Logarithmic and Exponential Functions

Lessons

Senior 6 · Mathematics

Logarithmic and Exponential Functions

4 lessons

Lesson 2 of 4

Differentiating logarithms: the curve of ln x and logarithms to any base

25 min

Measuring the slope of $\ln x$

Estimate the gradient of $y = \ln x$ with a calculator, using a small step $h = 0.001$ : $\frac{\ln(x + h) - \ln x}{h}$ .

$x$	1	2	4
$\frac{\ln(x + 0.001) - \ln x}{0.001}$	0.9995	0.4999	0.2500

The gradient halves each time $x$ doubles, and looks like $\frac{1}{x}$ . The proof below confirms it. You already differentiate polynomial, rational, irrational (S4) and trigonometric functions (S5); this lesson adds the logarithms.

Important

Theorem. For $x \gt 0$ , $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ .

Proof from first principles. For $x \gt 0$ and small $h \ne 0$ , using the laws of logarithms,

\begin{aligned} \frac{\ln(x + h) - \ln x}{h} &= \frac{1}{h}\ln\left(1 + \frac{h}{x}\right) = \frac{1}{x}\cdot\frac{x}{h}\ln\left(1 + \frac{h}{x}\right) \\ &= \frac{1}{x}\ln\left[\left(1 + t\right)^{\frac{1}{t}}\right], \end{aligned}

where $t = \frac{h}{x}$ . As $h \to 0$ , $t \to 0$ , so $(1 + t)^{\frac{1}{t}} \to e$ (Lesson 1). Since $\ln$ is continuous, the quotient tends to $\frac{1}{x}\ln e = \frac{1}{x}$ . ∎

With the chain rule: if $u(x) \gt 0$ , then $\dfrac{d}{dx}\ln u = \dfrac{u'}{u}$ . If $u(x) \ne 0$ has either sign, $\dfrac{d}{dx}\ln|u| = \dfrac{u'}{u}$ as well: for $x \lt 0$ , $\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}$ .

Worked example 1: differentiating logarithms

Differentiate (a) $\ln(x^2 + 1)$ (b) $x\ln x$ (c) $\ln\frac{x - 1}{x + 2}$ , for $x \gt 1$ (d) $\ln(\cos x)$ , for $0 \lt x \lt \frac{\pi}{2}$ .

Solution

$u = x^{2} + 1$ , $u^{'} = 2 x$ : the derivative is $\frac{2x}{x^2 + 1}$ .
Product rule: $1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$ .
Split first: $\ln(x - 1) - \ln(x + 2)$ , whose derivative is $\frac{1}{x - 1} - \frac{1}{x + 2} = \frac{3}{(x - 1)(x + 2)}$ .
$\frac{-\sin x}{\cos x} = -\tan x$ .

Variation and curve of $\ln x$

For $f(x) = \ln x$ on $(0, +\infty)$ : $f'(x) = \frac{1}{x} \gt 0$ , so $f$ is strictly increasing; $f''(x) = -\frac{1}{x^2} \lt 0$ , so the curve is concave down, with no point of inflexion.

$x$	$0^{+}$		$1$		$e$		$+\infty$
$f'(x) = \frac{1}{x}$		+	1	+	$\frac{1}{e}$	+
$f(x) = \ln x$	$-\infty$	↗	0	↗	1	↗	$+\infty$

The tangent at $(1, 0)$ is $y = x - 1$ . A concave-down curve lies below its tangents, which gives a useful inequality. Proof: let $g(x) = x - 1 - \ln x$ . Then $g'(x) = 1 - \frac{1}{x}$ , negative on $(0, 1)$ and positive on $(1, +\infty)$ , so $g$ has its minimum $g (1) = 0$ . Hence

\ln x \le x - 1 \quad \text{for all } x \gt 0, \text{ with equality only at } x = 1 .

The tangent to $y = \ln x$ at $(1, 0)$ has gradient $\frac{1}{1} = 1$ . Because the curve is concave down, it stays below this tangent: $\ln x \le x - 1$ for all $x \gt 0$ .

Important

The two standard limits (promised in Lesson 1).

For $x \gt 1$ , apply $\ln t \lt t$ with $t = \sqrt{x}$ : $\ln x = 2\ln\sqrt{x} \lt 2\sqrt{x}$ . So $0 \lt \frac{\ln x}{x} \lt \frac{2}{\sqrt{x}} \to 0$ , and by the squeeze theorem (S4) $\lim_{x \to +\infty}\frac{\ln x}{x} = 0$ .
Put $x = \frac{1}{t}$ : as $x \to 0^+$ , $t \to +\infty$ and $x\ln x = \frac{-\ln t}{t} \to 0$ . So $\lim_{x \to 0^+} x\ln x = 0$ .

Worked example 2: study of $f(x) = \frac{\ln x}{x}$ , and which is larger, $e^\pi$ or $\pi^e$ ?

Solution Domain $(0, +\infty)$ . Limits: $\lim_{x \to 0^+} f(x) = -\infty$ (Lesson 1), so $x = 0$ is a vertical asymptote; $\lim_{x \to +\infty} f(x) = 0$ , so $y = 0$ is a horizontal asymptote.

Quotient rule: $f'(x) = \frac{\frac{1}{x}\cdot x - \ln x}{x^2} = \frac{1 - \ln x}{x^2}$ , which is positive for $x \lt e$ , zero at $x = e$ and negative for $x \gt e$ . So $f$ has a maximum $f(e) = \frac{1}{e} \approx 0.368$ , and the curve crosses the $x$ -axis at $x = 1$ .

Since $\pi \gt e$ and $e$ is the only maximum, $f(\pi) \lt f(e)$ : $\frac{\ln\pi}{\pi} \lt \frac{1}{e}$ , so $e\ln\pi \lt \pi$ , that is, $\ln\left(\pi^e\right) \lt \ln\left(e^\pi\right)$ . Hence $\pi^e \lt e^\pi$ . (A calculator agrees: $\pi^e \approx 22.46$ and $e^\pi \approx 23.14$ .)

$f(x) = \frac{\ln x}{x}$ , plotted from computed points (the vertical scale is four times the horizontal one). The maximum is at $x = e$ , so $f(\pi) \lt f(e)$ .

Definition

Logarithm to base $a$ ( $a \gt 0$ , $a \ne 1$ ): for $x \gt 0$ ,

\log_a x = \frac{\ln x}{\ln a}, \qquad \text{so that } y = \log_a x \iff a^y = x .

It obeys the same laws as $\ln$ . Your S4 decimal logarithm is $\log x = \log_{10} x = \frac{\ln x}{\ln 10}$ . Since $\ln a$ is a constant,

\frac{d}{dx}\log_a x = \frac{1}{x\ln a} .

The function is increasing if $a \gt 1$ ( $\ln a \gt 0$ ) and decreasing if $0 \lt a \lt 1$ . Example: $\frac{d}{dx}\log_{10}(3x) = \frac{d}{dx}\left(\log_{10} 3 + \log_{10} x\right) = \frac{1}{x\ln 10}$ .

Logarithmic differentiation

To differentiate a long product or quotient, or a function with the variable in the exponent, take $\ln$ of both sides first, simplify with the laws, then differentiate implicitly using $\frac{d}{dx}\ln y = \frac{y'}{y}$ .

A variable power. For $y = x^{x}$ ( $x \gt 0$ ): $\ln y = x\ln x$ , so $\frac{y'}{y} = \ln x + 1$ and $y' = x^x(\ln x + 1)$ . Neither the power rule nor the rule for $a^{x}$ applies here, because both the base and the exponent vary.

Worked example 3: a product and quotient

Find $y^{'} (2)$ for $y = \frac{(x - 1)^3\sqrt{x + 2}}{(x + 1)^2}$ , $x \gt 1$ .

Solution All factors are positive for $x \gt 1$ , so

\ln y = 3\ln(x - 1) + \frac{1}{2}\ln(x + 2) - 2\ln(x + 1) .

\frac{y'}{y} = \frac{3}{x - 1} + \frac{1}{2(x + 2)} - \frac{2}{x + 1} .

At $x = 2$ : $y = \frac{1^3 \times 2}{3^2} = \frac{2}{9}$ and $\frac{y'}{y} = 3 + \frac{1}{8} - \frac{2}{3} = \frac{59}{24}$ . So $y'(2) = \frac{2}{9} \times \frac{59}{24} = \frac{59}{108} \approx 0.546$ .

Warning

$\frac{d}{dx}x^x$ is not $x \cdot x^{x - 1}$ (the power rule needs a constant exponent), and not $x^x\ln x$ (that rule needs a constant base).
Take logarithms only of positive quantities. If $y$ can be negative, use $\ln|y|$ ; the result $\frac{y'}{y}$ is the same.

Tip

Successive derivatives of $\ln x$ follow a pattern: $\frac{1}{x}, -\frac{1}{x^2}, \frac{2}{x^3}, -\frac{6}{x^4}, \ldots$ , that is, $\frac{d^n}{dx^n}\ln x = \frac{(-1)^{n-1}(n - 1)!}{x^n}$ . You will need this pattern for the series of $\ln(1 + x)$ in Unit 3.

MasterWuMathematics: Proof: Derivative of ln(x) = 1/x by First Principles. Watch how the limit that defines e appears in the middle of the proof, as it does in the theorem above.

✦

Practice

Simplify with the laws of logarithms before differentiating whenever you can.

Differentiate: (a) $\ln(5x)$ (b) $\ln(x^3 - 2x)$ (c) $(\ln x)^2$ (d) $\ln\sqrt{1 + x^2}$ .
Differentiate: (a) $\log_2 x$ (b) $\log_{10}(x^2 + 1)$ .
Find the equation of the tangent to $y = \ln(2x - 1)$ at the point where $x = 1$ .
Find the stationary point of $f(x) = x\ln x$ and its nature.
Use logarithmic differentiation to differentiate $y = x^{\sin x}$ , $x \gt 0$ .
Find $\frac{y'}{y}$ for $y = \frac{(x + 1)^2(x - 3)}{\sqrt{x + 5}}$ , $x \gt 3$ .
Find $f^{'''} (x)$ and $f^{(4)}(x)$ for $f(x) = \ln x$ .
Using $\ln x \le x - 1$ , prove that $\ln x \ge 1 - \frac{1}{x}$ for all $x \gt 0$ .

Show answersHide answers

(a) $\ln 5 + \ln x$ , so $\frac{1}{x}$ (b) $\frac{3x^2 - 2}{x^3 - 2x}$ (c) $\frac{2\ln x}{x}$ (d) $\frac{1}{2}\ln(1 + x^2)$ , so $\frac{x}{1 + x^2}$ .
(a) $\frac{1}{x\ln 2}$ (b) $\frac{2x}{(x^2 + 1)\ln 10}$ .
$y(1) = \ln 1 = 0$ and $y' = \frac{2}{2x - 1} = 2$ at $x = 1$ , so $y = 2 x - 2$ .
$f'(x) = \ln x + 1 = 0$ at $x = \frac{1}{e}$ ; $f\left(\frac{1}{e}\right) = -\frac{1}{e}$ ; $f''(x) = \frac{1}{x} \gt 0$ , so it is a minimum.
$\ln y = \sin x\ln x$ , so $\frac{y'}{y} = \cos x\ln x + \frac{\sin x}{x}$ and $y' = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right)$ .
$\frac{y'}{y} = \frac{2}{x + 1} + \frac{1}{x - 3} - \frac{1}{2(x + 5)}$ .
$f'''(x) = \frac{2}{x^3}$ , $f^{(4)}(x) = -\frac{6}{x^4}$ .
Replace $x$ by $\frac{1}{x} \gt 0$ : $\ln\frac{1}{x} \le \frac{1}{x} - 1$ , so $-\ln x \le \frac{1}{x} - 1$ , that is, $\ln x \ge 1 - \frac{1}{x}$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the graphs (plotted from computed points), the worked examples, the practice items and the quiz questions.
Video: Proof: Derivative of ln(x) = 1/x by First Principles, MasterWuMathematics, embedded from YouTube.

1Measuring the slope of ln⁡x \ln x lnx

2Variation and curve of ln⁡x \ln x lnx

3Logarithmic differentiation

Measuring the slope of $\ln x$

Variation and curve of $\ln x$

Logarithmic differentiation