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  4. The set of all points where the function f(x) = (x/ 1 + |x |) is differentiable, is

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

Limits and Derivatives

Solution:

$f\left(x\right) = \frac{x}{1-x} $ if $x \le 0$ and $ f\left(x\right) = \frac{x}{1+x} $ if $x \ge 0$
At x = 0, left handed derivative
$ = \lim_{h\to0} \frac{f\left(0-h\right) -f\left(0\right)}{-h}$
$= \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} = \lim_{h \to0} \frac{\frac{-h}{1+h}-0}{-h}$
$ = \lim_{h\to0} \frac{1}{1+h} = 1 $
and right handed derivative
$= \lim_{h \to0} \frac{f\left(0+h\right)-f\left(0\right)}{h} $
$= \lim_{h \to0} \frac{f\left(h\right) -f\left(0\right)}{h} = \lim_{h \to0} \frac{\frac{h}{1+h}-0}{h}$
$ = \lim_{h\to0} \frac{1}{1+h} = 1$
$\therefore \, l .h. $derivative = r.h. derivative at x = 0
Hence f(x) is differentiable at x = 0
Further $f' (x) = \frac{1}{(1 - x)^2}$ , where $x \le 0$
and $f'(x) = \frac{1}{(1 + x)^2}$, where $x \ge 0 $
Thus $f'(x)$ exists for all values of x in the interval $( - \infty , \infty)$