Differentiability of $\log|x|$ It is known that
$$\dfrac{d}{dx}\log f(x) = \dfrac{1}{f(x)} f'(x)$$
but how does it is not applicable for $\log |x|$ as $|x|$ is not differentiable?
 A: Since $\log(|x|)$ is not defined at $x=0$, we may investigate the differentiability for $x\not=0$. We  distinguish two cases depending on the sign of $x$.
For $x>0$, we know that $|x|=x$ and
$$\frac{d}{dx}\left(\log(|x|)\right)=\frac{d}{dx}\left(\log(x)\right)=\frac{1}{x}.$$
On the other hand, for $x<0$, we have that $|x|=-x$ and therefore
$$\frac{d}{dx}\left(\log(|x|)\right)=\frac{d}{dx}\left(\log(-x)\right)=\frac{1}{(-x)}\cdot (-1)=\frac{1}{x}.$$
So we may conclude that $\log(|x|)$ is differentiable for $x\not=0$, and its derivative is
$$\frac{d}{dx}\left(\log(|x|)\right)=\frac{1}{x}.$$
A: $\frac d {dx} \log|x|=\frac 1 x$ if $x \neq0$ but the derivative at $0$ does not even makes sense since the function is not defined at $0$. 
A: For non-zero x, we could still use the formula $\frac{d}{dx}\log f(x)=\frac{f'(x)}{f(x)}$
Since $f'(x)=\begin{cases}1&x>0\\-1&x<0\end{cases}$ for $f(x)=|x|$
so $\frac{d}{dx}\log |x|=\begin{cases}\frac{1}{|x|}&x>0\\\frac{-1}{|x|}&x<0\end{cases}=\frac{1}{x}$ given $x\neq 0$
A: First things first, no expression will work at $x=0$, since $\log|x|$ is not defined there. Also, $|x|$ is differentiable everywhere except at $x=0$, so things should work fin elsewhere.
It's often easier to work with $|x|$ as a piecewise-defined function, so:


*

*If $x>0, |x|=x$, you can apply the chain rule there and get $f'(x)=\frac{1}{|x|}=\frac{1}{x}$

*If $x<0, |x|<x$, you can apply the chain rule there and get $f'(x)=\frac{-1}{|x|}=\frac{-1}{-x}=\frac{1}{x}$
Long story short, $$f'(x) = \frac{1}{x}, x \neq 0$$
A: The function $x\mapsto|x|$ is differentiable over $\mathbb{R}\setminus\{0\}$, so there is no problem in applying the chain rule: for $x\ne0$,
$$
\frac{d}{dx}\log\lvert x\rvert=\frac{1}{\lvert x\rvert}\frac{\lvert x\rvert}{x}=\frac{1}{x}
$$
