Is $|\ln|x||$ differentiable? 
Is $|\ln|x||$ differentiable for all $x$ is defined and continuous? 

I can see that on the graph that it is not differentiable at $-1$ and $1$, but how can I prove it?
So I look at $\lim_{h\to 0+} \frac{|\ln(1-h)|-|\ln(1)|}{h}=\lim_{h\to 0+} \frac{|\ln(1-h)|}{h}$ because it is positive $(0^{+})$ we can say $\lim_{h\to 0+} \frac{\ln(1-h)}{h}$ applying L'Hôpital $\lim_{h\to 0+} \frac{\frac{-1}{1-h}}{1}=-1$?
 A: At points where $f(x)\ne0$ there is no problem in differentiating $|f|$, because the function is positive/negative in a neighborhood of the point.
So all you need is to check where $f(x)=0$. In this case at $1$ and $-1$. However, since the function is clearly even, we can just look at $1$:
$$
\lim_{x\to1^+}\frac{|\log|x|\,|-|\log|1|\,|}{x-1}=
\lim_{x\to1^+}\frac{\log x}{x-1}=1
$$
whereas
$$
\lim_{x\to1^-}\frac{|\log|x|\,|-|\log|1|\,|}{x-1}=
\lim_{x\to1^-}\frac{-\log x}{x-1}=-1
$$
So the function is not differentiable at $1$ and $-1$, but it is everywhere else (provided it is defined to begin with).
Yes, your attempt is good.
A: Assuming $f(x) : \mathbb{R}\setminus \{0\} \to \mathbb{R}$.
We can rewrite $f(x) = \left|\ln\left|x\right|\right|$ as:
$$
  f(x) =
\begin{cases}
\ln(x),  & x \geq 1 \\
-\ln(x), & 1 \geq x > 0 \\
-\ln(-x), & 0 > x \geq -1 \\
\ln(-x), & -1 \geq x
\end{cases}
$$
We see that on each of the intervals $(-\infty, -1), (-1, 0), (0, 1), (1, \infty)$, $f(x)$ is both continuous and differentiable, due to the continuity and differentiability of $\ln(x)$ on $(0, 1)$.
We thus have 2 possible $x$ values where $f$ is not differentiable, $-1$ and $1$. Note that we don't have to consider $x=0$ as $f$ is not defined there. 
Lets look at $x = 1$.
For $x \geq 1$ we have $f'_+(x) = \frac{1}{x}$, and for $1 \geq x > 0$ we have $f'_{\vphantom+-} (x) = -\frac{1}{x}$. Thus:
$$\lim_{x \to 1+} f'_+(x) = 1 \neq \lim_{x \to 1-} f'_{\vphantom+-}(x) = -1$$
A: The facts:


*

*$|x|$ is not differentiable at $x = 0$

*The derivative of $\ln |x|$ vanishes no-where on $\mathbb{R}$


implies that $ |\ln |x| |$ is not differentiable at $x$ for which $\ln |x| = 0$ i.e. $x= \pm 1$
A: Assuming a function from reals to reals:
$$\frac{\text{d}}{\text{d}x}\left(\left|\ln|x|\right|\right)=\frac{\ln|x|\left(\frac{\text{d}}{\text{d}x}\left(\ln|x|\right)\right)}{|\ln|x||}=\frac{\ln|x|\cdot\frac{\frac{\text{d}}{\text{d}x}\left(|x|\right)}{|x|}}{|\ln|x||}=\frac{\ln|x|\cdot\frac{x}{|x|}}{|\ln|x||}=\frac{\ln|x|}{x|\ln|x||}$$
A: The first step would be to differentiate it
Check out the derivative here (hint: $|x| = (x^2)^{0.5}$ for reals)
$\frac{d}{dx} |ln|x|| = \frac{ln|x|}{x|ln|x||}$
$lim_{x -> 0^+} \frac{ln|x|}{x|ln|x||} = + \infty$
while
$lim_{x -> 0^-} \frac{ln|x|}{x|ln|x||} = - \infty$
So the function IS differentiable, but not at certain points. 
Note that the derivative contains the logarithm function which is not defined for zero. To understand why this is not defined at zero, you just need to know what a logarithm is. 
$x = ln(y)$ is the inverse of $y = e^x$
there isn't any number $x$ that will produce zero for $e^x$
so $ln(0)$ is not defined
also the function doesn't make any sense at $+1$ and $-1$, just plug those values into the derivative function and see what happens. 
