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} $

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