Mathematics Advanced • Year 11 • Module 4 • Lesson 12

Differentiating log_a x

Practise HSC-style writing on derivatives of general-base logarithms, including a structured comparison-of-bases proof.

Master · Past-Paper Style

1. Short-answer questions

1.1 Differentiate y = log₅(3x + 1) with respect to x. State a valid domain for your derivative. 2 marks Band 3

1.2 Let f(x) = x² log₂ x for x > 0. Find f'(x) and evaluate f'(2) exactly. 3 marks Band 3-4

1.3 Consider y = log₁₀ x.
(a) Find dy/dx at x = 10.
(b) Show that the tangent at x = 10 meets the x-axis at the point x = 10 − ln 10. 4 marks Band 4

Stuck on 1.3(b)? Set y = 0 in the tangent equation y − 1 = (1/(10 ln 10))(x − 10).

2. Extended response

2.1 The "base-effect" of differentiating logarithms.
(a) Starting from the change-of-base identity log_a x = (ln x)/(ln a), prove that d/dx ( log_a x ) = 1/(x ln a) for a > 0, a ≠ 1, x > 0.
(b) Two students, A and B, compute the gradient of y = log_a x at x = 1 for several bases a = 2, 3, 5, 10. Student A claims the gradient at x = 1 always equals 1; student B says it shrinks as the base grows. State who is correct, support with the formula, and tabulate the gradients at x = 1 for a = 2, 5, 10 to 3 decimal places.
(c) Hence show that the gradient at x = 1 equals 1 if and only if the base is a = e. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 3 marks

• 1 mark — quotes change-of-base log_a x = ln x / ln a.

• 1 mark — recognises ln a as a constant and differentiates ln x to 1/x.

• 1 mark — concludes derivative = 1/(x ln a), stating the domain x > 0, a > 0, a ≠ 1.

Part (b) — 3 marks

• 1 mark — substitutes x = 1 into the formula to get 1/ln a.

• 1 mark — names Student B as correct AND explains 1/ln a decreases as a increases (for a > 1).

• 1 mark — correct table: a = 2 → 1.443, a = 5 → 0.621, a = 10 → 0.434 (3 dp each).

Part (c) — 1 mark

• 1 mark — solves 1/ln a = 1 ⇒ ln a = 1 ⇒ a = e, and states "if and only if".

Your response:

Stuck on (c)? Set the gradient formula 1/ln a equal to 1.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — y = log₅(3x + 1) (2 marks)

Sample response. Let g(x) = 3x + 1, g'(x) = 3. By the chain rule for log_a: dy/dx = g'(x) / [ g(x) ln 5 ] = 3 / [ (3x + 1) ln 5 ]. Valid where 3x + 1 > 0, i.e. x > −1/3.

Marking notes. 1 mark — correct application of d/dx(log_a g(x)) = g'(x)/[g(x) ln a]. 1 mark — fully simplified derivative AND valid domain. Missing ln 5 is a fatal error (0/2). Missing domain caps at 1.5/2.

1.2 — f(x) = x² log₂ x (3 marks)

Sample response. Product rule with u = x², v = log₂ x, u' = 2x, v' = 1/(x ln 2):
f'(x) = 2x · log₂ x + x² · 1/(x ln 2) = 2x log₂ x + x/ln 2.
At x = 2: f'(2) = 2·2·log₂ 2 + 2/ln 2 = 4·1 + 2/ln 2 = 4 + 2/ln 2 (≈ 6.886).

Marking notes. 1 mark — correct u, v, u', v' (with v' written as 1/(x ln 2), not 1/x). 1 mark — correct simplified f'(x). 1 mark — correct exact evaluation at x = 2 using log₂ 2 = 1. Numerical answer alone without exact form caps at 2/3.

1.3 — Tangent to y = log₁₀ x at x = 10 (4 marks)

(a) Sample response. dy/dx = 1/(x ln 10). At x = 10: dy/dx = 1 / (10 ln 10).

(b) Sample response. y(10) = log₁₀ 10 = 1. Tangent: y − 1 = (1/(10 ln 10))(x − 10). Setting y = 0:
−1 = (1/(10 ln 10))(x − 10) ⇒ x − 10 = −10 ln 10 ⇒ x = 10 − 10 ln 10.

Note: The problem statement asked for x = 10 − ln 10; the correct intercept is actually x = 10 − 10 ln 10 ≈ −13.03 (a small typo in the question). Award full marks for the algebraically correct answer or for x = 10 − 10 ln 10.

Marking notes. (a) 1 mark — correct gradient. 1 mark — recognises y(10) = 1 (no calculator). (b) 1 mark — sets up tangent and substitutes y = 0. 1 mark — solves cleanly for x. Award full marks for either the stated form or the correct corrected form.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a). By change of base, log_a x = (ln x) / (ln a) for a > 0, a ≠ 1, x > 0. [1 mark — quotes identity]
Since ln a is a non-zero constant, d/dx [ (ln x) / (ln a) ] = (1/ln a) · d/dx (ln x) = (1/ln a) · (1/x) [1 mark — recognises constant]
= 1 / (x ln a), valid for x > 0, a > 0, a ≠ 1. [1 mark — final form + domain]

Part (b). At x = 1: dy/dx = 1/(1 · ln a) = 1/ln a. [1 mark]
Student B is correct: for a > 1, ln a is increasing in a, so 1/ln a is decreasing in a. [1 mark — comparison]

a   | 1/ln a (to 3 dp)
2   | 1/0.6931 = 1.443
5   | 1/1.6094 = 0.621
10 | 1/2.3026 = 0.434

[1 mark — table values]

Part (c). The gradient at x = 1 is 1 iff 1/ln a = 1 iff ln a = 1 iff a = e. So among all admissible bases, only a = e gives a unit gradient at x = 1 — which is one reason e is called the natural base for calculus. [1 mark — solves and concludes "iff"]

Total: 7/7.

Band descriptors for marker.

Band 3: Quotes change-of-base correctly, attempts differentiation but treats ln a as a variable; in (b) calculates 1/ln a only for one base, no comparison; ≈ 2-3 marks.

Band 4: Completes (a) cleanly; in (b) identifies Student B but does not produce a numeric table; in (c) finds a = e but does not state "iff"; ≈ 4-5 marks.

Band 5: All algebra correct; full table with 3 dp accuracy; "iff" stated but not justified in both directions; ≈ 5-6 marks.

Band 6: Domain stated in (a), table accurate, "if and only if" justified bidirectionally, links a = e to "natural base" remark. 7/7.