The function f(x) = log(log x) , x gt 1 , is

The correct answer is Option (1) → strictly increasing on \((1, \infty)\)

By the chain rule,

    \( \displaystyle f'(x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x\log x} \)

For x > 1, both x and log x are positive, so

    \(f'(x) \gt 0\) for every \(x \gt 1\)

The derivative never changes sign, so f is strictly increasing on the whole of \((1, \infty)\); nothing special happens at x = e.