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.
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
You know $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$. Without using a formulawhat single extra factor would you expect in $\dfrac{d}{dx}(\log_2 x)$, and why? Think about the relationship between $\log_2 x$ and $\ln x$.
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 $\frac{1}{\ln a}$.
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
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 $e$
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
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:
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.
$\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\ln a}$, 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: $\dfrac{d}{dx}(\log_a x) = \dfrac{\ln a}{x}$.
Worked examples · 3 in a row, reveal as you go
Differentiate $y = \log_2 x$.
Differentiate $y = \log_3(x^2 + 1)$.
Differentiate $y = \log_2 x \cdot \log_3 x$.
Quick check: Which is the correct derivative of $y = \log_3(x^2+1)$?
Common errors, the 3 traps that cost marks
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.
Quick-fire practice · 5 problems
Differentiate $y = \log_5 x$.
Differentiate $y = \log_2(3x)$.
Differentiate $y = x\log_{10} x$.
Differentiate $y = \dfrac{\log_3 x}{x}$.
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 .
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.
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Differentiate $y = \log_7(2x + 3)$. Show working. (2 marks)
Q2. Differentiate $y = x^2\log_2 x$. Show full product rule working. (3 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)
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].
Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Climb platforms by answering general log differentiation questions. Lighter alternative to the boss.
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