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The set of all points where the function $f(x)=\large\frac{x}{(1+|x|)}$ is differentiable is

$\begin{array}{1 1}(a)\;[-\infty,\infty]&(b)\;[0,\infty]\\(c)\;[-\infty,0]\cup [0,\infty]&(d)\;[0,\infty]\end{array}$

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The given function is $f(x)=\large\frac{x}{1+|x|}$
$\Rightarrow \left\{\begin{array}{1 1}\large\frac{x}{1-x},&x < 0\\\large\frac{x}{1+x},&x \geq 0\end{array}\right.$
For $x < 0$
$f'(x)=\large\frac{1(1-x)-(-1)x}{(1-x)^2}$
$\qquad=\large\frac{1}{(1-x)^2}$
For $x > 0$
$f'(x)=\large\frac{1(1+x)-1x}{(1-x)^2}$
$\qquad=\large\frac{1}{(1+x)^2}$
For $x = 0$
$Lf'(0)=\lim\limits_{h\to 0}\large\frac{f(0+h)-f(0)}{-h}$
$\qquad\;\;\;=\lim\limits_{h\to 0}\large\frac{\Large\frac{-h}{1+h}-\normalsize 0}{-h}$$=1$
$Rf'(0)=\lim\limits_{h\to 0}\large\frac{f(0+h)-f(0)}{h}$
$\qquad=\lim\limits_{h\to 0}\large\frac{\Large\frac{h}{1+h}-\normalsize 0}{h}$$=1$
$\Rightarrow Lf'(0)=Rf'(0)$
$\Rightarrow f$ is differentiable at x=0
Hence f is differentiable in $[-\infty,\infty]$
Hence (a) is the correct answer.
answered Jan 4, 2014 by sreemathi.v
 

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