STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

$\Rightarrow f(x)=\left\{\frac{x}{1+x} ; x \geq 0 \quad \frac{x}{1-x} ; x \leq 0\right\}$

To check for differentiability of the above function we start off by checking for differentiability

of $f(x)$ at $x=0$ since the functional definition is altered at $x=0 .$ So let's find the right hand

derivative (RHD) and left hand derivative (내D) at $x=0$

$RHD$

$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$\lim _{x \rightarrow 0^{+}}\left(\frac{\frac{x}{1+x}-\frac{0}{1+0}}{x}\right)=\lim _{x \rightarrow 0^{+}} \frac{x}{(1+x) x}=\lim _{x \rightarrow 0^{+}} \frac{1}{1+x}=1$

LHD :

$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \lim _{x \rightarrow 0^{-}}\left(\frac{\frac{x}{1-x}-\frac{0}{1-0}}{x}\right)=\lim _{x \rightarrow 0^{-}} \frac{x}{(1-x) x}=\lim _{x \rightarrow 0^{-}} \frac{1}{1-x}=1$

$LHD = RHD$

The function is differentiable at $x=0$

Also, $f(x)=\frac{x}{1+x}$ is a well defined rational function of ${ }^{\prime} x^{\prime}$ for all $x$ greater than zero and hence will

be continuous and differentiable through out its domain. similarly, $f(x)=\frac{x}{1+x}$ is a well defined rational function of $^{\prime} x^{\prime}$ for all $^{\prime} x^{\prime}$ less than zero and hence

will be continuous ans differentiable through out its domain. So, we can conclude from the above calculation and discussion that $f(x)=\frac{x}{1+|x|}$ is

differentiable everywhere. $\therefore f(x)=\frac{x}{1+|x|}$ is differentiable in $x \in(-\infty, \infty)$