Differentiability with the modulus function in denominator

Question

Prove that for all $x\neq 0$, $f(x)=\frac{1}{1+|x|}$ is differentiable, and that for $x=0, f$ is not differentiable.

My attempt: Using the definition of differentiability, $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{|x|-|x+h|}{h(1+|x+h|)(1+|x|)}$. However I'm not sure where to go from here.

Answer 1

On $\Bbb R\setminus\{0\}$, $f$ is the quotient of two differentiable functions, and therefore it is differentiable there.

On the other hand$$\lim_{x\to0^+}\frac{\frac1{1+|x|}-1}x=\lim_{x\to0^+}\frac1{1+|x|}\cdot\frac{|x|}x=1,$$whereas$$\lim_{x\to0^-}\frac{\frac1{1+|x|}-1}x=\lim_{x\to0^-}\frac1{1+|x|}\cdot\frac{|x|}x=-1.$$Therefore, $\lim_{x\to0}\frac{\frac1{1+|x|}-1}{x-1}$ doesn't exist, which means that $f$ is not differentiable at $0$.

Answer 2

Notice that:

If $x>0$, then $f(x) = \dfrac{1}{1+x}$ and so $f$ is differentiable if $x>0$

If $x<0$, then $f(x) = \dfrac{1}{1-x}$ and so $f$ is differentiable if $x<0$

For $x=0$, we use the lateral derivatives.

$$f'_+(0) = \lim_{x\to 0^+} \dfrac{f(x) - f(0)}{x-0} = \lim_{x\to 0^+} \dfrac{\dfrac{1}{1+x}- 1}{x} = \lim_{x\to 0^+} \dfrac{-x}{x(x+1)} = -1$$

$$f'_-(0) = \lim_{x\to 0^-} \dfrac{f(x) - f(0)}{x-0} = \lim_{x\to 0^-} \dfrac{\dfrac{1}{1-x}- 1}{x} = \lim_{x\to 0^-} \dfrac{x}{x(x+1)} = 1$$

Thus, $f'_+(0)\neq f'_-(0)$, then $f$ isn't differentiable at $x=0$