5.14 Derivatives of Logarithmic Functions

Table of Contents

Differentiation of a Logarithm

Let $y = \ln x$ . To differentiate, we follow the steps:

Step 1: $y + Δ y = \ln (x + Δ x)$ Step 2:
$\begin{aligned} Δ y & = \ln (x + Δ x) - \ln x \\ = \ln (\frac{x + Δ x}{x}) \\ = \ln (1 + \frac{Δ x}{x}) \end{aligned}$
Step 3:
$\begin{aligned} \frac{Δ y}{Δ x} & = \frac{1}{Δ x} \ln (1 + \frac{Δ x}{x}) \\ = \ln {(1 + \frac{Δ x}{x})}^{1 / Δ x} \\ = \frac{1}{x} \ln {(1 + \frac{Δ x}{x})}^{\frac{x}{Δ x}} \end{aligned}$
Dividing the logarithm by $x$ and at the same time multiplying the exponent of the parentheses by $x$ changes the form of the expression but not its value. That is, $\ln u = \frac{1}{a} \ln u^{a}$ .] Step 4:
$\begin{aligned} \frac{d y}{d x} & = lim_{Δ x \to 0} \frac{Δ y}{Δ x} \\ = lim_{Δ x \to 0} [\frac{1}{x} \ln {(1 + \frac{Δ x}{x})}^{\frac{x}{Δ x}}] \\ = \frac{1}{x} lim_{Δ x \to 0} [\ln {(1 + \frac{Δ x}{x})}^{\frac{x}{Δ x}}] \\ = \frac{1}{x} \ln [lim_{Δ x \to 0} {(1 + \frac{Δ x}{x})}^{\frac{x}{Δ x}}] \\ = \frac{1}{x} \ln e \\ = \frac{1}{x} . \end{aligned}$
[Note than when $Δ x \to 0$ , $\frac{Δ x}{x} \to 0$ . Therefore, $lim_{Δ x \to 0} {(1 + \frac{Δ x}{x})}^{\frac{x}{Δ x}} = e$ from placing $u = \frac{Δ x}{x}$ in $lim_{u \to 0} (1 + u)^{1 / u} = e$ (see here)].

Hence
$\begin{matrix} (a) & \frac{d}{d x} \ln x = \frac{1}{x}, (x > 0) . \end{matrix}$

Now consider $y = \log_{a} x$ . What is $d y / d x$ ?
We write $y = \log_{a} x = \frac{\ln x}{\ln a} .$ Therefore
$\begin{aligned} \frac{d y}{d x} & = \frac{1}{\ln a} \frac{d}{d x} \ln x \\ = \frac{1}{\ln a} \frac{1}{x} . \end{aligned}$
$\begin{matrix} (b) & \frac{d}{d x} \log_{a} x = \frac{1}{x \ln a}, (x > 0) . \end{matrix}$

If $u (x) > 0$ is a differentiable function, applying the chain rule to (a) produces: $\frac{d}{d x} \ln u = \frac{1}{u} \frac{d u}{d x}$
When possible, to differentiate a function involving logarithms, first use the properties of logarithms to convert products to sums, quotients to differences and exponents to constant multiples then differentiate the result.

Example 1

Find the equation of the tangent line to the curve of $y = \ln (x^{2} + 1)$ when $x = 0$ .

Solution

The slope of the tangent is $d y / d x$
$\begin{aligned} \frac{d y}{d x} & = \frac{d}{d x} \ln (\overset{u}{\overset{⏞}{x^{2} + 1}}) \\ = \frac{1}{u} \frac{d u}{d x} \\ = \frac{1}{x^{2} + 1} (2 x) \\ = \frac{2 x}{x^{2} + 1} \end{aligned}$
The slope of the tangent $m_{tan}$ when $x = 0$ is obtained by substituting 0 for $x$ in the above equation: $m_{tan} = {\frac{2 x}{x^{2} + 1} |}_{x = 0} = 0$ Because when $x = 0$ , $y = \ln (0 + 1) = 0$ , the horizontal line $y = 0$ is tangent to the graph of $y = \ln (x^{2} + 1)$ at the origin. The graph of $y = \ln (x^{2} + 1)$ is shown below.

Fig 1: The tangent line to the graph of

y = \ln (1 + x^{2})

(0, 0)

is horizontal.

Example 2

Find $\frac{d}{d x} [\ln \frac{\sqrt{x + 1} (2 + \cos x)}{x^{2}}]$

Solution

$\begin{aligned} \frac{d}{d x} [\ln \frac{\sqrt{x + 1} (2 + \cos x)}{x^{2}}] & = \frac{d}{d x} [\ln \sqrt{x + 1} + \ln (2 + \cos x) - \ln x^{2}] \\ = \frac{d}{d x} [\frac{1}{2} \ln (x + 1) + \ln (2 + \cos x) - 2 \ln x] \\ = \frac{1}{2 (x + 1)} + \frac{1}{2 + \cos x} (\frac{d}{d x} \cos x) - \frac{2}{x} \\ = \frac{1}{2 (x + 1)} - \frac{\sin x}{2 + \cos x} - \frac{2}{x} \end{aligned}$

Now an important example:

Example 3

Find $\frac{d}{d x} \ln | x |$ .

Solution

We consider two cases:

(a) $x > 0$ . Then $| x | = x$ and $\frac{d}{d x} \ln | x | = \frac{d}{d x} \ln x = \frac{1}{x} .$ (b) $x < 0$ . Then $| x | = - x$ and
$\frac{d}{d x} \ln | x | = \frac{d}{d x} \ln (- x) = \frac{1}{- x} \frac{d}{d x} (- x) = \frac{1}{- x} (- 1) = \frac{1}{x} .$
Because we get the same results in both cases, we conclude
$\frac{d}{d x} \ln | x | = \frac{1}{x} .$

From the above example, we conclude

$\begin{matrix} (c) & \frac{d}{d x} \ln | x | = \frac{1}{x} . \end{matrix}$

Example 4

Find $\frac{d}{d x} \ln | \cos x |$ .

Solution

Let $u = \cos x$ . Then $\frac{d}{d x} \ln | \cos x | = \frac{d}{d x} \ln | u |$ and it follows from Equation (c) that

$\begin{aligned} \frac{d}{d x} \ln | \cos x | & = \frac{1}{u} \frac{d u}{d x} \\ = \frac{1}{\cos x} (- \sin x) \\ = - \tan x . \end{aligned}$

Logarithmic Differentiation

To differentiate a function $y = f (x)$ with respect to $x$ and $f$ composed of products, quotients, and exponents, it is sometimes easier to:

Take the natural logarithm of both sides of $y = f (x)$
Expand the right side using the properties of logarithms.
Eake derivative of both sides
Isolate $d y / d x$ and replace $y$ with the original function.

Example 5

Find $y^{'}$ given
$y = \frac{x \sin x}{(2 - x) \sqrt{1 + x^{4}}}$

Solution

This would be really messy to differentiate $y$ directly. Instead we can use logarithmic differentiation.

First we take the natural logarithm of both sides and exapnd the right hand side using the properties of logarithms:
$\begin{aligned} \ln y & = \ln \frac{x \sin x}{(2 - x) \sqrt{1 + x^{4}}} \\ = \ln x + \ln \sin x - \ln (2 - x) - \frac{1}{2} \ln (1 + x^{4}) . \end{aligned}$
Now we can differentiate both sides:
$\begin{aligned} \frac{1}{y} y^{'} & = \frac{1}{x} + \frac{1}{\sin x} (\frac{d}{d x} \sin x) \\ - \frac{1}{2 - x} (\frac{d}{d x} (2 - x)) - \frac{1}{2 (1 + x^{4})} \frac{d}{d x} (1 + x^{4}) \end{aligned}$
or
$\frac{y^{'}}{y} = \frac{1}{x} + \frac{\cos x}{\sin x} - \frac{(- 1)}{2 - x} - \frac{4 x^{3}}{2 (1 + x^{4})}$
Multiplying both sides by $y$ :
$\begin{aligned} y^{'} & = y [\frac{1}{x} + \frac{\cos x}{\sin x} + \frac{1}{2 - x} - \frac{2 x^{3}}{1 + x^{4}}] \\ = \frac{x \sin x}{(2 - x) \sqrt{1 + x^{4}}} [\frac{1}{x} + \frac{\cos x}{\sin x} + \frac{1}{2 - x} - \frac{2 x^{3}}{1 + x^{4}}] \end{aligned}$
We have found the solution, but we may want to simplify it further:
$\begin{aligned} y^{'} & = \frac{\sin x}{(2 - x) \sqrt{1 + x^{4}}} + \frac{x \cos x}{(2 - x) \sqrt{1 + x^{4}}} \\ + \frac{x \sin x}{(2 - x)^{2} \sqrt{1 + x^{4}}} - \frac{2 x^{4} \sin x}{(2 - x) \sqrt{(1 + x^{4})^{3}}} \\ = \frac{\sin x + x \cos x}{(2 - x) \sqrt{1 + x^{4}}} + \frac{x \sin x}{(2 - x)^{2} \sqrt{1 + x^{4}}} \\ - \frac{2 x^{4} \sin x}{(2 - x) \sqrt{(1 + x^{4})^{3}}} \end{aligned}$