Mini-Lesson

Differentiation

This mini-lesson covers differentiation from first principles, the standard derivatives, the chain, product and quotient rules, implicit and parametric differentiation, stationary points (with the second derivative test), tangents and normals and connected rates of change.

Where this sits: Unit AS 1: Pure Mathematics — differentiation from first principles, differentiating powers of x, gradients, tangents and normals, stationary points. Unit A2 1 — chain/product/quotient rules, trig, exponential and log derivatives, implicit and parametric differentiation, rates of change.

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.

Foundations

Differentiation from first principles

f′(x) = lim(h→0) [f(x + h) − f(x)] / hthe gradient of the chord becomes the gradient of the tangent as h → 0

Worked example — f(x) = x²

[f(x + h) − f(x)]/h = [(x + h)² − x²]/h = [x² + 2xh + h² − x²]/h

= (2xh + h²)/h = 2x + h

As h → 0 this tends to 2x. So f′(x) = 2x. ∎

Standard results: d/dx(xⁿ) = nxⁿ⁻¹ · d/dx(eᵏˣ) = keᵏˣ · d/dx(ln x) = 1/x · d/dx(sin x) = cos x · d/dx(cos x) = −sin x · d/dx(tan x) = sec²x. Trig derivatives require radians.

Calculate

Gradient of a cubic

1y = x³ − 4x. Find the gradient of the curve at x = 2.

Hint: dy/dx = 3x² − 4. At x = 2: 3(4) − 4 = 12 − 4.

Rules

Chain, product and quotient

chain: dy/dx = dy/du × du/dxproduct: (uv)′ = u′v + uv′ · quotient: (u/v)′ = (u′v − uv′)/v²

Chain rule — y = (3x² + 1)⁴

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 × 4³ = 24 × 64 = 1536

Quotient rule — y = (2x + 1)/(x − 3)

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

Calculate

Chain rule

2y = (3x² + 1)⁴. Find dy/dx when x = 1.

Hint: dy/dx = 4(3x² + 1)³ × 6x = 24x(3x² + 1)³. At x = 1: 24 × 1 × 4³ = 24 × 64.

Calculate

Quotient rule

3y = (2x + 1)/(x − 3). Find dy/dx when x = 5.

Hint: dy/dx = [2(x − 3) − (2x + 1)]/(x − 3)² = −7/(x − 3)². At x = 5: −7/4 = −1.75.

Calculate

Product rule

4y = x²eˣ. Find dy/dx when x = 1, to 3 significant figures.

Hint: dy/dx = 2xeˣ + x²eˣ = eˣ(x² + 2x). At x = 1: e(1 + 2) = 3e = 8.1548…

Applications

Stationary points and the second derivative

At a stationary point, dy/dx = 0. Classify it with the second derivative:

d²y/dx² > 0 → minimum · d²y/dx² < 0 → maximum · = 0 → inconclusive, test either sideconvex (curve bends upward) when f″ > 0 · concave when f″ < 0 · point of inflection where f″ changes sign

Worked example — y = x³ − 6x² + 9x + 2

dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3

d²y/dx² = 6x − 12. At x = 3 this is +6 > 0 ⇒ minimum. At x = 1 it is −6 < 0 ⇒ maximum.

Minimum value: y(3) = 27 − 54 + 27 + 2 = 2

Calculate

Minimum value

5The curve y = x³ − 6x² + 9x + 2 has a local minimum. Find the y-value at that minimum.

Hint: dy/dx = 3x² − 12x + 9 = 0 at x = 1 and x = 3. d²y/dx² = 6x − 12 is positive at x = 3, so that is the minimum: y = 27 − 54 + 27 + 2 = 2.

A2 skill

Implicit and parametric differentiation

Implicit — x² + y² = 25

Differentiate every term with respect to x, using the chain rule on y terms:

2x + 2y (dy/dx) = 0 ⇒ dy/dx = −x/y

At (3, 4): dy/dx = −3/4 — which matches "tangent ⟂ radius" from coordinate geometry ✓

Parametric — x = t², y = t³

dy/dx = (dy/dt) ÷ (dx/dt) = 3t² ÷ 2t = 3t/2

For a product of x and y (like xy or x²y) you need the product rule: d/dx(xy) = y + x(dy/dx).

Quick check

Which rule?

?Which rule differentiates y = x sin x?

Sort it

Which rule do you need?

Tap a function, then tap the differentiation rule it needs.

✖️ Product rule

➗ Quotient rule

🔗 Chain rule

Applications

Tangents, normals and connected rates

Tangent at (x₁, y₁): gradient m = dy/dx there, so y − y₁ = m(x − x₁). The normal is perpendicular: gradient −1/m.

dy/dt = dy/dx × dx/dtthe chain rule connects rates: use it whenever two quantities change with time

Worked example — connected rates

A sphere's radius grows at dr/dt = 0.2 cm/s. How fast is the volume growing when r = 5 cm?

V = (4/3)πr³ ⇒ dV/dr = 4πr²

dV/dt = dV/dr × dr/dt = 4π(25)(0.2) = 20π ≈ 62.8 cm³/s

Quick check

Second derivative test

?At a stationary point, f″(x) = −4. What is it?

Match it

Match the derivative

Tap the function on the left, then its derivative on the right.

Statement

Answer

Quick check

Implicit differentiation

?Differentiating y² with respect to x gives:

Quick check

Increasing function

?For which values of x is y = x² − 6x + 5 an increasing function?

Examiner traps

Calculus pitfalls (differentiation)

Multiplying derivatives. (uv)′ ≠ u′v′. Use the product rule.
Quotient rule order. The numerator is u′v − uv′ — the wrong way round flips the sign.
Implicit differentiation: d/dx(y²) = 2y·dy/dx. The dy/dx factor is compulsory.
f″ = 0 is inconclusive — it does not prove a point of inflection; check the sign of f″ either side.
Trig derivatives require radians. d/dx(sin x) = cos x only in radians.
Normal, not tangent: gradient of the normal = −1/(dy/dx).

Recap

The big ideas to know

First principles: f′(x) = lim(h→0)[f(x+h) − f(x)]/h

Standard: xⁿ → nxⁿ⁻¹ · eᵏˣ → keᵏˣ · ln x → 1/x · sin → cos · cos → −sin · tan → sec²x

Rules: chain (function inside function) · product u′v + uv′ · quotient (u′v − uv′)/v²

Implicit: d/dx(y²) = 2y·dy/dx · Parametric: dy/dx = (dy/dt)/(dx/dt)

Stationary points: f′ = 0, then f″ > 0 min, f″ < 0 max, f″ = 0 inconclusive

Rates: dV/dt = dV/dr × dr/dt

Differentiation is half of calculus — integration reverses it, and mechanics uses both. Press Finish to see your score.

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