What is the derivative of $\ln|x|$? I am somewhat confused about finding the derivative of $\ln|x|$. Is it $1/|x|$ or $1/x$? 
What if instead of $x$ we have a general function $f(x)$? Most importantly: why?
 A: It's $\,\dfrac 1x$. Indeed, if $x<0$, $\bigl(\ln\lvert x \rvert\bigr)'=\bigl(\ln(-x)\bigr)'=\dfrac1{-x}\times -1$ by the chain rule.
A: Consider 2 cases:
$x>0$ then $|x|=x$ so $(\ln x)'=\dfrac1x$. On the other hand
$x<0$ then $|x|=-x$ so $(\ln -x)'=\dfrac1{-x} \cdot (-1)=\dfrac1x$.
So $(\ln |x|)'=\dfrac1x$.
Now to find $(\ln |f(x)|)'$ just use chain rule and get $\dfrac{1}{f(x)} \cdot f'(x)=\dfrac{f'(x)}{f(x)}$
A: on negative points the $\ln |x|$ is decreasing, so its derivative is negative.  $\frac1x$ is the correct choice.
$$\frac{d}{dx}\ln|f(x)|=\frac{df(x)}{dx}\times \frac1{f(x)}$$
If you agree that on positive points:
$$\frac{d}{dx}\ln(x)=\frac1x$$
for negative points:
$$\frac{d}{dx}\ln|x|=\frac{d}{dx}\ln(-x)=\frac{-1}{-x}=\frac1x$$
A: It's $\frac{1}{x}$ (for positive $x$ value, for negative $x$ you would need the absolute value) , the derivative itself can be derived from the fact that derivative of $e^{x}$ is also $e^{x}$ (by letting $y = ln(x)$, so $x = e^{y}$, then you can derive the derivative of $\ln(x)$ with implicit differentiation), you can also do the formal definition of derivative to get it, using logarithm properties and also the limit definition of $e^{x}$ (you will need those two thing if you're using the formal definition of derivative)
