Prove Differentiability of $(\ln(1+|x|))^2$

Question

I have been asked to determine in which points $x \in \mathbb R$,

$f: \mathbb R \to \mathbb R, x \to (\ln(1+|x|))^2$ is differentiable.

I know for $x>0$ $f(x)$ is differentiable according to the chain rule, and the same goes for $x<0$. In my attempt to show differentiability (or lack thereof) in $0$ I got stuck.

$\lim_{x \rightarrow 0^{+}} \frac{(\ln(1+x))^2-f(0)}{x}$ = $\lim_{x \rightarrow 0^{+}} \frac{(\ln(1+x))^2}{x}$. My first guess now would be l'Hôpital but I do not have the $\frac{\infty}{\infty} or \frac{0}{0}$ condition. Any suggestions on how I could proceed? Or, is it more appropriate to use the $h \to 0$ limit definition here? Generally speaking I am a bit perplexed why we use two definitions, or not even that - just alternatives - for differentiability , namely $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ and $\lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x-0}$. Is one more suitable than the other in certain circumstances, or is the use of either equivalently suitable? Your help is greatly appreciated.

Answer 1

There are at least two ways to do this. First, you can use the difference quotient definition as you've been doing. You've done the right-hand limit correctly, but when you say you don't have the form $\frac{0}{0},$ you're mistaken. I think you're being misled by the $f(0)$ term, but notice that unless $f(0)=0,$ the function is not continuous at zero, and surely not differentiable there. One application of Hopital's rule gives 0. Then you can compute the left-hand limit in the same way, remembering to put $-x$ in for $|x|$. This also turns out to be $0$, so the answer is yes, the function is differentiable and $f'(0)=0.$

The second way to do it is to differentiate $f$ in $(0,\infty)$ and $(-\infty, 0)$ and observe that both derivates approach $0$ at $0$. It is a consequence of the Mean Value Theorem that under these circumstances, $f$ is continuously differentiable at $0$.

Answer 2

I think this function is not differentiable in $x=0$.

Because for $x>0$ we have $$f'(x)=\frac{2}{1+x}$$ and for $x<0$ we have $$f'(x)=\frac{2}{x-1}$$ and $2\neq-2.$

Answer 3

For $x> 0,$

$$ \frac{(\ln (1+|x|))^2}{x} = x\frac{(\ln (1+x))^2}{x^2} = x\left (\frac{\ln (1+x)}{x}\right )^2.$$

Now $\dfrac{\ln (1+x)}{x} \to 1$ as $x\to 0.$ This follows from the definition of $\ln'(1),$ which equals $1.$ Thus your limit from the right is $0\cdot 1 = 0.$ That $\dfrac{(\ln (1+|x|))^2}{x}\to 0$ from the left follows from this.