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MATH1061 · Mathematics 1a

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Chapter 2 of 7 · MATH1061

Differentiation

The derivative is the slope of the tangent line — defined as the limit of the difference quotient, f′(a) = limh→0 [f(a+h) − f(a)]/h — and it runs the whole computational core of the calculus stream. In practice you almost never differentiate from the limit; you apply the rules (power, product, quotient, chain) and read off the standard table (trig, exp, log, inverse-trig). The exam then pushes into implicit and logarithmic differentiation and the three applications that carry marks: L'Hôpital's rule for indeterminate limits, optimisation (set f′ = 0, classify the critical point), and curve sketching (monotonicity from f′, concavity and inflections from f″). Every step is by hand, and showing the chain-rule and product-rule working is where the method marks live.

In this chapter

What this chapter covers

  • 013.1 The derivative as a limit — the difference quotient
  • 023.2 The differentiation rules (power, product, quotient, chain)
  • 033.3 The standard-derivative table — memorise cold
  • 043.4 Implicit and logarithmic differentiation
  • 053.5 Application I — L'Hôpital's rule for 0/0 and ∞/∞
  • 063.6 Application II — optimisation (f′ = 0, then classify)
  • 07Application III — curve sketching (monotonicity, concavity, inflections)
Worked example · free

Worked example: chain rule + a turning point

Q [6 marks]. Let f(x) = (2x − 1)³. (a) Find f′(x). (b) Let g(x) = x³ − 3x. Find and classify every critical point of g.
  • +1(a) Chain rule: outer (·)³ gives 3(2x − 1)²; inner (2x − 1)′ = 2. So f′(x) = 3(2x − 1)² · 2 = 6(2x − 1)².
  • +1(b) Differentiate g: g′(x) = 3x² − 3.
  • +1(b) Solve g′ = 0: 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = ±1 (the critical points).
  • +1(b) Second derivative: g″(x) = 6x.
  • +1(b) Classify x = 1: g″(1) = 6 > 0 ⇒ local minimum, value g(1) = −2.
  • +1(b) Classify x = −1: g″(−1) = −6 < 0 ⇒ local maximum, value g(−1) = 2.
(a) f′(x) = 6(2x − 1)². (b) Critical points at x = ±1: x = 1 is a local minimum (g″ > 0, value −2) and x = −1 is a local maximum (g″ < 0, value 2).
Glossary

Key terms

Derivative
The instantaneous rate of change, f′(a) = limh→0 [f(a+h) − f(a)]/h — geometrically the slope of the tangent line at x = a. The whole differentiation toolkit is shortcuts for this limit.
Chain rule
To differentiate a composition f(g(x)), multiply the outer derivative (evaluated at the inner function) by the inner derivative: [f(g(x))]′ = f′(g(x)) · g′(x). The most-used and most-forgotten rule in the course.
Implicit differentiation
Differentiating an equation in x and y without solving for y first: differentiate both sides with respect to x, treating y as a function of x (so every y term gains a dy/dx), then solve for dy/dx.
L'Hôpital's rule
For a 0/0 or ∞/∞ limit, lim f/g = lim f′/g′ provided the latter exists. It converts an indeterminate limit into a derivative quotient — only valid once you have confirmed the indeterminate form.
Critical point
A point where f′(x) = 0 (or is undefined). Candidate maxima, minima and saddle/inflection points; classify with the second-derivative test (f″ > 0 minimum, f″ < 0 maximum) or a sign chart of f′.
FAQ

Differentiation FAQ

When do I use the chain rule versus the product rule?

Product rule is for a product of two functions, (uv)′ = u′v + uv′. Chain rule is for a composition — one function inside another, like (2x − 1)³ or sin(x²). Many problems need both: differentiate the outside with the chain rule, and if the inside is itself a product, use the product rule there. The test is structural: 'multiplied together' → product; 'plugged inside' → chain.

How do I classify a critical point?

Set f′(x) = 0 to find the candidates, then use the second-derivative test: f″ > 0 means the curve is concave up there, so it's a local minimum; f″ < 0 means concave down, a local maximum. If f″ = 0 the test is inconclusive — fall back to a first-derivative sign chart (does f′ change + to − or − to +?).

When is L'Hôpital's rule actually allowed?

Only after you've confirmed the limit is 0/0 or ∞/∞ by substitution. Then differentiate the numerator and denominator separately (not as a quotient) and take the new limit; repeat if it's still indeterminate. Applying it to a limit that isn't indeterminate gives wrong answers — always state the form first.

What's the point of implicit differentiation?

It finds dy/dx for curves you can't (or don't want to) solve for y, like x² + y² = 25. Differentiate every term with respect to x, remembering that each y carries a hidden dy/dx by the chain rule, then collect the dy/dx terms and solve. It's the standard tool for tangent lines to circles, ellipses and other implicit curves.

Study strategy

Exam move

Lock the standard-derivative table cold — trig, exp, log and inverse-trig — because every rule application starts from it. Then drill the rule-selection reflex: product, quotient or chain, and when nested, which order. For applications, the optimisation chain is fixed: differentiate, solve f′ = 0, classify with f″, and read back the value the question wants (often the maximum value, not just where it occurs). Confirm the indeterminate form before any L'Hôpital step. Because the exam is computational and partly no-calculator, write every line of the chain and product rule — examiners award method marks for correct structure even when the arithmetic slips.

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