Mathematics Advanced • Year 11 • Module 4 • Lesson 11
Differentiating ln x
Practise HSC-style writing on derivatives of natural logarithms, including a structured stationary-point analysis.
1. Short-answer questions
1.1 Differentiate y = ln(3x² − 1) with respect to x. State the domain on which your derivative is valid. 2 marks Band 3
1.2 Let f(x) = (x + 1) ln x for x > 0. Find f'(x) and use it to determine f'(1) exactly. 3 marks Band 3-4
1.3 The curve y = ln x is graphed below (no graph required to answer).
(a) Find the equation of the tangent to the curve at the point where x = 1.
(b) Hence show that the tangent passes through the origin only if a constant term were added to ln x. State that constant. 4 marks Band 4
2. Extended response
2.1 Consider the function f(x) = (ln x) / x, defined for x > 0.
(a) Show that f'(x) = (1 − ln x) / x².
(b) Find the stationary point of f and classify it (use either the sign-of-f' test or the second-derivative test, stating which).
(c) Sketch a quick sign table of f'(x) on (0, ∞), and describe the long-run behaviour of f as x → ∞ in one sentence.
(d) Hence determine the largest value of (ln x) / x for x > 0, giving an exact answer. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — names quotient rule with u = ln x, v = x, u' = 1/x, v' = 1.
• 1 mark — algebraically simplifies (u'v − uv')/v² to (1 − ln x)/x².
Part (b) — 2 marks
• 1 mark — solves f'(x) = 0, recognising x² ≠ 0, to obtain x = e.
• 1 mark — states which test is used and concludes "maximum" with brief justification.
Part (c) — 2 marks
• 1 mark — sign table or written statement: f' > 0 on (0, e), f' < 0 on (e, ∞).
• 1 mark — describes f(x) → 0⁺ as x → ∞ (and notes f → −∞ as x → 0⁺ for fuller credit).
Part (d) — 1 mark
• 1 mark — states largest value = f(e) = 1/e, with units / exact form, linking back to part (b).
Your response:
Stuck on (c)? For large x, ln x grows but x grows faster — ratio → 0.How did this worksheet feel?
What I'll revisit before next class:
1.1 — y = ln(3x² − 1) (2 marks)
Sample response. Let g(x) = 3x² − 1, so g'(x) = 6x. By the chain rule, dy/dx = g'(x)/g(x) = 6x / (3x² − 1). Valid where 3x² − 1 > 0, i.e. x < −1/√3 or x > 1/√3.
Marking notes. 1 mark — correct application of d/dx(ln g(x)) = g'(x)/g(x). 1 mark — correct simplified derivative AND a valid domain statement. Missing domain caps at 1.5/2.
1.2 — f(x) = (x + 1) ln x (3 marks)
Sample response. Product rule with u = x + 1, v = ln x, u' = 1, v' = 1/x:
f'(x) = 1 · ln x + (x + 1) · (1/x) = ln x + 1 + 1/x.
At x = 1: f'(1) = ln 1 + 1 + 1/1 = 0 + 1 + 1 = 2.
Marking notes. 1 mark — correct u, v, u', v'. 1 mark — correct simplified f'(x). 1 mark — correct evaluation f'(1) = 2 (showing ln 1 = 0).
1.3 — Tangent to y = ln x (4 marks)
(a) Sample response. dy/dx = 1/x, so at x = 1 the gradient is m = 1, and y = ln 1 = 0. Tangent: y − 0 = 1·(x − 1), i.e. y = x − 1.
(b) Sample response. For the tangent at x = 1 to pass through the origin, we would need y = 0 at x = 0; substituting gives 0 = 0 − 1 = −1, a contradiction. If we instead use y = ln x + c, the tangent at x = 1 becomes y − c = 1·(x − 1), i.e. y = x − 1 + c. For this to pass through (0, 0): 0 = −1 + c ⇒ c = 1. So adding the constant +1 makes the tangent at x = 1 go through the origin.
Marking notes. (a) 1 mark — gradient 1 from dy/dx = 1/x. 1 mark — correct tangent equation. (b) 1 mark — recognises original tangent misses origin (substitution showing −1 ≠ 0). 1 mark — finds c = 1 with justification.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Let u = ln x and v = x, so u' = 1/x and v' = 1.
By the quotient rule, f'(x) = (u'v − uv') / v² = [ (1/x)·x − (ln x)·1 ] / x² = (1 − ln x) / x². [2 marks — rule + simplification]
Part (b). Since x > 0, the denominator x² is never zero. So f'(x) = 0 iff 1 − ln x = 0, i.e. ln x = 1, i.e. x = e. [1 mark]
Testing the sign of f': for 0 < x < e we have ln x < 1, so 1 − ln x > 0 and f'(x) > 0 (increasing). For x > e we have ln x > 1, so 1 − ln x < 0 and f'(x) < 0 (decreasing). The sign goes + then −, so x = e is a local maximum. [1 mark — test named and conclusion]
Part (c).
x: 0⁺ … e … ∞
f': + 0 −
f: ↗ max ↘
[1 mark — sign table] As x → ∞, the denominator x grows much faster than the numerator ln x, so f(x) = (ln x)/x → 0⁺. [1 mark — long-run behaviour]
Part (d). Combining (b) and (c), the global maximum of f on x > 0 occurs at x = e, with value f(e) = (ln e) / e = 1/e. [1 mark — exact value linked to (b)]
Total: 7/7.
Band descriptors for marker.
Band 3: Names quotient rule, attempts simplification but errors in algebra; finds x = e by guessing or partial work; no test of nature; ≈ 2-3 marks.
Band 4: Completes (a) cleanly; obtains x = e and classifies as max without a rigorous test; states 1/e but no long-run analysis; ≈ 4-5 marks.
Band 5: All algebra correct; classification uses sign test or f''; states f → 0 as x → ∞; minor omissions in sign table; ≈ 5-6 marks.
Band 6: Fully rigorous: quotient rule labelled, sign table or f'' test stated explicitly, exact value 1/e justified, long-run behaviour mentioned. 7/7.