G: Differentiation

AQA A-Level Mathematics: G Differentiation (3.8)

Overview

Differentiation is the mathematics of instantaneous change. It turns a function into a new function that tells you the gradient of the original curve at every point. In AQA 7357 this begins with tangents and rates of change, but it quickly becomes the tool behind maxima and minima, normals, graph behaviour, parametric curves, connected rates and simple differential equations. It is an important exam topic because the same method can be assessed as a short AO1 skills question, a multi-step pure problem or a contextual AO3 modelling question. A candidate who can choose a rule, show the link between their derivative and the question, and interpret the result gains marks far more reliably than one who only memorises formulas. Differentiation links directly to functions, trigonometry, exponentials and logarithms, coordinate geometry, kinematics and later integration.

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Key concepts

1. The derivative: gradient and rate of change

For y = f(x), f'(x) or dy/dx is the gradient of the tangent at a general point on the curve. A secant joins two points. As the second point moves towards the first, its gradient approaches the tangent gradient. This limiting idea is why the derivative is an instantaneous, rather than average, rate of change.

If s is displacement in metres and t is time in seconds, ds/dt is velocity in metres per second. The notation tells you both what changes and what it changes with respect to. Read it as “rate of change of s with respect to t”. Positive derivative means increasing; negative derivative means decreasing; zero derivative means horizontal tangent.

First principles. For a small positive integer n, the derivative can be established from

f'(x) = lim(h→0) [f(x+h) - f(x)] / h.

For example, with f(x) = x²,

[(x+h)² - x²]/h = [2xh + h²]/h = 2x + h, so the limit is 2x.

Credit is given for expanding accurately, cancelling the factor h, and only then applying the limit. Never substitute h = 0 at the beginning because that produces division by zero.

2. Standard derivatives and the power rule

The power rule is d/dx(x^n) = nx^(n-1) for rational n. It works for roots and reciprocals because these are powers: √x = x^(1/2) and 1/x² = x^(-2). Rewrite awkward fractions as powers before differentiating where this reduces risk.

Example. y = 5x^(3/2) - 4x^(-1) gives dy/dx = (15/2)x^(1/2) + 4x^(-2). The last sign is positive because -4 × -1 = +4.

The AQA specification also requires exponential, logarithmic and trigonometric derivatives. The standard forms are shown in the formula bank below. For sin(kx), cos(kx), tan(kx), e^(kx) and a^(kx), the factor k appears because the inside function has gradient k.

3. Choosing and executing a differentiation rule

Use the chain rule for a function inside a function. Differentiate the outer function, leave the inner expression in place, then multiply by the derivative of the inner expression. For y = (3x² - 5)^4,

dy/dx = 4(3x² - 5)^3 × 6x = 24x(3x² - 5)^3.

Use the product rule when two expressions are multiplied: d(uv)/dx = u'v + uv'. For y = x²sin x, dy/dx = 2x sin x + x² cos x. Do not multiply the two derivatives together.

Use the quotient rule when one expression is divided by another: d(u/v)/dx = (vu' - uv')/v². Candidates should label numerator u and denominator v before substitution. This makes the order of the subtraction unambiguous and exposes method marks.

Rule-selection check. One expression inside another? Chain. Two expressions multiplied? Product. One divided by another? Quotient. Otherwise, differentiate each term using standard results. Simplify first only when it genuinely makes the structure safer.

4. Tangents, normals and stationary points

To find a tangent at x = a, calculate m = f'(a) and the point (a, f(a)). Then use point-gradient form:

y - y₁ = m(x - x₁).

For a normal, the gradient is the negative reciprocal: m_normal = -1/m_tangent, provided the tangent is neither horizontal nor vertical. A normal question needs the point as well as its gradient.

Stationary points satisfy dy/dx = 0. Solve for all x-values, then obtain y-values from the original function. The second derivative classifies them:

• d²y/dx² > 0: local minimum, because the curve is concave up.
• d²y/dx² < 0: local maximum, because the curve is concave down.
• d²y/dx² = 0: inconclusive. Test the sign of dy/dx on either side.

A stationary point of inflection has a horizontal tangent but dy/dx does not change sign. A point of inflection need not be stationary: it is where concavity changes.

5. Graph behaviour and the derivative function

f'(x) is itself a function. Sketch it above the x-axis where f is increasing and below the x-axis where f is decreasing. A root of f' is a candidate stationary point. If f' changes from positive to negative, f has a local maximum. If it changes from negative to positive, f has a local minimum. This is often faster and more informative than looking only at the original graph.

For graphical questions, label relevant intercepts, stationary coordinates and the sign of the gradient. A sketch does not need a scale unless one is supplied, but it must show the correct overall behaviour. Use the second derivative to describe concavity: positive means gradients are increasing; negative means gradients are decreasing.

6. Implicit and parametric differentiation

For an implicit relation, differentiate every term with respect to x. Whenever you differentiate a term containing y, multiply by dy/dx because y depends on x. From x² + y² = 25,

2x + 2y(dy/dx) = 0, hence dy/dx = -x/y.

At (3, 4), the gradient is -3/4. The common error is treating y as a constant and writing the derivative of y² as 2y.

For parametric equations x = f(t) and y = g(t), use

dy/dx = (dy/dt) / (dx/dt).

Differentiate x and y separately with respect to t, divide, then substitute a particular t-value. For a tangent or normal, also substitute that t-value into the original parametric equations to locate the point. AQA G5 requires first derivative only.

7. Connected rates and simple differential equations

In a connected-rates problem, variables are linked and change with time. Differentiate the relationship with respect to time before inserting values. For a sphere, V = (4/3)πr³ gives

dV/dt = 4πr²(dr/dt).

If r = 5 cm and dr/dt = 0.4 cm s⁻¹, then dV/dt = 40π cm³ s⁻¹. Units matter: a rate of volume must be cubic units per unit time.

To construct a simple differential equation, state variables and translate the relationship precisely. “The rate of increase of P is proportional to P” becomes dP/dt = kP. The sign and units of constants should make sense in context. AQA may use kinematics, population growth or price-demand modelling. Constructing the equation is the G6 requirement; solving separable equations belongs to integration.

Formula bank: learn these, do not wait for a formula sheet

All differentiation results below should be treated as must memorise for AQA A-Level Mathematics. The specification’s formulae appendix lists required knowledge rather than a set of routine differentiation formulas supplied in the examination.

Practical applications and modelling

Differentiation is used to optimise a design, predict velocity, quantify marginal cost and model population change. In a modelling answer, define variables, preserve units, make any reasonable domain restrictions explicit and interpret the sign or size of the derivative. For example, a negative dp/dt means price is falling, not that price is negative. If a maximum occurs at an endpoint of a restricted interval, compare endpoint values as well as stationary values.

There is no required practical for this pure mathematics topic and there are no Foundation/Higher tiers at A-Level. The transferable skills are representing a situation algebraically, differentiating accurately, testing whether a mathematical result is valid in its context, and communicating a conclusion with the correct units.

Examiner-style final check

Before moving on, ask: have I selected the rule from the structure, used brackets around a composite inner function, kept every minus sign, supplied a coordinate as well as a gradient where needed, classified stationary points rather than merely found them, and given units for a contextual rate? Those checks protect the most accessible marks.