AQA A-level Mathematics 7357, section G (Differentiation), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: algebraic manipulation and function notation from section B. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
A car's distance-time curve keeps bending, so no single "speed" fits the whole journey — yet the speedometer shows one number now. What does that number mean geometrically?
Start with a prediction. State how a tangent gradient differs from an average speed between two times.
Build the model
The derivative 导数 is the limit of the difference quotient [f(x+h) − f(x)]/h as h → 0: the tangent gradient and instantaneous rate. For y = axⁿ, dy/dx = anxⁿ⁻¹; trigonometric and exponential forms follow: (sin x)′ = cos x, (cos x)′ = −sin x, (eˣ)′ = eˣ, (ln x)′ = 1/x, in radians.
A stationary point 驻点 has dy/dx = 0; classify it by the sign change of the derivative or the sign of the second derivative 二阶导数 (positive: local minimum; negative: local maximum). The chain, product and quotient rules handle composite forms, with every inner-rate factor retained.
derivative: The instantaneous rate of change, the tangent gradient; stationary point: A point where the derivative is zero; second derivative: The derivative of the derivative, measuring how the gradient changes.
Choose evidence that can test it
A zero derivative is only a candidate: y = x³ is stationary at 0 yet keeps rising, so classification needs the sign change or f″. On a restricted interval, compare boundary values too — the global extreme can sit at an endpoint.
After differentiating a product such as xeˣ, check at an easy input: the product rule gives eˣ + xeˣ, which at x = 0 is 1, not 0. Multiplying the separate derivatives would give the wrong value there.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: y = x³ − 3x, so dy/dx = 3x² − 3. Stationary where 3x² − 3 = 0: x = ±1. The second derivative is 6x: at x = 1 it is 6 > 0, so (1, −2) is a local minimum; at x = −1 it is −6 < 0, so (−1, 2) is a local maximum.
Example:
For y = x³ − 3x at x = 2, find the tangent and normal equations using the same sequence: point from f → gradient from f′ → line equations. (Answer: y = 2 at x = 2 and the gradient is 9, so the tangent is y − 2 = 9(x − 2) and the normal is y − 2 = −(1/9)(x − 2).)
Check the conclusion and its limits
The derivative supplies the gradient, not the y-coordinate — the point always comes from the original curve. A horizontal tangent (gradient 0) has a vertical normal, stated as x = constant, since the negative reciprocal of zero does not exist.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
Every point with zero derivative is a local maximum or minimum. This claim is false: y = x³ has dy/dx = 0 at x = 0 but the derivative 3x² stays positive on both sides — a stationary point of inflection, with no extremum at all.
Key:
Differentiation: the derivative is the limit of the secant gradient, giving the tangent gradient and rate. Stationary candidates come from f′ = 0 and are classified by the sign change or f″; tangents and normals use the point from f and the gradient from f′.
