Your weak spots

Insights load after your first practice round.

Module 4 · L11 of 15 ~35 min ⚡ +90 XP available

Differentiating $\ln x$

The derivative of $\ln x$ is $\frac{1}{x}$, the simplest reciprocal function. Like a thermostat that responds inversely to temperature, the logarithm's rate of change decreases as $x$ grows. Master this rule and you unlock composite log differentiation through the chain rule.

Today's hook, If $e^x$ is the function that grows at its own rate, what is the function that slows down exactly as fast as the reciprocal of where it is? The answer is $\ln x$, and its derivative $\frac{1}{x}$ is the most elegant rate you'll ever differentiate.

0/5QUESTS

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

Recall, your gut answer first

+5 XP warm-up

If $\dfrac{d}{dx}(e^x) = e^x$ and $\ln x$ undoes $e^x$, what shape do you expect $\dfrac{d}{dx}(\ln x)$ to have? Without using a formulaa line? A hyperbola? A curve that flattens?

auto-saved

The two moves

+5 XP to read

There are only two core moves in this lesson. Lock $\frac{d}{dx}(\ln x) = \frac{1}{x}$ into memory, then extend it to composite functions using the chain rule: $\frac{d}{dx}(\ln g(x)) = \frac{g'(x)}{g(x)}$.

Every log derivative in this lesson uses one of two roads: the basic rule $\frac{1}{x}$ for $\ln x$ itself, or the chain rule form $\frac{g'(x)}{g(x)}$ when there's a function inside the log.

$$\frac{d}{dx}(\ln x) = \frac{1}{x} \qquad \frac{d}{dx}(\ln g(x)) = \frac{g'(x)}{g(x)}$$

Basic rule

$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$. Worth memorising alongside $\dfrac{d}{dx}(e^x) = e^x$.

Chain rule form

$\dfrac{d}{dx}(\ln g(x)) = \dfrac{g'(x)}{g(x)}$. Numerator is always the derivative of what's inside.

Log laws shortcut

Use log laws first: $\ln(kx) = \ln k + \ln x$, so $\frac{d}{dx}(\ln kx) = \frac{1}{x}$.

What you'll master

Know

Key facts

$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ for $x > 0$
The chain rule form for $\ln(g(x))$
Domain restriction: $\ln x$ requires $x > 0$

Understand

Concepts

Why the derivative follows from implicit differentiation of $e^y = x$
How log laws can simplify differentiation before applying rules
The connection between $\ln x$ and $\ln|x|$ for $x \neq 0$

Can do

Skills

Differentiate $\ln(kx)$, $\ln(ax + b)$, and composite logs
Differentiate products and quotients involving $\ln x$
Find stationary points of functions containing $\ln x$

Key terms

Derivative of $\ln x$$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ for $x > 0$.

Chain rule for logs$\dfrac{d}{dx}(\ln g(x)) = \dfrac{g'(x)}{g(x)}$, numerator is $g'(x)$, denominator is $g(x)$.

Logarithmic differentiationA technique that takes $\ln$ of both sides to simplify products, quotients, or powers before differentiating.

$\ln|x|$ extension$\dfrac{d}{dx}(\ln|x|) = \dfrac{1}{x}$ for $x \neq 0$. Covers both positive and negative $x$.

Why $\frac{d}{dx}(\ln x) = \frac{1}{x}$

core concept

Since $y = \ln x$ is the inverse of $y = e^x$, the derivative follows from implicit differentiation. Let $y = \ln x$, so $x = e^y$. Differentiating both sides with respect to $x$: $1 = e^y \cdot \dfrac{dy}{dx}$, which gives $\dfrac{dy}{dx} = \dfrac{1}{e^y} = \dfrac{1}{x}$.

$$\frac{d}{dx}(\ln x) = \frac{1}{x} \qquad \frac{d}{dx}(\ln g(x)) = \frac{g'(x)}{g(x)}$$

Also: $\dfrac{d}{dx}(\ln|x|) = \dfrac{1}{x}$ for $x \neq 0$

For composite functions like $\ln(g(x))$, apply the chain rule: differentiate the outer function ($\frac{1}{\text{inside}}$) and multiply by the derivative of the inside. This gives $\dfrac{g'(x)}{g(x)}$, which is especially useful because log laws can convert messy products and quotients into sums and differences before you differentiate.

Log laws shortcut. When you see $\ln(kx)$ or $\ln\!\left(\tfrac{f}{g}\right)$, expand using log laws first. For example, $\ln(3x) = \ln 3 + \ln x$, so $\frac{d}{dx}(\ln 3x) = \frac{1}{x}$ with no chain rule needed. This shortcut saves time and avoids errors in exams.

$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$, derived from implicit differentiation of $e^y = x$; Chain rule: $\dfrac{d}{dx}(\ln g(x)) = \dfrac{g'(x)}{g(x)}$, numerator is always $g'(x)$

Pause, copy the rule $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ and its chain rule extension $\dfrac{d}{dx}(\ln g(x)) = \dfrac{g'(x)}{g(x)}$ into your book.

Did you get this? True or false: $\dfrac{d}{dx}(\ln x) = \dfrac{1}{\ln x}$.

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

PROBLEM 1 · BASIC CHAIN RULE

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

$\dfrac{dy}{dx} = \dfrac{3}{3x} = \dfrac{1}{x}$

Chain rule: derivative of $3x$ is $3$, divided by $3x$. Simplifies to $\frac{1}{x}$.

PROBLEM 2 · COMPOSITE FUNCTION

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

$g(x) = x^2 + 1$, so $g'(x) = 2x$.

Identify the inside function and differentiate it.

PROBLEM 3 · PRODUCT RULE + STATIONARY POINT

Differentiate $y = x\ln x$ and find its stationary point.

Product rule: $u = x$, $v = \ln x$, so $u' = 1$, $v' = \dfrac{1}{x}$.

Product of a linear and a logarithmic function, apply product rule.

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

Common errors, the 3 traps that cost marks

Trap 01

Writing $\frac{d}{dx}(\ln x) = \frac{1}{\ln x}$

The derivative is $\frac{1}{x}$, not $\frac{1}{\ln x}$. Think: $\ln x$ grows slowly, so its derivative $\frac{1}{x}$ gets smaller as $x$ increases, it's a hyperbola, not another log.

Trap 02

Forgetting $g'(x)$ in the numerator

$\frac{d}{dx}(\ln(x^2+1)) = \frac{2x}{x^2+1}$, not $\frac{1}{x^2+1}$. The chain rule always puts the derivative of the inside on top.

Trap 03

Differentiating $\ln x$ for $x \le 0$

$\ln x$ is only defined for $x > 0$. For $x < 0$, use $\ln|x|$ which still has derivative $\frac{1}{x}$. The HSC typically works with $x > 0$ unless absolute value is specified.

Fill the gap: The function $\ln x$ is only defined for $x$ . Its derivative is $\frac{d}{dx}(\ln x) =$ .

Match up: Connect each function on the left with its correct derivative on the right.

$y = \ln(5x)$
$y = \ln(x^2+1)$
$y = x\ln x$
$y = \ln(ax+b)$

$\dfrac{a}{ax+b}$
$\ln x + 1$
$\dfrac{2x}{x^2+1}$
$\dfrac{1}{x}$

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Differentiate $y = \ln(2x)$.

Differentiate $y = \ln(x^3)$.

Differentiate $y = x^2 \ln x$.

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

Find the gradient of $y = \ln(x^2 + 4)$ at $x = 2$.

Teach to learn: In your own words, explain to a classmate why $\dfrac{d}{dx}(x^2 \ln x) = 2x \ln x + x$.

auto-saved

Revisit your thinking

Earlier you were asked what shape $\dfrac{d}{dx}(\ln x)$ might have. The answer is $\dfrac{1}{x}$, a hyperbola that starts steep (fast change near $x = 0$) and flattens out as $x$ grows. For composite logs, the chain rule gives $\dfrac{g'(x)}{g(x)}$, which often simplifies otherwise messy differentiations.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

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

Short answer

ApplyBand 42 marks

Q1. Differentiate $y = \ln(5x + 2)$. Show working. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. Differentiate $y = (x + 1)\ln x$. Show full product rule working. (3 marks)

auto-saved

AnalyseBand 54 marks

Q3. Find the stationary point of $y = \dfrac{\ln x}{x}$ and determine its nature. (4 marks)

auto-saved

Comprehensive answers (click to reveal)

Drill 1: $\frac{1}{x}$ · 2: $\frac{3}{x}$ · 3: $2x\ln x + x$ · 4: $\frac{1-\ln x}{x^2}$ · 5: $\frac{4}{8} = \frac{1}{2}$ (gradient at $x=2$ is $\frac{2(2)}{4+4}=\frac{4}{8}=\frac{1}{2}$)

Q1 (2 marks): Chain rule, $g(x)=5x+2$, $g'(x)=5$. $\frac{dy}{dx}=\frac{5}{5x+2}$ [2]

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

Q3 (4 marks): Quotient rule: $\frac{dy}{dx} = \frac{\frac{1}{x}\cdot x - \ln x\cdot 1}{x^2} = \frac{1-\ln x}{x^2}$ [1.5]. Set $=0$: $\ln x = 1 \Rightarrow x = e$ [1]. $y = \frac{1}{e}$, point $(e,\frac{1}{e})$ [0.5]. For $x < e$: $\frac{dy}{dx} > 0$; for $x > e$: $\frac{dy}{dx} < 0$. Local maximum. [1]

Boss battle · The Reciprocal

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Arena coming soon

Science Jump · platform challenge

Climb platforms by answering log differentiation questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 10 Lesson 12 · Differentiating logₐ x →

Module overview · Maths Advanced