Lesson 15: Differentiation (A-level Intro)

Differentiation is the mathematics of change — how fast is something growing, falling, accelerating? It answers questions like: at what speed is a reservoir filling in Derbyshire? When is a business's profit at its peak? This lesson introduces the derivative, the power rule, gradients of curves, tangents and normals, and stationary points — the first chapter of calculus, and the single most important bridge from GCSE to A-level mathematics.

Rates of change and the gradient of a curve

On a straight line, the gradient is constant: rise ÷ run. On a curve, the gradient varies — a cyclist climbing the Kirkstone Pass in the Lake District finds the slope steepening and easing. The derivative dy/dx measures the gradient at a point: the rate of change of y with respect to x there. Geometrically it is the gradient of the tangent — the line that just touches the curve at that point.

Approximating from first principles: the gradient between two nearby points approaches the derivative as the points merge. For y = x² at x = 3: at x = 3.001, y = 9.006001; gradient ≈ (9.006001 − 9)/0.001 = 6.001 ≈ 6. The exact derivative of x² is 2x, giving 2 × 3 = 6 at that point. That limit process — the gradient of an ever-shorter chord — is what 'differentiation from first principles' means, and Higher-tier students should be able to narrate it.

Units matter: if s is distance (m) and t is time (s), ds/dt is speed (m/s). If V is volume (litres) and t is time (minutes), dV/dt is a filling rate (litres/min). A derivative always carries 'per unit' units — state them in context answers.

The power rule and differentiating polynomials

The rule: d/dx (xⁿ) = n·xⁿ⁻¹ — bring the power down, reduce the power by 1. Constants vanish (a constant doesn't change), and terms differentiate separately:

d/dx (x³) = 3x²        d/dx (x²) = 2x        d/dx (x) = 1        d/dx (7) = 0
d/dx (3x² − 5x + 2) = 6x − 5
d/dx (2x⁴ + x³ − 4x) = 8x³ + 3x² − 4

Worked example: find the gradient of y = x³ − 4x at x = 2. dy/dx = 3x² − 4, so at x = 2: 3(4) − 4 = 8. The tangent there is steep and rising. At x = 0 the gradient is −4 (falling); the gradient changes sign where 3x² − 4 = 0, i.e. x = ±2/√3 ≈ ±1.15 — previewing stationary points below.

Roots and fractions first need rewriting as powers: √x = x^(1/2), so d/dx (√x) = ½x^(−1/2) = 1/(2√x). And 1/x² = x^(−2), so d/dx (1/x²) = −2x^(−3) = −2/x³. Worked example: y = 4√x + 3/x = 4x^(1/2) + 3x^(−1). dy/dx = 2x^(−1/2) − 3x^(−2) = 2/√x − 3/x². At x = 4: 2/2 − 3/16 = 1 − 0.1875 = 0.8125. Rewrite-then-differentiate is the routine move for anything that isn't a plain polynomial (A-level lives on it).

Tangents and normals

The tangent at a point has the curve's gradient there; the normal is perpendicular to it (gradient = negative reciprocal: if the tangent gradient is m, the normal's is −1/m).

Worked example: find the tangent and normal to y = 2x² − 3x + 1 at x = 2. The point: y = 8 − 6 + 1 = 3, so (2, 3). dy/dx = 4x − 3 → at x = 2, gradient = 5. Tangent: y − 3 = 5(x − 2) → y = 5x − 7. Normal gradient = −1/5: y − 3 = −(1/5)(x − 2) → multiply by 5: 5y − 15 = −x + 2 → 5y = −x + 17 → y = (−x + 17)/5. Check the tangent at x = 2: 5(2) − 7 = 3. ✓ Tangents model local straight-line behaviour — engineers use them to approximate curves near an operating point.

Worked example: find where the tangent to y = x² − 6x + 10 is horizontal. Horizontal means gradient 0: dy/dx = 2x − 6 = 0 → x = 3, where y = 9 − 18 + 10 = 1. The point (3, 1) is the vertex — the parabola's minimum. Tangents being horizontal is exactly the stationary-point condition of the next section.

Stationary points — maxima, minima, optimisation

A stationary point is where dy/dx = 0. Classify it with the second derivative d²y/dx²: positive → minimum (the curve is ∪-shaped there); negative → maximum (∩-shaped). (If d²y/dx² = 0, further tests are needed — A-level territory.)

Worked example: y = 2x³ − 3x² − 12x + 5. dy/dx = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1). So stationary points at x = 2 and x = −1. Values: at x = −1, y = −2 − 3 + 12 + 5 = 12; at x = 2, y = 16 − 12 − 24 + 5 = −15. Second derivative d²y/dx² = 12x − 6: at x = −1, −18 < 0 → maximum (−1, 12); at x = 2, 18 > 0 → minimum (2, −15). Complete answer: a local maximum of 12 at x = −1 and a local minimum of −15 at x = 2.

Optimisation is stationary points in disguise. Worked example: a farmer has 100 m of fencing for a rectangular pen built against a long wall (fence on three sides only). If the two perpendicular sides have length x, the parallel side is 100 − 2x and the area A = x(100 − 2x) = 100x − 2x². dA/dx = 100 − 4x = 0 → x = 25 m. Then the parallel side is 100 − 50 = 50 m, and A = 25 × 50 = 1,250 m². Second derivative −4 < 0 confirms a maximum. This is identical mathematics to the Devon farmer of Lesson 4 — differentiation is the general machine that replaces completing the square.

Worked example: a company models profit (£ thousands) on x thousand units sold as P = −x² + 30x − 100. dP/dx = −2x + 30 = 0 → x = 15 (thousand units), where P = −225 + 450 − 100 = £125 thousand — the maximum profit. Break-even points solve P = 0. Calculus turns 'how much should we make?' into a solvable equation.

Worked example — velocity and acceleration

A particle moves with displacement s = t³ − 6t² + 9t metres (t in seconds). Velocity is the derivative: v = ds/dt = 3t² − 12t + 9. Acceleration is the derivative of velocity: a = dv/dt = 6t − 12. At t = 1: s = 1 − 6 + 9 = 4 m; v = 3 − 12 + 9 = 0 m/s (the particle is instantaneously at rest — a turning point of the motion); a = −6 m/s². At t = 3: s = 27 − 54 + 27 = 0 m (back at the start); v = 27 − 36 + 9 = 0 again. When is the particle at rest? 3t² − 12t + 9 = 0 → t² − 4t + 3 = 0 → t = 1 or t = 3. Differentiation turns one formula into the whole story of the motion — position, stillness, and acceleration — which is why mechanics (Lesson 18) is calculus with units.

Second derivatives classify in motion too: a = 0 at t = 2 is where the velocity is most negative (v = 12 − 24 + 9 = −3 m/s at t = 2 — the fastest return speed). Reading meaning out of derivatives, not just computing them, is what separates A-level readiness from rule-following. Quick check habit: after differentiating, substitute a second point and compare with a numerical gradient estimate (rise over a tiny run on the original function). If 6x − 5 at x = 3 says 13, then checking (f(3.001) − f(3))/0.001 ≈ 13 on the original confirms the algebra — thirty seconds that rescue whole questions.

Common mistakes

• Dropping the power correctly. d/dx (x⁴) = 4x³ — reduce the exponent by exactly 1, and multiply by the old power.
• Differentiating a constant to 1. Constants vanish: d/dx (9) = 0. (But d/dx (9x) = 9 — the x matters.)
• Forgetting to find y. A stationary point is a point: report (x, y), not just x.
• Classifying with the first derivative's value. The second derivative's sign classifies; plugging x into dy/dx gives 0 by construction.
• Normal gradient errors. Perpendicular means negative reciprocal: m = 5 → normal −1/5, not −5.
• Differentiating products term-wise wrongly. (x + 1)(x + 2) must be expanded to x² + 3x + 2 before differentiating at GCSE level: 2x + 3.

Exam Experience

Question 1. y = 3x² − 5x + 2. (a) Find dy/dx. (b) Work out the gradient of the curve at x = 3. [2 marks]

Question 2. A curve has equation y = x³ − 3x² − 9x + 2. Find the coordinates of the two stationary points and determine the nature of each. [3 marks]

Question 3. Show, by differentiation from first principles, that the derivative of x² + 3x is 2x + 3. [3 marks]

Mark scheme.

• Q1: B1 for (a) dy/dx = 6x − 5. B1 for (b) 6 × 3 − 5 = 13.
• Q2: M1 for dy/dx = 3x² − 6x − 9 = 0 and solving (divide by 3: x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0) giving x = 3, x = −1. M1 for finding both y-values: at x = −1, y = −1 − 3 + 9 + 2 = 7; at x = 3, y = 27 − 27 − 27 + 2 = −25. A1 for classification with justification: d²y/dx² = 6x − 6 gives −12 at x = −1 (maximum, (−1, 7)) and +12 at x = 3 (minimum, (3, −25)).
• Q3: M1 for the difference quotient: [(x + h)² + 3(x + h) − (x² + 3x)] / h. M1 for expanding and simplifying: [2xh + h² + 3h] / h = 2x + h + 3. A1 for taking the limit as h → 0, giving 2x + 3 as required.