# Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

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

A. 0

B. 1

C. $1 / \mathrm{e}$

D $1 / 2 \mathrm{e}$

Solution:

$f(x)=\log _{x} 2(\log x)$

Changing the base, we get

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\log (\log \mathrm{x})}{\log \mathrm{x}^{2}}$

$\because \log _{b} a=\frac{\log a}{\log b}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\log (\log \mathrm{x})}{2 \cdot \log \mathrm{x}}$

So, $f^{\prime}(x)=\frac{1}{2}\left\{\frac{1}{\log x}\left[\frac{d}{d x}\{\log (\log x)\}\right]+\log (\log x)\left[\frac{d}{d x}\left\{\frac{1}{\log x}\right\}\right]\right\}$

$\Rightarrow f^{\prime}(x)=\frac{1}{2}\left\{\frac{1}{\log x}\left[\frac{1}{\log x} \cdot \frac{1}{x}\right]+\log (\log x)\left[-\left(\frac{1}{\log x}\right)^{2} \cdot \frac{1}{x}\right]\right\}$

Putting $x=e$, we get

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2}\left\{\frac{1}{\log \mathrm{e}}\left[\frac{1}{\log \mathrm{e}} \cdot \frac{1}{\mathrm{e}}\right]+\log (\log \mathrm{e})\left[-\left(\frac{1}{\log \mathrm{e}}\right)^{2} \cdot \frac{1}{\mathrm{e}}\right]\right\}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2}\left\{\left[\frac{1}{(\log \mathrm{e})^{2}} \cdot \frac{1}{\mathrm{e}}\right]+\log (\log \mathrm{e})\left[-\left(\frac{1}{\log \mathrm{e}}\right)^{2} \cdot \frac{1}{\mathrm{e}}\right]\right\}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2}\left\{\left[\frac{1}{1^{2}} \cdot \frac{1}{\mathrm{e}}\right]+\log (1)\left[-\left(\frac{1}{1}\right)^{2} \cdot \frac{1}{\mathrm{e}}\right]\right\}(\because \log \mathrm{e}=1)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2}\left\{\left[\frac{1}{1^{2}} \cdot \frac{1}{e}\right]+0 \cdot\left[-\left(\frac{1}{1}\right)^{2} \cdot \frac{1}{e}\right]\right\}(\because \log 1=0$

$\therefore f^{\prime}(e)=\frac{1}{2 e}$