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The set of all points, where the derivative of the functions $\mathrm{f}(x)=\frac{x}{1+|x|}$ exists, is
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Verified Answer
The correct answer is:
$(-\infty, \infty)$
$\mathrm{f}(x)$ can be written as
$\begin{aligned}
& \mathrm{f}(x)=\left\{\begin{array}{l}
\frac{x}{1-x}, x \leq 0 \\
\frac{x}{1+x}, x>0
\end{array}\right. \\
& \mathrm{f}^{\prime}(x)= \begin{cases}\frac{(1-x)+x}{(1+x)^2}, & x \leq 0 \\
\frac{(1+x)-x}{(1+x)^2}, & x>0\end{cases} \\
& \mathrm{f}^{\prime}(x)=\frac{1}{(1+x)^2} \forall x \in(-\infty, \infty)
\end{aligned}$
$\therefore \quad$ Derivative of $\mathrm{f}(x)$ exists $\forall x \in(-\infty, \infty)$
$\begin{aligned}
& \mathrm{f}(x)=\left\{\begin{array}{l}
\frac{x}{1-x}, x \leq 0 \\
\frac{x}{1+x}, x>0
\end{array}\right. \\
& \mathrm{f}^{\prime}(x)= \begin{cases}\frac{(1-x)+x}{(1+x)^2}, & x \leq 0 \\
\frac{(1+x)-x}{(1+x)^2}, & x>0\end{cases} \\
& \mathrm{f}^{\prime}(x)=\frac{1}{(1+x)^2} \forall x \in(-\infty, \infty)
\end{aligned}$
$\therefore \quad$ Derivative of $\mathrm{f}(x)$ exists $\forall x \in(-\infty, \infty)$
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