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hscscience Maths Adv · Y11
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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
1

Get oriented

Set up your goals and key terms for differentiating the natural logarithm.

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

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 formula a line? A hyperbola? A curve that flattens?

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02
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.

Comparison of the two differentiation rules for the natural logarithm.
$$\frac{d}{dx}(\ln x) = \frac{1}{x} \qquad \frac{d}{dx}(\ln g(x)) = \frac{g'(x)}{g(x)}$$
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 $\frac{d}{dx}(\ln kx) = \frac{1}{x}$.
03
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 $\ln x$
04
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 $x$.
2

See where the rule comes from

Learn why the derivative of the natural log is one over x.

05
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: $\dfrac{d}{dx}(\ln x) = \dfrac{1}{\ln x}$.

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 CHAIN RULE

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

1
$\dfrac{dy}{dx} = \dfrac{3}{3x} = \dfrac{1}{x}$
Chain rule: derivative of $3x$ is $3$, divided by $3x$. Simplifies to $\frac{1}{x}$.
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PROBLEM 2 · COMPOSITE FUNCTION

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

1
$g(x) = x^2 + 1$, so $g'(x) = 2x$.
Identify the inside function and differentiate it.
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PROBLEM 3 · PRODUCT RULE + STATIONARY POINT

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

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

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.

  • $y = \ln(5x)$
  • $y = \ln(x^2+1)$
  • $y = x\ln x$
  • $y = \ln(ax+b)$
  • $\dfrac{a}{ax+b}$
  • $\ln x + 1$
  • $\dfrac{2x}{x^2+1}$
  • $\dfrac{1}{x}$
5

Drill the derivatives

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

Quick-fire practice · 5 problems

1

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

2

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

3

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

4

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

5

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

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

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6

Revisit and consolidate

Revisit your first answer and confirm what stuck.

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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.

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7

Practise the natural log rule

Answer the questions on differentiating the natural logarithm.

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

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ApplyBand 43 marks

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

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AnalyseBand 54 marks

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

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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]