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hscscience Maths Adv · Y11
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Module 4 · L12 of 15 ~30 min ⚡ +90 XP available

Differentiating $\log_a x$

For logarithms with bases other than $e$, differentiation introduces a correction factor of $\frac{1}{\ln a}$. Like converting between measurement systems, changing the log base requires scaling by the natural log of that base. Once you see the pattern, all log derivatives unify.

Today's hook, You already know $\frac{d}{dx}(\ln x) = \frac{1}{x}$. But what about $\log_2 x$ or $\log_{10} x$? There's a single correction factor, $\frac{1}{\ln a}$, that bridges every base back to the natural one. Today you'll see exactly why.
0/5QUESTS
1

Get oriented

Set up your goals and key terms for differentiating a logarithm to any base.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

You know $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$. Without using a formula what single extra factor would you expect in $\dfrac{d}{dx}(\log_2 x)$, and why? Think about the relationship between $\log_2 x$ and $\ln x$.

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02
The two moves
+5 XP to read

There are only two core moves in this lesson. Either rewrite using change of base ($\log_a x = \frac{\ln x}{\ln a}$) and then differentiate, or use the memorised formula $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$ directly.

The change of base formula converts every log into a multiple of $\ln x$, then $\ln a$ in the denominator is a constant. Every general log derivative just multiplies $\frac{1}{x}$ by $\frac{1}{\ln a}$.

Two-step vertical flow from the change of base formula to the resulting derivative.
$$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \qquad \frac{d}{dx}(\log_a g(x)) = \frac{g'(x)}{g(x)\ln a}$$
$\ln a$ is a constant
For a fixed base $a$, $\ln a$ is a number, it just scales the $\frac{1}{x}$ result.
Chain rule extension
For $\log_a(g(x))$: $\dfrac{g'(x)}{g(x)\ln a}$. The $\ln a$ goes in the denominator with $g(x)$.
Base $e$ check
When $a = e$: $\ln e = 1$, so $\frac{1}{x \ln e} = \frac{1}{x}$, recovering the $\ln x$ rule perfectly.
03
What you'll master
Know

Key facts

  • $\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}$ for any valid base $a$
  • Change of base: $\log_a x = \dfrac{\ln x}{\ln a}$
  • $\ln a$ in the denominator, never the numerator
Understand

Concepts

  • Why the formula follows directly from change of base + $\frac{d}{dx}(\ln x)$
  • Why setting $a = e$ recovers $\frac{1}{x}$ exactly
  • How $\ln a$ measures the "distance" of base $a$ from $e$
Can do

Skills

  • Differentiate $\log_a x$ and $\log_a(g(x))$ for any base
  • Use change of base to simplify products of different-base logs
  • Apply general log differentiation in HSC-style problems
04
Key terms
General log derivative$\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x \ln a}$ for any base $a > 0$, $a \neq 1$.
Change of base formula$\log_a x = \dfrac{\ln x}{\ln a}$, converts any base to natural log.
Correction factor$\dfrac{1}{\ln a}$, the multiplicative factor introduced by a non-$e$ base.
Chain rule for $\log_a$$\dfrac{d}{dx}(\log_a g(x)) = \dfrac{g'(x)}{g(x)\ln a}$.
2

See where the rule comes from

Derive the general formula using change of base.

05
Deriving the general formula
core concept

Using the change of base formula, $\log_a x = \dfrac{\ln x}{\ln a}$. Since $\ln a$ is a constant for a given base $a$, we can factor it out:

$$\frac{d}{dx}(\log_a x) = \frac{d}{dx}\!\left(\frac{\ln x}{\ln a}\right) = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$$

Change of base: $\log_a x = \dfrac{\ln x}{\ln a}$. When $a = e$: $\ln e = 1$, so the formula gives $\frac{1}{x}$ as expected.

For composite functions $\log_a(g(x))$, the chain rule gives $\dfrac{g'(x)}{g(x)\ln a}$, the same structure as the $\ln$ case but with $\ln a$ in the denominator alongside $g(x)$. This means every log derivative you already know still works; you just add $\ln a$ to the denominator.

Why the factor is $\frac{1}{\ln a}$ and not $\ln a$. The change of base gives $\log_a x = \frac{\ln x}{\ln a}$, so $\ln a$ divides $\ln x$. When you differentiate, that constant divisor stays in the denominator. If you instead wrote $\ln a \cdot \frac{1}{x}$ you'd be multiplying, that would mean $\log_{10} x$ changes 2.3 times faster than $\ln x$, which is backwards. $\log_{10} x$ grows much more slowly, so its derivative must be smaller than $\frac{1}{x}$.

$\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}$, the $\ln a$ always goes in the denominator; Derivation: use change of base $\log_a x = \frac{\ln x}{\ln a}$, then differentiate $\frac{1}{\ln a}\cdot\ln x$

Pause, copy the rule $\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}$, derived via change of base: $\log_a x = \tfrac{\ln x}{\ln a}$, into your book.

Did you get this? True or false: $\dfrac{d}{dx}(\log_a x) = \dfrac{\ln a}{x}$.

3

Work through examples

Follow three worked problems, revealing each step as you go.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC FORMULA

Differentiate $y = \log_2 x$.

1
Apply $\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}$ with $a = 2$.
Identify the base: $a = 2$. The formula applies directly.
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PROBLEM 2 · CHAIN RULE

Differentiate $y = \log_3(x^2 + 1)$.

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$g(x) = x^2 + 1$, $g'(x) = 2x$, base $a = 3$.
Identify inside function and its derivative; note the base.
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PROBLEM 3 · PRODUCT OF TWO DIFFERENT BASES

Differentiate $y = \log_2 x \cdot \log_3 x$.

1
Rewrite: $y = \dfrac{\ln x}{\ln 2} \cdot \dfrac{\ln x}{\ln 3} = \dfrac{(\ln x)^2}{\ln 2 \cdot \ln 3}$
Change of base both logs, now it's a single function with a constant denominator.
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4

Avoid the common traps

Meet the three errors that cost marks, and check yourself.

Quick check: Which is the correct derivative of $y = \log_3(x^2+1)$?

Common errors, the 3 traps that cost marks

Trap 01
Forgetting $\ln a$ in the denominator
$\frac{d}{dx}(\log_2 x) = \frac{1}{x\ln 2}$, not $\frac{1}{x}$. The correction factor $\frac{1}{\ln a}$ is essential for bases other than $e$, omitting it is the single most common error.
Trap 02
Putting $\ln a$ in the numerator
Writing $\frac{\ln a}{x}$ instead of $\frac{1}{x\ln a}$. Remember: change of base puts $\ln a$ in the denominator so the derivative also has $\ln a$ in the denominator.
Trap 03
Confusing $\log$ (base 10) with $\ln$ (base $e$)
In HSC, $\log x$ without a stated base usually means $\log_{10} x$. Its derivative is $\frac{1}{x\ln 10}$, not $\frac{1}{x}$. Always check the base before differentiating.

Odd one out: Three of these derivatives are correct for their function. Which one is the incorrect pairing?

Teach to learn: In your own words, explain to a classmate why $\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}$, starting from the change-of-base formula.

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5

Drill the derivatives

Choose your work mode, then work five quick-fire derivatives.

Quick-fire practice · 5 problems

1

Differentiate $y = \log_5 x$.

2

Differentiate $y = \log_2(3x)$.

3

Differentiate $y = x\log_{10} x$.

4

Differentiate $y = \dfrac{\log_3 x}{x}$.

5

Find the gradient of $y = \log_4(x^2 + 1)$ at $x = 1$.

Fill the gap: Using change of base, $\log_a x = \dfrac{\ln x}{\ln a}$, so $\dfrac{d}{dx}(\log_a x) =$ . When $a = e$ this simplifies to .

6

Revisit and consolidate

Revisit your first answer and confirm what stuck.

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Revisit your thinking

Earlier you predicted the extra factor in $\dfrac{d}{dx}(\log_2 x)$. The answer is $\dfrac{1}{x\ln 2}$, the $\dfrac{1}{\ln 2}$ factor comes from the change-of-base formula, which converts $\log_a x$ to $\dfrac{\ln x}{\ln a}$. The $\ln a$ in the denominator measures how far the base $a$ sits from $e$: a large base means a smaller correction factor and a slower-growing log.

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7

Practise the general log rule

Answer the questions on differentiating a logarithm to any base.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 42 marks

Q1. Differentiate $y = \log_7(2x + 3)$. Show working. (2 marks)

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ApplyBand 43 marks

Q2. Differentiate $y = x^2\log_2 x$. Show full product rule working. (3 marks)

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AnalyseBand 53 marks

Q3. Differentiate $f(x) = \ln x$ and $g(x) = \log_2 x$. Show that $f'(x) = g'(x) \cdot \ln 2$ and explain why this relationship holds. (3 marks)

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Comprehensive answers (click to reveal)

Drill 1: $\frac{1}{x\ln 5}$ · 2: $\frac{3}{3x \cdot \ln 2} = \frac{1}{x\ln 2}$ · 3: $\log_{10} x + \frac{1}{\ln 10}$ · 4: $\frac{1 - \ln_3 x \cdot \ln 3}{x^2 \ln 3}$ simplified: $\frac{1-\log_3 x \cdot \ln 3}{x^2\ln 3}$ · 5: $\frac{2(1)}{(1+1)\ln 4} = \frac{2}{2\ln 4} = \frac{1}{\ln 4}$

Q1 (2 marks): Chain rule, $g(x)=2x+3$, $g'(x)=2$. $\frac{dy}{dx}=\frac{2}{(2x+3)\ln 7}$ [2]

Q2 (3 marks): Product rule [0.5]. $\frac{dy}{dx} = 2x\log_2 x + x^2 \cdot \frac{1}{x\ln 2}$ [1.5]. $= 2x\log_2 x + \frac{x}{\ln 2}$ [1]

Q3 (3 marks): $f'(x)=\frac{1}{x}$ [0.5]. $g'(x)=\frac{1}{x\ln 2}$ [0.5]. $g'(x)\cdot\ln 2 = \frac{1}{x\ln 2}\cdot\ln 2 = \frac{1}{x} = f'(x)$ [1]. This holds because $\ln x = \log_2 x \cdot \ln 2$ by change of base, so their derivatives have the same scaling relationship [1].