AQA A-Level Mathematics: Differentiation — Revision Notes

Section G is Paper 1 content, and Papers 2 (adding mechanics) and 3 (adding statistics) can each assess any Paper 1 content, so fluency here underpins marks across the whole exam, not just one paper. These notes are condensed for the final weeks — build understanding first from full teaching material.

G1 — What a derivative actually is

The derivative of f(x) is the gradient of the tangent to y = f(x) at a point (x, y), defined formally as the limit of the gradient of a chord as the two points on the curve move together. Two interpretations to keep separate:

• Geometric: gradient of the tangent line.
• Physical: instantaneous rate of change.

You are expected to sketch a gradient function from a given curve (where the original is increasing, the gradient function is positive; where it has a turning point, the gradient function crosses zero) and to use the second derivative as the rate of change of gradient — this connects directly to identifying convex/concave sections and points of inflection. First-principles differentiation is required for small positive integer powers of x and for sin x and cos x.

f'(x) = lim(h→0) [f(x+h) - f(x)] / h

G2 — Standard derivatives to know cold

Function	Derivative
xⁿ (n rational)	nxⁿ⁻¹
eᵏˣ	k eᵏˣ
aᵏˣ	k aᵏˣ ln a
sin kx	k cos kx
cos kx	−k sin kx
tan kx	k sec²kx
ln x	1/x

These extend to constant multiples, sums and differences by linearity — differentiate term-by-term. The aᵏˣ result (needing an extra ln a factor compared with eᵏˣ) is the one most often mis-recalled under exam pressure.

G3 — Using derivatives: gradients, tangents, turning points

Applying differentiation, you’re expected to find:

• Gradients of tangents and normals at a given point (normal gradient = −1 ÷ tangent gradient).
• Maxima, minima and stationary points — set f’(x) = 0 and solve; classify using the second derivative (f’‘(x) > 0 → minimum, f’‘(x) < 0 → maximum, f’‘(x) = 0 → test further, e.g. via the sign of f’ either side).
• Points of inflection — where f’’(x) = 0 and concavity changes.
• Where a function is increasing (f’(x) > 0) or decreasing (f’(x) < 0).

A frequent exam-trap: finding f’(x) = 0 and stopping, without classifying which stationary point type it is — always follow through with the second-derivative (or sign) test unless the question only asks for the location.

G4 — Product, quotient and chain rules

Rule	Formula
Product	(uv)’ = u’v + uv’
Quotient	(u/v)’ = (u’v − uv’) / v²
Chain	dy/dx = dy/du × du/dx

These extend to connected rates of change (using the chain rule to relate dV/dt to dV/dr × dr/dt, for example) and to differentiating inverse functions. Connected-rates-of-change problems are a classic source of marks lost purely through not identifying which two rates the chain rule needs to link — write out explicitly which variable each given rate is with respect to before attempting the chain-rule step.

G5 — Implicit and parametric differentiation (first derivative only)

• Implicit: differentiate both sides of an equation like x² + y² = r² term by term with respect to x, treating y as a function of x (so d(y²)/dx = 2y dy/dx via the chain rule), then rearrange for dy/dx.
• Parametric: given x = f(t) and y = g(t), find dy/dx via dy/dx = (dy/dt) ÷ (dx/dt).

Only the first derivative is required for implicit/parametric relations at this level — do not spend revision time on second derivatives for these two techniques specifically.

G6 — Forming differential equations

You’re expected to construct simple differential equations, both in pure contexts and in modelling contexts the specification names explicitly: kinematics, population growth, and the relationship between price and demand. The skill tested is translating a verbal rate-of-change statement (“the rate of increase of population is proportional to the population size”) into a correct equation (dP/dt = kP) — not necessarily solving it, since differential equation solving methods sit in Section H (Integration).

Exam traps

• Confusing which rule applies: product rule needs both factors differentiated and summed with cross-terms; chain rule needs a composite function unpacked into outer/inner parts first.
• Missing the extra ln a factor when differentiating aᵏˣ (as opposed to eᵏˣ, which needs no such factor).
• Finding a stationary point but never classifying it as max/min/point of inflection when the question requires this.
• Applying implicit differentiation but forgetting the chain-rule factor of dy/dx when differentiating a y-term.
• Attempting to solve a constructed differential equation from G6 using G-section techniques, when solving belongs to Section H.

Self-test

1. What are the two interpretations of a derivative at a point?
2. What extra factor appears when differentiating aᵏˣ compared with eᵏˣ, and why?
3. How do you classify a stationary point using the second derivative?
4. What is the key difference between implicit and parametric differentiation at this level in terms of what’s required?
5. Name the three modelling contexts the specification explicitly names for constructing differential equations.

Answers: 1. The gradient of the tangent to the curve at that point, and the instantaneous rate of change. 2. A factor of ln a, because aᵏˣ = e^(kx ln a), so the chain rule introduces ln a when differentiating. 3. f’‘(x) > 0 indicates a minimum, f’‘(x) < 0 indicates a maximum, and f’‘(x) = 0 requires further investigation (e.g. checking the sign of f’ on either side). 4. Both use the chain rule, but only the first derivative is required for implicit and parametric relations at this level — second derivatives are not expected. 5. Kinematics, population growth, and the relationship between price and demand.