Differentiating logarithms and composite logarithms | Year 12

01 / LEARN

Understand the idea.

The reciprocal rate of a logarithm

The derivative of $ln x$ is $\frac{1}{x}$ for $x > 0$ . For a positive differentiable input $u (x)$ , the chain rule multiplies this reciprocal by the inner rate, giving $\frac{u^{'} (x)}{u (x)}$ .

For a fixed valid base $b$ , change of base writes ${log}_{b} u = \frac{ln u}{ln b}$ . Since $ln b$ is constant, it remains in the denominator of the derivative. The original positive-input restriction still applies.

\begin{aligned} \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} \end{aligned}

Domain first, then choose an efficient method

Determine the domain before simplifying. Although $ln (x^{2})$ has derivative $\frac{2}{x}$ for $x \neq 0$ , writing it as $2 ln x$ would lose the entire negative branch. An identity must be valid on every domain interval being used.

A logarithmic factor does not replace other derivative rules. For a product such as $x ln x$ , differentiate each factor in turn. For a tangent, evaluate the original function for the point and the derivative for the gradient.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Chain rule with a positive input

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

Worked solution

1
Check the logarithm input.
$1 + x^{2} \geq 1 > 0 for all x \in ℝ$
The logarithm exists for every real x.
2
Find the inner derivative.
$u (x) = 1 + x^{2}, u^{'} (x) = 2 x$
The chain rule needs the rate of the full logarithm input.
3
Divide the inner rate by its input.
$f^{'} (x) = \frac{u^{'} (x)}{u (x)} = \frac{2 x}{1 + x^{2}}, x \in ℝ$
The denominator never vanishes, so the derivative has the same real domain as the function.

Answer

$f^{'} (x) = 2 x / (1 + x^{2})$ , $x \in ℝ$ .

EXAMPLE 02

A tangent involving a logarithm

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

Worked solution

1
Find the curve point.
$x > 0, y (1) = 1 ln 1 = 0$
The stated input is valid and the original function supplies the contact point.
2
Differentiate the product.
$y^{'} = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1$
Both x and ln x change, so both product-rule contributions are required.
3
Evaluate the gradient.
$y^{'} (1) = ln 1 + 1 = 1$
The logarithm vanishes at input 1.
4
Write the tangent equation.
$y - 0 = 1 (x - 1) ⟹ y = x - 1$
Point-gradient form combines the point (1,0) with gradient 1.

Answer

$y = x - 1$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The input condition is $5 - 2 x > 0$ , so $x < \frac{5}{2}$ . Its derivative is $- 2$ , hence $\frac{d}{d x} ln (5 - 2 x) = - \frac{2}{5 - 2 x}$ throughout that domain.

Keep practising

6 worksheet questions · 16 marks

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

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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 ↗

In this lesson

Understand the idea.

The reciprocal rate of a logarithm

Domain first, then choose an efficient method

See the method in action.

Chain rule with a positive input

Check the logarithm input.

Find the inner derivative.

Divide the inner rate by its input.

A tangent involving a logarithm

Find the curve point.

Differentiate the product.

Evaluate the gradient.

Write the tangent equation.

Try it yourself.

Check your understanding

Keep practising