The set of all points, where the derivative of the functions $\mathrm{f}(x)=\frac{x}{1+|x|}$ exists, is

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Aicharya · Accepted Answer

$\mathrm{f}(x)$ can be written as $\begin{aligned} & \mathrm{f}(x)=\left\{\begin{array}{l} \frac{x}{1-x}, x \leq 0 \ \frac{x}{1+x}, x>0 \end{array}ight. \ & \mathrm{f}^{\prime}(x)= \begin{cases}\frac{(1-x)+x}{(1+x)^2}, & x \leq 0 \ \frac{(1+x)-x}{(1+x)^2}, & x>0\end{cases} \ & \mathrm{f}^{\prime}(x)=\frac{1}{(1+x)^2} \forall x \in(-\infty, \infty) \end{aligned}$ $	herefore \quad$ Derivative of $\mathrm{f}(x)$ exists $\forall x \in(-\infty, \infty)$

The set of all points, where the derivative of the functions $\mathrm{f}(x)=\frac{x}{1+|x|}$ exists, is

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