Differentiating logarithms and composite logarithms | Year 12

01 / LEARN

The idea, step by step

The reciprocal rate of a logarithm

The derivative of $ln x$ is $1 / x$ on $x > 0$ . For a positive differentiable input $u (x)$ , the chain rule gives $u^{'} (x) / u (x)$ . For a fixed base $b$ , use change of base first: ${log}_{b} u = ln u / ln b$ . The constant $ln b$ remains in the denominator.

\frac{d}{d x} ln u (x) = \frac{u^{'} (x)}{u (x)}; \frac{d}{d x} {log}_{b} u (x) = \frac{u^{'} (x)}{u (x) ln b}

Domain first, then choose an efficient method

A derivative formula does not restore points missing from the original function. For example, the derivative of $ln (x^{2})$ is $2 / x$ on $x \neq 0$ , whereas $2 ln x$ only describes its positive branch. Products such as $x ln x$ still require the product rule. Simplify only with identities valid on the domain you are using.

A common trap

$\frac{d}{d x} ln (3 x + 1) = 1 / (3 x + 1)$ .

Instead: The chain rule also multiplies by the derivative of $3 x + 1$ : the answer is $3 / (3 x + 1)$ on $x > - 1 / 3$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Chain rule with a positive input

Differentiate $f (x) = ln (1 + x^{2})$ and state its domain.

The input $1 + x^{2}$ is positive for every real $x$ .
Use $u = 1 + x^{2}$ , so $u^{'} = 2 x$ .
Divide the input derivative by the input.

f^{'} (x) = 2 x / (1 + x^{2})

x \in ℝ

Example 2

A tangent involving a logarithm

Find the tangent to $y = x ln x$ at $x = 1$ .

The point is $(1, 0)$ .
Product rule gives $y^{'} = ln x + 1$ .
At $x = 1$ , the gradient is $1$ , so $y - 0 = 1 (x - 1)$ .

y = x - 1

Check your understanding

Differentiate $ln (5 - 2 x)$ on its real domain.

Show answer

$- 2 / (5 - 2 x)$ , for $x < 5 / 2$ .

The input derivative is $- 2$ ; divide by the positive input $5 - 2 x$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Differentiate $ln (7 x)$ , retaining its domain.
2 marks
Differentiate ${log}_{3} (2 x + 5)$ .
2 marks
Differentiate ${(ln x)}^{2}$ .
2 marks
Differentiate $y = ln (x^{2} - 9)$ and state the domain.
3 marks
Find the stationary point of $f (x) = ln x - x / 2$ and classify it.
4 marks
Find the tangent to $y = ln (2 x - 1)$ at $x = 1$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Differentiate $ln (7 x)$ , retaining its domain.
2 marks
$u^{'} / u = 7 / (7 x) = 1 / x$ .
$1 / x$ , $x > 0$
Differentiate ${log}_{3} (2 x + 5)$ .
2 marks
Use change of base, then apply the chain rule.
$2 / ((2 x + 5) ln 3)$ , $x > - 5 / 2$
Differentiate ${(ln x)}^{2}$ .
2 marks
The outer square contributes $2 ln x$ and the inner derivative is $1 / x$ .
$2 ln x / x$ , $x > 0$
Differentiate $y = ln (x^{2} - 9)$ and state the domain.
3 marks
Solve $x^{2} - 9 > 0$ before applying the chain rule.
$y^{'} = 2 x / (x^{2} - 9)$ ; $x < - 3$ or $x > 3$
Find the stationary point of $f (x) = ln x - x / 2$ and classify it.
4 marks
$f^{'} = 1 / x - 1 / 2 = 0$ gives $x = 2$ .
$f^{''} = - 1 / x^{2} < 0$ .
$(2, ln 2 - 1)$ , a local maximum.
Find the tangent to $y = ln (2 x - 1)$ at $x = 1$ .
3 marks
The point is $(1, 0)$ and the gradient is $2 / (2 - 1) = 2$ .
$y = 2 x - 2$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: logarithmic functions and their applications; reciprocal-linear antiderivatives are taught in the Integration chapter.

Official curriculum source ↗

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