Solve this

(a) $f^{\prime}\left(1^{+}\right)=1$ and (b) $f^{\prime}(1)=-1$

$f(x)=\left|\log _{e} x\right|,=\left\{\begin{array}{l}-\log _{e} x, \text { for } 0

Differentiablity at $x=1$,

we have,

$(L H D$ at $x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$

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

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

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

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

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

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

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