If f(x) = logx(logx), then f'(x) at x = e is equal to:

$y=\log _x(\log x)=\frac{\log (\log x)}{\log x}=\frac{\log t}{t}$,

where t = log x

∴ when x = e, t = 1

$\frac{d y}{d t}=\frac{t . \frac{1}{t}-(\log t)}{t^2}-1=\frac{1-\log t}{t^2}$

$\Rightarrow \frac{d y}{d x}=\frac{d y}{d t} \cdot \frac{d t}{d x}=\frac{1-\log t}{t^2} \times \frac{1}{x}$

$\left.\frac{d y}{d x}\right|_{x=e}=\frac{1-\log 1}{(1)^2} \times \frac{1}{e}=\frac{1}{e}$

Hence (2) is correct answer.