles

Differentiate $f(x) = x^2$ from first principles.

1. $f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$.
2. $\dfrac{f(x + h) - f(x)}{h} = \dfrac{2xh + h^2}{h} = 2x + h$.
3. Let $h \to 0$: the expression approaches $2x$.

Answer: $f'(x) = 2x$, agreeing with the power rule.

Example 2: Using the rules

Differentiate $y = 4x^3 - 5x^2 + 7x - 2$ and find the gradient at $x = 2$.

1. Differentiate term by term: $\dfrac{dy}{dx} = 12x^2 - 10x + 7$.
2. Substitute $x = 2$: $12(4) - 10(2) + 7 = 48 - 20 + 7 = 35$.

Answer: $\dfrac{dy}{dx} = 12x^2 - 10x + 7$; gradient $35$ at $x = 2$.

Example 3: Turning points

Find the turning points of $y = x^3 - 3x$ and say whether each is a maximum or minimum.

1. Differentiate: $\dfrac{dy}{dx} = 3x^2 - 3$.
2. At a turning point the gradient is zero: $3x^2 - 3 = 0$, so $x = \pm 1$.
3. Find $y$: at $x = 1$, $y = -2$; at $x = -1$, $y = 2$.
4. Second derivative: $\dfrac{d^2y}{dx^2} = 6x$. At $x = 1$ it is positive, so this is a minimum; at $x = -1$ it is negative, so this is a maximum.

Answer: maximum at $(-1, 2)$, minimum at $(1, -2)$.

Rewrite roots and fractions as powers before differentiating: $\sqrt{x} = x^{1/2}$ and $\frac{1}{x^2} = x^{-2}$.

Why it matters

Derivatives are everywhere change is measured. Velocity is the derivative of position and acceleration is the derivative of velocity. Economists talk about marginal cost, the derivative of total cost. Engineers find the dimensions that minimise material, and machine-learning algorithms train by following derivatives (gradients) downhill to reduce error.