What is the set of all points, where the function $$f\left( x \right) = \frac{x}{{1 + \left| x \right|}}$$    is differentiable ?

Solution :
Given $$f\left( x \right) = \frac{x}{{1 + \left| x \right|}}$$
\[ = \left\{ \begin{array}{l} \frac{x}{{1 - x}},\,\,\,\,x < 0\\ \frac{x}{{1 + x}},\,\,\,\,x \ge 0 \end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\because\left| x \right| = \left\{ \begin{array}{l} \,\,\,x\,\,\,\,{\rm{if}}\,x \ge 0\\ - x\,\,\,\,{\rm{if}}\,x < 0 \end{array} \right.} \right)\]
$$\eqalign{ & \therefore {\text{ L}}{\text{.H}}{\text{.D}}{\text{.}} = f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( { - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{1 + \left| { - h} \right|}} - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{1 + h}} - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + h}} \cr & = 1 \cr & {\text{and R}}{\text{.H}}{\text{.D}}{\text{.}} = f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{h}{{1 + h}} - 0}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + h}} \cr & = 1 \cr & {\text{Since, L}}{\text{.H}}{\text{.D}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr & \therefore \,f\left( x \right){\text{ is differentiable at }}x = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is differentiable in }}\left( { - \infty ,\,\infty } \right){\text{. }} \cr} $$