Topic 7 starts from first principles and builds the whole toolkit: the chain, product and quotient rules, the derivatives of eˣ, ln x and the trigonometric functions, implicit and parametric differentiation, connected rates of change, stationary points, and convex / concave curves and points of inflection.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
The derivative is the limit of the gradient of a chord as the chord shrinks to a point. This is a proof (Topic 1) as much as a calculation.
[f(x+h) − f(x)]/h = [(x + h)³ − x³]/h
= [x³ + 3x²h + 3xh² + h³ − x³]/h = [3x²h + 3xh² + h³]/h
= 3x² + 3xh + h². As h → 0 this tends to 3x². ∎
h = 0.1: chord gradient = (2.1³ − 2³)/0.1 = (9.261 − 8)/0.1 = 12.61
h = 0.01: (2.01³ − 8)/0.01 = 12.0601. h = 0.001: 12.006. The limit is f′(2) = 3(2²) = 12. ✓
Exam trap: you must keep the lim (h→0) notation on every line until you take the limit, and you may only cancel the h after factorising. Setting h = 0 too early gives 0/0.
f(2.1) = 2.1³ = 9.261, f(2) = 8.
Gradient = (9.261 − 8) ÷ 0.1 = 1.261 ÷ 0.1 = 12.61.
Compare with the true derivative f′(2) = 3(2)² = 12 — the chord is close, and gets closer as h shrinks.
Let u = 3x² − 1, so y = u⁴. dy/du = 4u³, du/dx = 6x.
dy/dx = 4(3x² − 1)³ × 6x = 24x(3x² − 1)³. At x = 1: 24 × 1 × (2)³ = 24 × 8 = 192.
u = x² (u′ = 2x), v = e^(3x) (v′ = 3e^(3x)).
dy/dx = 2x e^(3x) + 3x² e^(3x) = e^(3x)(2x + 3x²). At x = 1: e³(2 + 3) = 5e³ ≈ 100.43.
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: −7/4 = −1.75.
Exam trap: the quotient rule numerator is u′v minus uv′ — the order matters, unlike the product rule. Get it backwards and every sign flips.
dy/dx = 24x(3x² − 1)³.
At x = 1: 24 × 1 × (3 − 1)³ = 24 × 8 = 192.
dy/dx = e^(3x)(2x + 3x²). At x = 1: e³(2 + 3) = 5e³.
5 × 20.0855 = 100.4277 → 100.4 (1 d.p.)
Numerator: 2(x − 3) − (2x + 1)(1) = 2x − 6 − 2x − 1 = −7.
dy/dx = −7/(x − 3)². At x = 5: −7/2² = −7/4 = −1.75.
Tap a function, then the rule you would use to differentiate it.
These must be instant. All angles in radians.
Exam trap: d/dx (sin x) = cos x is only true in radians. In degrees the derivative would carry an ugly π/180 factor — which is exactly why A-level calculus abandons degrees.
When the curve is not in the form y = f(x), differentiate every term with respect to x, remembering that y is a function of x — so d/dx (y²) = 2y dy/dx (chain rule), and products like 3xy need the product rule.
Check the point is on the curve: 1 + 3(1)(2) + 4 = 1 + 6 + 4 = 11 ✓
Differentiate: 2x + 3y + 3x (dy/dx) + 2y (dy/dx) = 0
Collect: (dy/dx)(3x + 2y) = −(2x + 3y)
dy/dx = −(2x + 3y)/(3x + 2y). At (1, 2): −(2 + 6)/(3 + 4) = −8/7 ≈ −1.14.
Exam trap: d/dx(3xy) = 3y + 3x dy/dx — it is a product. Writing just 3x dy/dx loses the mark and wrecks the answer.
(dy/dx)(3x + 2y) = −(2x + 3y)
At (1, 2): dy/dx = −(2 + 6)/(3 + 4) = −8/7 = −1.1428…
If x and y are both given in terms of t, chain them together:
dx/dt = 2, dy/dt = 2t.
dy/dx = 2t/2 = t. At t = 5 (i.e. the point (10, 22)) the gradient is 5.
Check with the Cartesian form y = x²/4 − 3: dy/dx = x/2 = 10/2 = 5 ✓
Exam trap: the answer to a parametric gradient question is usually left in terms of t. Only substitute a number if you are given the point or the value of t.
At a stationary point dy/dx = 0. The second derivative tells you which kind:
dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3.
d²y/dx² = 6x − 12. At x = 1: −6 < 0 → maximum. At x = 3: +6 > 0 → minimum.
y-values: at x = 1, y = 1 − 6 + 9 + 2 = 6; at x = 3, y = 27 − 54 + 27 + 2 = 2.
Inflection where 6x − 12 = 0 ⇒ x = 2 (and the concavity really does change there).
Exam trap: d²y/dx² = 0 does not automatically mean a point of inflection — the concavity must actually change sign. For y = x⁴ at x = 0 the second derivative is 0 but the point is a minimum.
Chain the rates together — write down what you know and what you want, then bridge with the chain rule.
V = (4/3)πr³ and the radius grows at dr/dt = 0.2 cm s⁻¹. Find dV/dt when r = 5 cm.
dV/dr = 4πr² = 4π(25) = 100π ≈ 314.159
dV/dt = 100π × 0.2 = 20π = 62.8 cm³ s⁻¹ (1 d.p.)
Exam trap: quote units, and never substitute r = 5 before differentiating — you would be differentiating a constant and get zero.
dV/dr = 4πr² = 4π × 25 = 100π = 314.159…
dV/dt = 314.159 × 0.2 = 62.83 = 62.8 cm³ s⁻¹ (1 d.p.)
Tap a function on the left, then its derivative.
Link: differentiation from first principles is a Topic 1 proof; parametric differentiation needs Topic 3; and every rate-of-change model reappears as a differential equation in Topic 8.
First principles: f′(x) = lim(h→0) [f(x+h) − f(x)]/h — keep the limit notation
Rules: chain: dy/du × du/dx · product: u′v + uv′ · quotient: (u′v − uv′)/v²
Standard: eˣ → eˣ · ln x → 1/x · sin x → cos x · cos x → −sin x · tan x → sec²x
Implicit: d/dx(y²) = 2y dy/dx; d/dx(xy) = y + x dy/dx
Parametric: dy/dx = (dy/dt) ÷ (dx/dt)
Stationary points: dy/dx = 0, then use the sign of d²y/dx²; rates chain together
Topic 7 is the largest source of method marks on Paper 1 — the set-up is worth more than the arithmetic. Press Finish to see your score.
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