Question
यदि $f(x) = {\log _x}(\log x),$ तो $x = e$ पर $f'(x)$है
==> $f'(x) = \frac{{\frac{1}{x} - \frac{1}{x}\log (\log x)}}{{{{(\log x)}^2}}} \Rightarrow f'(e) = \frac{{\frac{1}{e} - 0}}{1} = \frac{1}{e}$.
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