Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 40

Differentiate $f(x) = \ln \ln \ln x$ and find the domain of $f$.



$
\begin{equation}
\begin{aligned}

\text{if } f(x) =& \ln \ln \ln x, \text{ then}
\\
\\
f'(x) =& \frac{\displaystyle \frac{d}{dx} (\ln \ln (x)) }{\ln \ln (x)}
\\
\\
f'(x) =& \frac{\displaystyle \frac{\displaystyle \frac{d}{dx} (\ln x)}{\ln (x)}}{\ln \ln (x)}
\\
\\
f'(x) =& \frac{\displaystyle \frac{1}{x \ln (x)}}{\ln [\ln (x)]}
\\
\\
f'(x) =& \frac{1}{x \ln (x) \ln [\ln (x)]}

\end{aligned}
\end{equation}
$


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 $(e, \infty)$