## Differentiation of a Logarithm

#### Show the derivation of the derivative of the natural logarithm

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

Step 1: $y+\Delta y=\ln(x+\Delta x)$ Step 2:
\begin{align} \Delta y & =\ln(x+\Delta x)-\ln x\\ & =\ln\left(\frac{x+\Delta x}{x}\right)\\ & =\ln\left(1+\frac{\Delta x}{x}\right)\end{align}
Step 3:
\begin{align} \frac{\Delta y}{\Delta x} & =\frac{1}{\Delta x}\ln\left(1+\frac{\Delta x}{x}\right)\\ & =\ln\left(1+\frac{\Delta x}{x}\right)^{1/\Delta x}\\ & =\frac{1}{x}\ln\left(1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}\end{align}
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{align} \frac{dy}{dx} & =\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}\\ & =\lim_{\Delta x\to0}\left[\frac{1}{x}\ln\left(1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}\right]\\ & =\frac{1}{x}\lim_{\Delta x\to0}\left[\ln\left(1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}\right]\\ & =\frac{1}{x}\ln\left[\lim_{\Delta x\to0}\left(1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}\right]\\ & =\frac{1}{x}\ln e\\ & =\frac{1}{x}.\end{align}
[Note than when $$\Delta x\to0$$, $$\frac{\Delta x}{x}\to0$$. Therefore, $${\displaystyle \lim_{\Delta x\to0}\left(1+\frac{\Delta x}{x}\right)^{\frac{x}{\Delta x}}=e}$$ from placing $$u=\frac{\Delta x}{x}$$ in $${\displaystyle \lim_{u\to0}(1+u)^{1/u}=e}$$
(see here)].

Hence
$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\ln x=\frac{1}{x},\qquad(x>0).\tag{a}}$

Now consider $$y=\log_{a}x$$. What is $$dy/dx$$?

We write $y=\log_{a}x=\frac{\ln x}{\ln a}.$ Therefore
\begin{align} \frac{dy}{dx} & =\frac{1}{\ln a}\frac{d}{dx}\ln x\\ & =\frac{1}{\ln a}\frac{1}{x}.\end{align}
$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\log_{a}x=\frac{1}{x\ln a},\qquad(x>0).\tag{b}}$

• If $$u(x)>0$$ is a differentiable function, applying the chain rule to (a) produces: $\dfrac{d}{dx}\ln u=\frac{1}{u}\frac{du}{dx}$
• 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 $$dy/dx$$
\begin{align} \frac{dy}{dx} & =\frac{d}{dx}\ln(\overbrace{x^{2}+1}^{u})\\ & =\frac{1}{u}\frac{du}{dx}\\ & =\frac{1}{x^{2}+1}(2x)\\ & =\frac{2x}{x^{2}+1}\end{align}
The slope of the tangent $$m_{\text{tan}}$$ when $$x=0$$ is obtained by substituting 0 for $$x$$ in the above equation: $m_{\text{tan}}=\left.\frac{2x}{x^{2}+1}\right|_{x=0}=0$ 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.

Example 2

Find $$\dfrac{d}{dx}\left[\ln\dfrac{\sqrt{x+1}(2+\cos x)}{x^{2}}\right]$$

Solution

\begin{align} \dfrac{d}{dx}\left[\ln\dfrac{\sqrt{x+1}(2+\cos x)}{x^{2}}\right] & =\dfrac{d}{dx}\left[\ln\sqrt{x+1}+\ln(2+\cos x)-\ln x^{2}\right]\\ & =\dfrac{d}{dx}\left[\frac{1}{2}\ln(x+1)+\ln(2+\cos x)-2\ln x\right]\\ & =\frac{1}{2(x+1)}+\frac{1}{2+\cos x}\left(\dfrac{d}{dx}\cos x\right)-\frac{2}{x}\\ & =\frac{1}{2(x+1)}-\frac{\sin x}{2+\cos x}-\frac{2}{x}\\\end{align}

Now an important example:

Example 3

Find $$\dfrac{d}{dx}\ln|x|$$.

Solution

We consider two cases:

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

From the above example, we conclude

$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\ln|x|=\frac{1}{x},\tag{c}}$

More generally, if $u$ is a differentiable function of $x$ using the chain rule we have

$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\ln|u|=\frac{1}{u}\frac{du}{dx}.\tag{d}}$

Example 4

Find $$\dfrac{d}{dx}\ln|\cos x|$$.

Solution

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

\begin{align} \dfrac{d}{dx}\ln|\cos x| & =\frac{1}{u}\frac{du}{dx}\\ & =\frac{1}{\cos x}(-\sin x)\\ & =-\tan x.\end{align}

## 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:

1. Take the natural logarithms of both sides of $$y=f(x)$$: $\ln y=\ln f(x).$
2. Expand the right side using the properties of logarithms.
3. Take the derivatives of both sides.
4. Isolate $$dy/dx$$ and replace $$y$$ by $$f(x)$$.

Notice that if $$f(x)<0$$ for some $$x$$, then $$\ln f(x)$$ is not defined for those values of $$x$$. So to use logarithmic differentiation, we should technically start with $|y|=\left|f(x)\right|$, and then take the natural logarithm and derivative of each side. However, because the derivative of $\ln |u|$  is the same as the derivative of $\ln u$ (when $u>0$), we can ignore the fact the $\ln u$ is not defined for $u\leq 0$.

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{align} \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{align}
Now we can differentiate both sides:
\begin{align} \frac{1}{y}y’ & =\frac{1}{x}+\frac{1}{\sin x}\left(\frac{d}{dx}\sin x\right)\\ & \hspace{1em}-\frac{1}{2-x}\left(\frac{d}{dx}(2-x)\right)-\frac{1}{2(1+x^{4})}\frac{d}{dx}(1+x^{4})\end{align}
or
$\frac{y’}{y}=\frac{1}{x}+\frac{\cos x}{\sin x}-\frac{(-1)}{2-x}-\frac{4x^{3}}{2(1+x^{4})}$
Multiplying both sides by $$y$$:
\begin{align} y’ & =y\left[\frac{1}{x}+\frac{\cos x}{\sin x}+\frac{1}{2-x}-\frac{2x^{3}}{1+x^{4}}\right]\\ & =\frac{x\sin x}{(2-x)\sqrt{1+x^{4}}}\left[\frac{1}{x}+\frac{\cos x}{\sin x}+\frac{1}{2-x}-\frac{2x^{3}}{1+x^{4}}\right]\end{align}
We have found the solution, but we may want to simplify it further:
\begin{align} y’ & =\frac{\sin x}{(2-x)\sqrt{1+x^{4}}}+\frac{x\cos x}{(2-x)\sqrt{1+x^{4}}}\\ & \qquad+\frac{x\sin x}{(2-x)^{2}\sqrt{1+x^{4}}}-\frac{2x^{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}}}\\ & \qquad-\frac{2x^{4}\sin x}{(2-x)\sqrt{(1+x^{4})^{3}}}\end{align}