Question

If $f\left(x\right)=\frac{x}{1+|x|}$ for $xϵR$ , then $f^{\prime }\left(0\right)$ is :

• A. $0$
• B. $1$
• C. $2$
• D. Does not exist

Answer 1

Answer: (B)

$1$

Given : $f\left(x\right)=\frac{x}{1+|x|}$

If $x<0\Rightarrow |x|=-x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1-x-x(-1)}{(1-x{)}^2}=\frac{1}{(1-x{)}^2}$
$\Rightarrow \left[f^{\prime }\right(x){]}_{x=0}=1$

Again, if $x>0\Rightarrow |x|=x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x}{1+x}\right)=\frac{1}{(1+x{)}^2}$
$\Rightarrow \left[f^{\prime }\right(x){]}_{x=0}=1$

$\Rightarrow f^{\prime }\left(0\right)=1$