If $\mathrm{f}(x)=\log _{x^2}\left(\log _{\mathrm{e}} x\right)$, then $\mathrm{f}^{\prime}(x)$ at $x=\mathrm{e}$ is

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Aicharya · Accepted Answer

$\begin{aligned} f(x) & =\log _{x^2}\left(\log _{\mathrm{e}} xight) \ & =\frac{\log \left(\log _{\mathrm{e}} xight)}{\log x^2} \ & =\frac{\log \left(\log _{\mathrm{e}} xight)}{2 \log x}\end{aligned}$ $	herefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{2}\left[\frac{\log x \cdot \frac{1}{\log _{\mathrm{e}} x} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}(\log x)-\log \left(\log _{\mathrm{e}} xight) \cdot \frac{1}{x}}{(\log x)^2}ight]$ $	herefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{2}\left[\frac{\frac{1}{x}-\frac{\log \left(\log _{\mathrm{e}} xight)}{x}}{(\log x)^2}ight]$ $	herefore \quad \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2}\left[\frac{\frac{1}{\mathrm{e}}-0}{(1)^2}ight]=\frac{1}{2 \mathrm{e}}$

If $\mathrm{f}(x)=\log _{x^2}\left(\log _{\mathrm{e}} x\right)$, then $\mathrm{f}^{\prime}(x)$ at…

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