Question

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

• A. $1$
• B. $\frac{1}{e}$
• C. $\frac{1}{2e}$
• D. $0$

Answer 1

The correct option is C

$\frac{1}{2e}$

Explanation for the correct option:

Step 1: Simplifying the given equation:

Given that,

$\begin{array}{rcl}f\left(x\right)& =& {\log }_{x^2}\left({\log }_{e}x\right)\\ & =& \frac{1}{2}\left[{\log }_x\left({\log }_{e}x\right)\right]\left[\because {\log }_{a^n}b=\frac{1}{n}{\log }_{a}b\right]\\ & =& \frac{1}{2}\left[\frac{{\log }_e\left({\log }_{e}x\right)}{{\log }_{e}x}\right]\left[\because {\log }_{a}b=\frac{\log b}{\log a}\right]\end{array}$

Step 2: Finding the value of $f\text{'}\left(x\right)$ at $x=e$:

Differentiate the above equation with respect to $x$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{2}\left[\frac{{\log }_{e}x\left({\displaystyle \frac{1}{lo{g}_{e}x}}\left({\displaystyle \frac{1}{x}}\right)-{\log }_e\left({\log }_{e}x\right)\left({\displaystyle \frac{1}{x}}\right)\right)}{{\left({\log }_{e}x\right)}^2}\right]\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]\\ & =& \frac{1}{2}\left[\frac{{\displaystyle \frac{1}{x}}-{\log }_e\left({\log }_{e}x\right)\left({\displaystyle \frac{1}{x}}\right)}{{\left({\log }_{e}x\right)}^2}\right]\end{array}$

Substitute $x=e$ in the above differentiation, we get

$\begin{array}{rcl}f\text{'}\left(e\right)& =& \frac{1}{2}\left[\frac{{\displaystyle \frac{1}{e}}-{\log }_e\left({\log }_{e}e\right){\displaystyle \frac{1}{e}}}{{\left({\log }_{e}e\right)}^2}\right]\\ & =& \frac{1}{2e}\left[\because {\log }_{a}a=1,{\log }_{e}1=0\right]\end{array}$

Hence, the correct option is C.