How does one take the derivative of $\ln|\ln|x||$? I have seen a solution based on the following substitution: $t = \ln|\ln|x||$ $\Leftrightarrow$ $e^{t}$ = $\ln|x|$. I can however not understand why the absolute value drops, I mean should it not be: $t = \ln|\ln|x||$ $\Leftrightarrow$ $e^{t}$ = $|\ln|x||$?
 A: You can distinguish four cases:


*

*$ x \in (-\infty ,-1) \Rightarrow \ln |\ln|x|| = \ln (\ln(-x))$

*$x \in (-1,0) \Rightarrow \ln |\ln|x|| = \ln(-\ln(-x))$

*$x \in (0,1) \Rightarrow \ln |\ln|x|| = \ln(-\ln(x))$

*$x \in (1,\infty) \Rightarrow \ln |\ln|x|| = \ln(\ln(x))$.


Then you know how to take derivatives in the individual intervals.
A: If $f$ is a differentiable function and non-zero on an interval $I$ then $f(x)$ has constant sign on that interval: $|f(z)| = s f(x)$ with $s=1$ or $s =-1$. It follows that
$$
 (\ln|f(x)|)' = (\ln (s f(x)))' = \frac{s f'(x)}{sf(x)} = \frac{f'(x)}{f(x)}
$$
on $I$. Applying this reasoning twice (first with  $f(x) = \ln |x|$ and then with $f(x) = x$) gives
$$
  (\ln| \ln |x||)' = \frac{(\ln |x|)'}{\ln |x|} = \frac{1}{x \ln |x|}
$$
on each interval where the function is defined, that is on $\Bbb R \setminus \{-1,0,1 \}$.
A: The function is even, so its enough to consider $x>0$; then $f'(-x) = -f'(x)$. Thus we can dispense with the inner absolute value. So really its just
$$ \ln|\ln x| = \begin{cases} \ln \ln x & x > 1 \\ \ln \ln (1/x) & 0<x<1\end{cases}$$
by chain rule
$$ (\ln|\ln x|)' =  \begin{cases}  \frac1{x\ln x} & x > 1 \\ \frac{-1}{x \ln (1/x)} & 0<x<1\end{cases} =  \frac{1}{x \ln x}$$
Note that  $e^t =  \ln x$ implies $e^t = | \ln x|$, but not the other way round: $x=1/e, t=0$ is a solution to $e^t = | \ln x|$, but not  to $e^t =  \ln x$.
(The same thing happens in the $\ln$ world: If you had $x$ and $t$ such that $t = \ln  \ln x$, then since $\ln$ only takes in positive values, this actually implies $t = \ln |\ln x|$.)
