Mathematics Advanced • Year 11 • Module 4 • Lesson 12

Differentiating log_a x

Build procedural fluency in differentiating logarithms of arbitrary base, using the formula 1 / (x ln a) and the change-of-base method.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete each derivative:

d/dx ( log_a x ) = ____________ (for a > 0, a ≠ 1, x > 0)

d/dx ( log_a g(x) ) = ____________ (chain rule form)

Q1.2 Write each logarithm in terms of natural logarithms using change of base:

log₂ x = __________________

log₁₀ x = __________________

Q1.3 When a = e, the factor 1 / ln a becomes ____________, so d/dx ( log_e x ) = ____________ (the familiar derivative of ln x).

Stuck? Revisit lesson § Concept — Change of base.

2. Worked example — differentiate y = log₃(x² + 1)

Follow each line. Every step has a reason on the right.

Problem. Differentiate y = log₃(x² + 1).

Step 1 — Identify inside function and base.

g(x) = x² + 1, g'(x) = 2x, a = 3

Reason: chain rule form uses g'/g with an extra 1/ln a factor.

Step 2 — Apply d/dx ( log_a g(x) ) = g'(x) / [ g(x) · ln a ].

dy/dx = 2x / [ (x² + 1) · ln 3 ]

Reason: the ln 3 in the denominator is the change-of-base factor.

Step 3 — Domain check.

x² + 1 > 0 for all real x ✓

Reason: the argument of log must be positive.

Conclusion. dy/dx = 2x / [ (x² + 1) ln 3 ], valid for all real x.

3. Faded example — fill in the missing steps

Differentiate y = x log₂ x. Use the product rule with v = log₂ x. 4 marks

Step 1 — Identify u and v for the product rule:

u = ____________ , v = ____________

Step 2 — Differentiate each factor:

u' = ____________ , v' = d/dx ( log₂ x ) = ____________

Step 3 — Apply product rule:

dy/dx = ________ · log₂ x + ________ · ________

Step 4 — Simplify:

dy/dx = log₂ x + ________

Conclusion. dy/dx = log₂ x + 1/ln 2.

Stuck? Revisit lesson § Practice SAQ — Product with general log.

4. Graduated practice — differentiate each function

For each function, find dy/dx. Show working for Standard and Extension items.

Foundation — direct rule (4 questions)

Q	Function	dy/dx
4.1 1	y = log₂ x
4.2 1	y = log₅ x
4.3 1	y = log₁₀ x
4.4 1	y = log_e x (= ln x)

Standard — chain, product, quotient (6 questions)

For chain-rule items, use g'(x) / [ g(x) ln a ]. For product/quotient, name u, v, u', v'.

4.5 y = log₂(3x) 2 marks

4.6 y = log₇(2x + 3) 2 marks

4.7 y = x² log₂ x 2 marks

4.8 y = (log₃ x) / x 2 marks

4.9 y = log₁₀(x² + 4) 2 marks

4.10 Find the gradient of y = log₄(x² + 1) at x = 1. 2 marks

Extension — combine ideas (2 questions)

4.11 Differentiate y = log₂ x · log₃ x. (Hint: rewrite both logs using change of base.) 3 marks

4.12 Let f(x) = ln x and g(x) = log₂ x. Show that f'(x) = ln 2 · g'(x) and explain in one sentence why this multiplicative relationship matches the relationship between f(x) and g(x) themselves. 3 marks

Stuck on 4.12? Start from ln x = log₂ x · ln 2 and differentiate both sides.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I wrote 1 / (x ln 2), not 1/x. The ln 2 in the denominator is essential.

For 4.3 I wrote 1 / (x ln 10), not 1 / (x log 10). ln 10 is the natural log of 10, not 1.

For 4.4 the factor 1/ln e becomes 1/1 = 1, so the answer reduces to 1/x.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Derivative rules

d/dx ( log_a x ) = 1 / (x ln a). d/dx ( log_a g(x) ) = g'(x) / [ g(x) ln a ].

Q1.2 — Change of base

log₂ x = ln x / ln 2. log₁₀ x = ln x / ln 10.

Q1.3 — When the base is e

ln e = 1, so 1/ln a = 1/1 = 1, and d/dx (log_e x) = 1/x.

Q3 — Faded example y = x log₂ x

Step 1: u = x, v = log₂ x. Step 2: u' = 1, v' = 1/(x ln 2).
Step 3: dy/dx = 1 · log₂ x + x · 1/(x ln 2).
Step 4: dy/dx = log₂ x + 1/ln 2.

Q4.1 — y = log₂ x

dy/dx = 1 / (x ln 2).

Q4.2 — y = log₅ x

dy/dx = 1 / (x ln 5).

Q4.3 — y = log₁₀ x

dy/dx = 1 / (x ln 10) ≈ 0.4343 / x.

Q4.4 — y = log_e x = ln x

dy/dx = 1 / (x · ln e) = 1 / (x · 1) = 1/x. (Reduces to the natural-log derivative.)

Q4.5 — y = log₂(3x)

Chain rule: dy/dx = 3 / [ 3x · ln 2 ] = 1 / (x ln 2). (Same as log₂ x, because log₂(3x) = log₂ 3 + log₂ x and the constant differentiates to 0.)

Q4.6 — y = log₇(2x + 3)

g = 2x + 3, g' = 2. dy/dx = 2 / [ (2x + 3) ln 7 ].

Q4.7 — y = x² log₂ x

Product rule with u = x², v = log₂ x, u' = 2x, v' = 1/(x ln 2):
dy/dx = 2x · log₂ x + x² · 1/(x ln 2) = 2x log₂ x + x / ln 2.

Q4.8 — y = (log₃ x) / x

Quotient rule with u = log₃ x, v = x, u' = 1/(x ln 3), v' = 1:
dy/dx = [ (1/(x ln 3)) · x − log₃ x · 1 ] / x² = (1/ln 3 − log₃ x) / x².

Q4.9 — y = log₁₀(x² + 4)

g = x² + 4, g' = 2x. dy/dx = 2x / [ (x² + 4) ln 10 ].

Q4.10 — Gradient of y = log₄(x² + 1) at x = 1

dy/dx = 2x / [ (x² + 1) ln 4 ]. At x = 1: 2 / (2 · ln 4) = 1 / ln 4 ≈ 0.721.

Q4.11 — y = log₂ x · log₃ x

Rewrite: y = (ln x / ln 2)(ln x / ln 3) = (ln x)² / (ln 2 · ln 3). Differentiate:
dy/dx = 2 ln x · (1/x) / (ln 2 · ln 3) = 2 ln x / [ x · ln 2 · ln 3 ].

Q4.12 — Relating f' and g'

f(x) = ln x, so f'(x) = 1/x. g(x) = log₂ x, so g'(x) = 1/(x ln 2).
Then ln 2 · g'(x) = ln 2 · 1/(x ln 2) = 1/x = f'(x). ✓ Why this works: ln x = log₂ x · ln 2, so the function f is just g scaled by the constant ln 2; constants survive differentiation, so the derivatives satisfy the same scaling.