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If $f(x)=\log _{x^{2}}(\log x)$, then $f^{\prime}(x)$ at $x=e$ is
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Verified Answer
The correct answer is:
$\frac{1}{2 e}$
We have, $f(x)=\log _{x^{2}}(\log x)$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[\log _{x}(\log x)\right] \quad\left[\because \log _{a^{n}} b=\frac{1}{n} \log _{a} b\right]$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[\frac{\log (\log x)}{\log x}\right] \quad\left(\because \log _{a} b=\frac{\log b}{\log a}\right)$
$\Rightarrow f^{\prime}(x)=\frac{1}{2}\left[\frac{\log (x) \frac{d}{d x}[\log (\log (x))]-\log (\log x) \frac{d}{d x} \log x}{(\log x)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\log (x) \frac{1}{\log x} \times \frac{1}{x}-\left(\log (\log x) \frac{1}{x}\right)}{(\log x)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\frac{1}{x}-\frac{\log (\log (x))}{x}}{(\log x)^{2}}\right]$
$\Rightarrow \quad f^{\prime}(x)=\frac{1}{2}\left[\frac{1-\log \log (x)}{x(\log x)^{2}}\right]$
$\Rightarrow \quad f^{\prime}(e)=\frac{1}{2}\left[\frac{1-\log \log (e)}{e(\log e)^{2}}\right]=\frac{1}{2}\left[\frac{1-0}{e}\right]=\frac{1}{2 e}$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[\log _{x}(\log x)\right] \quad\left[\because \log _{a^{n}} b=\frac{1}{n} \log _{a} b\right]$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[\frac{\log (\log x)}{\log x}\right] \quad\left(\because \log _{a} b=\frac{\log b}{\log a}\right)$
$\Rightarrow f^{\prime}(x)=\frac{1}{2}\left[\frac{\log (x) \frac{d}{d x}[\log (\log (x))]-\log (\log x) \frac{d}{d x} \log x}{(\log x)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\log (x) \frac{1}{\log x} \times \frac{1}{x}-\left(\log (\log x) \frac{1}{x}\right)}{(\log x)^{2}}\right]$
$=\frac{1}{2}\left[\frac{\frac{1}{x}-\frac{\log (\log (x))}{x}}{(\log x)^{2}}\right]$
$\Rightarrow \quad f^{\prime}(x)=\frac{1}{2}\left[\frac{1-\log \log (x)}{x(\log x)^{2}}\right]$
$\Rightarrow \quad f^{\prime}(e)=\frac{1}{2}\left[\frac{1-\log \log (e)}{e(\log e)^{2}}\right]=\frac{1}{2}\left[\frac{1-0}{e}\right]=\frac{1}{2 e}$
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