For the function $f(x)=\left\{\begin{array}{cc}\frac{x}{1+e^{\frac{1}{x}}}, & x
eq 0 \ 0, & x=0\e

Question

For the function $f(x)=\left\{\begin{array}{cc}\frac{x}{1+e^{\frac{1}{x}}}, & x 
eq 0 \ 0, & x=0\end{array} ;ight.$ the derivative from the right $f^{\prime}\left(0^{+}ight)=$ and the derivative from the left $f^{\prime}\left(0^{-}ight)=$

Professor PC Thomas Classes · Answer

$R f^{\prime}(0)=f^{\prime}\left(0^{+}ight)=\lim _{h ightarrow 0}\left[\frac{\frac{h}{1+e^{\frac{1}{h}}}-0}{h}ight]$ $$ =\lim _{h ightarrow 0} \frac{1}{1+e^{\frac{1}{h}}}=0 $$ $L f^{\prime}(0)=f^{\prime}\left(0^{-}ight)=\lim _{h ightarrow 0}\left(\frac{\frac{-h}{1+e^{-\frac{1}{h}}}-0}{-h}ight)$\ $=\lim _{h ightarrow 0}\left(\frac{1}{1+\frac{1}{e^{1 / h}}}ight)=\frac{1}{1+0}=1$\ $	herefore \quad f^{\prime}\left(0^{+}ight)=0$ and $f^{\prime}\left(0^{-}ight)=1$