Differentiability of $f (x)=\frac{x}{1+|x|}$ Discuss continuity and differentiability of following function
$$f (x)=\frac{x}{1+|x|}$$
Where x is any real number
My attempt:
Being a rational funtion it is continous everywhere. Also its dervative exists for all x. So it is differentiable everywhere. I am not sure so I need suggestion. Thanks
 A: Hint: by definition,
$$
f(x)=\cases{ \frac{x}{1-x} &if $x<0$ \cr
\frac{x}{1+x} &if $x \geq 0$.\cr}
$$
Now compute the derivative at $x=0$. now the two "pieces" of $f$ are rational functions; however the differentiability at the junction point $x=0$ should be investigated directly.
A: It's not a rational function. However, it's everywhere defined because $|x|\ne-1$ for all $x$.
The function is clearly continuous, as composition of continuous functions.
If you write the function as
$$
f(x)=\begin{cases}
\dfrac{x}{1+x}=1-\dfrac{1}{1+x} & \text{if $x>0$}\\[8px]
0 & \text{if $x=0$}\\[4px]
\dfrac{x}{1-x}=\dfrac{1}{1-x}-1 & \text{if $x<0$}
\end{cases}
$$
you can see that
$$
f'(x)=\begin{cases}
\dfrac{1}{(1+x)^2} & \text{if $x>0$}\\[4px]
\dfrac{1}{(1-x)^2} & \text{if $x<0$}
\end{cases}
$$
and it follows from this that the function is also differentiable at $x=0$. Why?
A: there is only point there should be any question of differentiability; that is at $x = 0.$ let us see how $f(x) = \dfrac{x}{1+|x|}$ behaves for $x = 0 + \cdots.$
we can see that $$f(x)= \frac{x}{1+|x|} = x + \cdots  \text{ for } x = 0+ \cdots$$
therefore $f$ is differentiable at $x = 0$ and $f^\prime(0) = 1.$
