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

Solution:

We have,

$f\left(x\right)=\frac{x}{1+\left|x\right|}=\frac{g\left(x\right)}{h\left(x\right)},\text{ where }g\left(x\right)=x,h\left(x\right)=1+\left|x\right|\text{. }$

$\text{ Tearly, }g\left(x\right)\text{ is differentiable on }(-\mathrm{\infty },\mathrm{\infty })\text{. The domain of }h\left(x\right)\text{ is }$

$-x,\mathrm{\infty }\left)\text{. Since }\right|x|\text{ is not differentiable at }x=0\text{. Therefore, }h(x)$

$\text{ not differentiable at }x=0\text{. Thus, }f\left(x\right)\text{ is differentiable on }$

$-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$

$\text{ At }x=0\text{, we have }$

$\underset{x\to 0}{lim} \frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{lim} \frac{\frac{x}{1+\left|x\right|}}{x}=\underset{x\to 0}{lim} \frac{1}{1+\left|x\right|}=1$

$\text{ Therefore, }f\left(x\right)\text{ is differentiable at }x=0\text{. }$

$\text{ Hence, }f\text{ is differentiable on }(-\mathrm{\infty },\mathrm{\infty })\text{. }$