This mini-lesson covers AQA section G — Differentiation: differentiation from first principles, the chain, product and quotient rules, the derivatives of sin, cos, tan, ex and ln x, implicit and parametric differentiation, stationary points and their nature, convex and concave curves and points of inflection, and connected rates of change.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
The derivative is the limit of a gradient of a chord as the chord shrinks to a point:
f(x + h) − f(x) = (x + h)³ − x³ = x³ + 3x²h + 3xh² + h³ − x³ = 3x²h + 3xh² + h³.
Divide by h: [f(x + h) − f(x)]/h = 3x² + 3xh + h².
Now let h → 0: the terms with h vanish, leaving f′(x) = 3x².
So at x = 2, the gradient is 3(2)² = 12.
You must divide by h before letting h → 0. Setting h = 0 first gives 0/0, which is meaningless. Write "lim as h → 0" at every line — the notation carries marks.
Learn these — they are the vocabulary of the whole topic.
Why radians? The proof of d/dx(sin x) = cos x from first principles relies on sin h ≈ h for small h — which only holds in radians. In degrees the derivative picks up a stray factor of π/180.
For a function of a function, differentiate the outside and multiply by the derivative of the inside.
y = (3x + 1)⁵: outside gives 5(3x + 1)⁴, inside gives 3. So dy/dx = 15(3x + 1)⁴.
y = ex²: dy/dx = ex² × 2x = 2x ex².
y = ln(3x + 4): dy/dx = 1/(3x + 4) × 3 = 3/(3x + 4).
The forgotten factor. Writing 5(3x + 1)⁴ and stopping is the commonest error in the whole of A-level calculus. Always ask: "what is the derivative of the inside?"
Tap a function, then tap the rule that differentiates it most efficiently.
Two more rules, for products and for fractions.
y = x² ex. Take u = x² (u′ = 2x) and v = ex (v′ = ex).
dy/dx = 2x ex + x² ex = ex(x² + 2x).
At x = 1: e(1 + 2) = 3e = 8.15 (3 s.f.).
y = (2x + 1)/(x − 3). Take u = 2x + 1 (u′ = 2), v = x − 3 (v′ = 1).
dy/dx = [2(x − 3) − (2x + 1)(1)] / (x − 3)² = (2x − 6 − 2x − 1)/(x − 3)² = −7/(x − 3)².
At x = 5: −7/(2)² = −7/4 = −1.75.
The quotient rule is not symmetric. u′v − uv′ (numerator derivative first) — swapping the order flips the sign of every answer.
When y is not given explicitly as a function of x (for example x² + y² = 25), differentiate every term with respect to x and apply the chain rule to any term in y.
x² + y² = 25. Differentiate term by term: 2x + 2y · dy/dx = 0.
Rearranging: dy/dx = −x/y.
At the point (3, 4): dy/dx = −3/4 = −0.75.
Sense check: the radius to (3, 4) has gradient 4/3, and the tangent must be perpendicular to it — and indeed (4/3) × (−3/4) = −1 ✓
Products of x and y need the product rule. d/dx(xy) = 1·y + x·dy/dx = y + x dy/dx. Then collect all the dy/dx terms on one side and factorise.
If x and y are both given in terms of a parameter t, chain the derivatives together:
x = t², y = t³ − 3t.
dx/dt = 2t · dy/dt = 3t² − 3.
dy/dx = (3t² − 3)/(2t).
At t = 2: (3 × 4 − 3)/(2 × 2) = 9/4 = 2.25.
Stationary points on a parametric curve occur where dy/dt = 0 (provided dx/dt ≠ 0). Here 3t² − 3 = 0 ⇒ t = ±1 — you find them from the parameter, not from x.
A stationary point is where f′(x) = 0 — the tangent is horizontal. The second derivative classifies it:
f″(a) = 0 does not mean inflection. For y = x⁴ at x = 0, f″(0) = 0 yet it is a clear minimum. The definition of an inflection requires a change of sign in f″, not merely a zero.
Tap an item on the left, then its partner on the right.
When several quantities change together, link their rates with the chain rule:
A spherical balloon has volume V = (4/3)πr³ and is inflated so that dV/dt = 20 cm³ s⁻¹. Find dr/dt when r = 5 cm.
dV/dr = 4πr² = 4π(25) = 100π.
dr/dt = (dV/dt) ÷ (dV/dr) = 20/(100π) = 0.0637 cm s⁻¹ (3 s.f.).
Method: write down every rate you are given with its correct symbol, write down the rate you want, and find the chain of derivatives connecting them. Half the marks are for setting it up.
First principles: f′(x) = lim_{h→0} [f(x + h) − f(x)]/h — divide by h before letting h → 0
Standard results: sin → cos · cos → −sin · tan → sec²x · e^x → e^x · ln x → 1/x
Chain rule: dy/dx = dy/du × du/dx — never forget the derivative of the inside
Product / quotient: (uv)′ = u′v + uv′ · (u/v)′ = (u′v − uv′)/v²
Implicit: d/dx(y²) = 2y dy/dx — then collect and factorise
Parametric: dy/dx = (dy/dt) ÷ (dx/dt)
Stationary points: f′ = 0; f″ > 0 min (convex), f″ < 0 max (concave), f″ = 0 inconclusive
Rates: chain the derivatives — dV/dt = dV/dr × dr/dt
That is AQA section G — Differentiation. Press Finish to see your score.
You have worked through Differentiation for AQA A-level Mathematics. 🎉
Your stars: 0 / 0
Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.