Lesson 10: Differentiation Rules

The derivative f′(x) = lim(h→0) [f(x + h) − f(x)]/h measures instantaneous rate of change and gives the gradient of the tangent. Nobody computes that limit from scratch in an exam except when a question explicitly demands it (Lesson 9's extended problem) — instead you deploy the rules, which are the limit pre-computed for whole families (AA Topic 5). This lesson is about deploying them flawlessly: the right rule for the structure in front of you, with the chain rule — the most used and most dropped rule in the course — at the centre.

The basic rules and the standard derivatives

Power rule: d/dx [xⁿ] = nxⁿ⁻¹, for any real n. Linearity: derivatives pass through sums and constant multiples. Standard results (formula booklet): d/dx [e^x] = e^x; d/dx [ln x] = 1/x (x > 0); d/dx [sin x] = cos x; d/dx [cos x] = −sin x; d/dx [tan x] = 1/cos²x. Read structure before differentiating: is it a sum (term-by-term), a product (product rule), a quotient (quotient rule), or a function of a function (chain rule)? Most expressions are two of these at once.

Chain, product, quotient

Chain rule: d/dx [f(g(x))] = f′(g(x)) × g′(x) — differentiate the outside, keep the inside, multiply by the derivative of the inside. Product rule: (uv)′ = u′v + uv′. Quotient rule: (u/v)′ = (u′v − uv′)/v². Second derivatives f″(x) are just the process repeated; HL adds implicit differentiation (differentiate both sides with respect to x, treating y as a function of x, so d/dx [y²] = 2y y′) and derivatives of inverse trig functions — the machinery Lesson 12's proofs and Lesson 25's differential equations will assume.

Worked Example 1 — chain rule, twice over. Differentiate (a) y = (3x² + 1)⁵; (b) y = e^(sin x); (c) y = ln(x² + 4).

Solution. (a) Outside: fifth power; inside: 3x² + 1 with derivative 6x. y′ = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴. (b) Outside: exponential (its own derivative); inside: sin x. y′ = e^(sin x) × cos x. (c) Outside: ln (derivative 1/(inside)); inside derivative 2x. y′ = 2x/(x² + 4). In every case the inside's derivative appears as a factor — if your answer lacks it, you have not finished the chain rule. Examiners plant a "derivative of the inside" mark in almost every chain-rule question; it is the most frequently lost single mark in Topic 5.

Worked Example 2 — product and quotient (past-paper style). Find f′(x) for (a) f(x) = x² e^x; (b) f(x) = ln x / x, and evaluate f′(1) for (b).

Solution. (a) u = x², v = e^x: f′ = 2x e^x + x² e^x = e^x(2x + x²) — factorising e^x is expected presentation. (b) u = ln x (u′ = 1/x), v = x (v′ = 1): f′ = [(1/x)(x) − ln x (1)]/x² = (1 − ln x)/x². Then f′(1) = (1 − 0)/1 = 1. Quotient rule sign order (u′v − uv′) is the trap; reversing it negates the whole answer.

Worked Example 3 — where is the gradient zero? Let f(x) = x³ − 3x² − 9x + 2. Find the x-coordinates where the tangent is horizontal.

Solution. f′(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1). Setting f′ = 0: x = 3 or x = −1. Horizontal tangents at x = −1 and x = 3 — Lesson 11 will classify these points (maximum and minimum respectively, since f″(−1) = −12 < 0 and f″(3) = 12 > 0) and turn them into optimisation answers. Note the workflow: differentiate, factor, solve — the derivative questions in Section B are usually algebra questions wearing a calculus coat.

By hand and by GDC

Paper 1: rules by hand, exact values. Paper 2: the GDC's numeric derivative (nDeriv) and graph of f′ verify your work, and questions may hand you the *graph* of f′ and ask about f — the derivative as an object, not just a computation. A reliable habit in both papers: differentiate, then sanity-check the sign of f′ against the graph's slope at one point.

Real-world modelling: marginal change

Economists define marginal cost as the derivative of the cost function: if C(x) = 0.02x² + 30x + 5000 dollars for x units, then C′(x) = 0.04x + 30 is the cost of "one more unit" at production level x — at x = 1000, about $70 per unit, a number managers actually use. Population growth rates, reaction rates in chemistry, and velocity from position (Lesson 14) are the same derivative wearing different clothes.

Common mistakes

• The dropped inner derivative in the chain rule — see above; it is error number one in every examiner report on Topic 5.
• Quotient rule reversed (uv′ − u′v on top), or the denominator left unsquared.
• Differentiating e^x as x·e^(x−1) (power rule on an exponential base) or x² whose derivative is written 2x¹ and left unsimplified in a chain — presentation matters for follow-through marks.
• d/dx [ln(5x)] answered 1/(5x) — the chain rule gives 5/(5x) = 1/x; conversely d/dx [ln(x²)] = 2/x, not 1/x².
• sin x differentiated to −cos x (that is cos x's derivative, with the sign pattern sin → cos → −sin → −cos).
• Forgetting the second pass: f″ requires differentiating f′, and "gradient is zero" questions are solved by f′(x) = 0, not f(x) = 0.

IB Exam Experience

Question 1 [2 marks]. Differentiate f(x) = 4x³ − 2/x + e^x. *Marking notes:* Rewrite 2/x as 2x⁻¹ first (M1 if needed as a step); A1 for f′(x) = 12x² + 2x⁻² + e^x, i.e. 12x² + 2/x² + e^x. Negative powers trip more candidates than any rule.

Question 2 [3 marks]. Let g(x) = (2x − 1)⁴. Find g′(x) and g′(1). *Marking notes:* M1 for the chain structure 4(2x − 1)³; A1 for the inner factor ×2: g′(x) = 8(2x − 1)³; A1 for g′(1) = 8. The classic wrong answer 4(2x − 1)³, g′(1) = 4 appears in every session — the inner-derivative mark is designed to catch it.

Question 3 [4 marks]. Let h(x) = sin x · e^(2x). Find h′(x) and the exact value of h′(0). *Marking notes:* M1 product rule; M1 chain rule on e^(2x) (factor 2); A1: h′(x) = e^(2x)(cos x + 2 sin x); A1: h′(0) = 1 × (1 + 0) = 1.

Extended problem (Paper 2 style) [9 marks]. Let f(x) = ln(x² + 1). (a) Find f′(x). [3] (b) Find the coordinates of the point(s) where the tangent to the graph of f is horizontal, and describe the behaviour of f there. [3] (c) The tangent at x = 1 is drawn. Find its equation. [3] *Marking notes:* (a) M1 chain rule structure (derivative of ln u is u′/u); A1: f′(x) = 2x/(x² + 1); A1 for correct domain remark if made (defined for all real x). (b) M1: horizontal tangent ⟺ f′(x) = 0 ⟺ 2x = 0; A1: point (0, 0); A1: f decreases for x < 0 and increases for x > 0 (sign of f′ either side), so (0, 0) is a minimum — accept GDC graph reasoning on Paper 2. (c) M1: f(1) = ln 2, f′(1) = 1; M1: point–gradient form y − ln 2 = 1(x − 1); A1: y = x − 1 + ln 2 (≈ x − 0.307 to 3 s.f.). Note how the question chains rule → application → equation: that is Section B's signature move.

Exploration (IA) idea: model a real cost or revenue dataset and interpret its derivative as marginal change; or track a bouncing ball / swinging pendulum and examine position–velocity–acceleration as iterated derivatives of measured data.

Going deeper: derivatives of inverse and reciprocal structures

HL stretches the rule kit: derivatives of arcsin, arccos, arctan (booklet: 1/√(1 − x²), −1/√(1 − x²), 1/(1 + x²)) and — more importantly for everyone — implicit differentiation, where y is trapped inside an equation and dy/dx is extracted by differentiating through, chain rule attached to every y-term.

Extra worked example. The curve x² + y² = 25 (a circle) has y as an implicit function of x. Find dy/dx and the tangent at (3, 4).

Solution. Differentiate both sides with respect to x: 2x + 2y·(dy/dx) = 0 — the factor dy/dx appears because y depends on x (chain rule). Solve: dy/dx = −x/y. At (3, 4): gradient −3/4, tangent y − 4 = −(3/4)(x − 3). Elegant check: the radius to (3, 4) has gradient 4/3; the tangent's gradient is its negative reciprocal ✓ — geometry confirming calculus.

Extra exam practice [4 marks]. Let y = e^(x²) × ln x for x > 0. Find dy/dx in factored form. *Marking notes:* M1 for product rule with chain on e^(x²) (factor 2x); M1 for derivative of ln x = 1/x; A1: dy/dx = 2x e^(x²) ln x + e^(x²)/x; A1 for factored form e^(x²)(2x ln x + 1/x) — presentation marks reward the clean factorisation, and at x = 1 this gives dy/dx = e, a free plausibility check.

Going further: layered rules and the second-derivative test

Real exam derivatives arrive layered: a quotient whose numerator needs the chain rule, a product inside a power. Work outside-in, labelling each layer. Example: differentiate f(x) = e^(2x)/(x² + 1). Quotient rule with u = e^(2x) (u′ = 2e^(2x), chain rule), v = x² + 1 (v′ = 2x): f′(x) = [2e^(2x)(x² + 1) − e^(2x)(2x)]/(x² + 1)² = 2e^(2x)(x² − x + 1)/(x² + 1)² — factor the e^(2x) immediately; it tidies every subsequent step. The second derivative then classifies stationary points without sign tables: for g(x) = x³ − 3x² − 9x, g′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1) gives stationary points at x = 3 and x = −1; g″(x) = 6x − 6 gives g″(3) = 12 > 0 (local minimum, value g(3) = −27) and g″(−1) = −12 < 0 (local maximum, value g(−1) = 5). When g″ = 0 at the point, the test is silent — fall back on the sign of g′ either side. Examiner reports repeat one plea: state *which* rule you are using at each layer; unnamed, tangled differentiation is where HL marks leak.

Extra exam practice [3 marks]. Differentiate ln(cos x) and state where the derivative exists. *Marking notes:* M1 for the chain rule (outer ln, inner cos): derivative = −sin x / cos x; A1: = −tan x; A1: exists where cos x ≠ 0 (and the function itself needs cos x > 0), e.g. x ∈ (−π/2, π/2).