Write the points where

Solution:

Given: $f(x)=\left|\log _{e} x\right|= \begin{cases}-\log _{e} x, & 0

Clearly $f(x)$ is differentiable for all $x>1$ and for all $x<1$. So, we have to check the differentiability at $x=1$.

We have,

(LHD at $x=1$ )

$\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$

$=\lim _{x \rightarrow 1^{-}} \frac{-\log x-\log 1}{x-1}$

$=-\lim _{x \rightarrow 1^{-}} \frac{\log x}{x-1}$

$=-\lim _{h \rightarrow 0} \frac{\log (1-h)}{1-h-1}$

$=-\lim _{h \rightarrow 0} \frac{\log (1-h)}{-h}$

$=-1$

(RHD at x=1)

$=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$

$=\lim _{x \rightarrow 1^{+}} \frac{\log x-\log 1}{x-1}$

$=\lim _{x \rightarrow 1^{+}} \frac{\log x}{x-1}$

$=\lim _{h \rightarrow 0} \frac{\log (1+h)}{1+h-1}$

$=\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}$

 

$=1$

Thus, $(\mathrm{LHD}$ at $x=1) \neq(\mathrm{RHD}$ at $x=1)$

 

So, $f(x)$ is not differentiable at $x=1$.