Limit Continuity And Differentiability Ques 18
- If $f(x)=\log _x(\log x)$, then $f^{\prime}(x)$ at $x=e$ is
$(1985,2 \mathrm{M})$
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Answer:
Correct Answer: 18.$(\frac{1}{e})$
Solution: Given, $f(x)=\log _x(\log x)$
$\therefore \quad f(x)=\frac{\log (\log x)}{\log x}$
On differentiating both sides, we get
$ \begin{aligned} f^{\prime}(x) & =\frac{(\log x)\left(\frac{1}{\log x} \cdot \frac{1}{x}\right)-\log (\log x) \cdot \frac{1}{x}}{(\log x)^2} \\ \therefore \quad f^{\prime}(e) & =\frac{1 \cdot\left(\frac{1}{1} \cdot \frac{1}{e}\right)-\log (1) \cdot \frac{1}{e}}{(1)^2} \\ \Rightarrow \quad f^{\prime}(e) & =\frac{1}{e} \end{aligned} $
BETA
