Differentiation

Differentiation answers one question: how does a quantity change as another quantity changes? The answer is the derivative, a function that gives the instantaneous rate of change at any point. Calculus accounts for five Paper 01 items and two Paper 02 questions.

The Derivative as a Gradient¶

For a straight line, the gradient is constant. For a curve, the gradient varies from point to point. The derivative $\dfrac{dy}{dx}$ (also written $f'(x)$) gives the gradient of the tangent to $y = f(x)$ at any $x$ value.

The derivative is defined as a limit:

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

This is the slope of the chord between $(x, f(x))$ and $(x+h, f(x+h))$ as $h$ approaches zero. The syllabus expects an intuitive understanding of this limit, not formal proofs.

The Power Rule¶

The most-used differentiation rule:

$\frac{d}{dx}(x^n) = nx^{n-1}$

Works for any real $n$, including fractions and negative values.

Linear rules: $\dfrac{d}{dx}[cf(x)] = cf'(x)$ and $\dfrac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$.

Differentiate polynomial functions term by term.

Derivatives of Sine and Cosine¶

$\frac{d}{dx}(\sin ax) = a\cos ax \qquad \frac{d}{dx}(\cos ax) = -a\sin ax$

The chain applied to a linear argument $ax$ simply multiplies by the inner derivative $a$.

The Chain Rule¶

For a composite function $y = f(g(x))$, written as $y = f(u)$ where $u = g(x)$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Informally: differentiate the outer function (leaving the inner unchanged), then multiply by the derivative of the inner function.

The Product Rule¶

For $y = u(x)\,v(x)$:

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

"First times derivative of second, plus second times derivative of first."

The Quotient Rule¶

For $y = \dfrac{u(x)}{v(x)}$:

$\frac{dy}{dx} = \frac{v\,u' - u\,v'}{v^2}$

"Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared."

Tangents and Normals to Curves¶

• The tangent at $x = a$ has gradient $f'(a)$.
• The normal at $x = a$ is perpendicular to the tangent; its gradient is $-1/f'(a)$.

Both use the point-slope form $y - y_0 = m(x - a)$ where $(a, y_0) = (a, f(a))$.

Stationary Points¶

A stationary point occurs where $\dfrac{dy}{dx} = 0$. At these points the tangent is horizontal.

Second derivative test:

• If $\dfrac{d^2y}{dx^2} > 0$ at a stationary point: minimum (curve concave up).
• If $\dfrac{d^2y}{dx^2} < 0$ at a stationary point: maximum (curve concave down).
• If $\dfrac{d^2y}{dx^2} = 0$: test is inconclusive — use the first derivative sign-change test instead.

First derivative sign-change test: evaluate $\dfrac{dy}{dx}$ at values just before and just after the stationary point.

• $+ \to -$: maximum (gradient goes from positive to negative).
• $- \to +$: minimum (gradient goes from negative to positive).
• Same sign on both sides: point of inflection (not a turning point).

Rates of Change and Kinematics¶

The derivative represents any rate of change, not just geometric gradient.

For a particle with displacement $s(t)$:

$v = \frac{ds}{dt} \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

• When $v = 0$: particle is instantaneously at rest.
• When $a = 0$: velocity is at a stationary value (often maximum velocity).
• Positive $v$: moving in the positive direction; negative $v$: moving in the negative direction.

Connected Rates of Change¶

When two quantities both change with time, the chain rule links their rates:

$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$

Applied Optimisation¶

Differentiation finds the value of a variable that maximises or minimises a given quantity. The standard exam scenario involves a profit or revenue function.

For any business model:

• Revenue $R(x)$: income from selling $x$ units.
• Cost $C(x)$: total cost to produce $x$ units (fixed costs plus variable costs).
• Profit $P(x) = R(x) - C(x)$.

To maximise profit: set $P'(x) = 0$, solve for $x$, then verify with the second derivative.