Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 34

Differentiate $f(x) = \ln \ln \ln x$ and find the domain of $f$
For the domain of $f$, we want $\ln \ln x > 0$

$
\begin{equation}
\begin{aligned}
\ln \ln x &> 0\\
\\
e^{\ln \ln x} &> e^0\\
\\
\ln x &> 1\\
\\
e^{\ln x} &> e^1\\
\\
x &> e^1
\end{aligned}
\end{equation}
$

Therefore, the domain of $f$ is $(e, \infty)$
Solving for $f'$

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{d}{dx} \ln \ln \ln x\\
\\
f'(x) &= \frac{1}{\ln\ln x} \cdot \frac{d}{dx} (\ln \ln x)\\
\\
f'(x) &= \frac{1}{\ln \ln x} \cdot \frac{1}{\ln x} \frac{d}{dx} \ln x\\
\\
f'(x) &= \frac{1}{\ln \ln x} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}\\
\\
f'(x) &= \frac{1}{ x \ln x \ln \ln x}
\end{aligned}
\end{equation}
$