Question

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

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

Answer 1

The correct option is A

$\frac{1}{e}$

Explanation for the correct option:

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

Given that,

$\begin{array}{rcl}f\left(x\right)& =& {\log }_x\left(\ln x\right)\\ & =& \frac{lo{g}_e\ln x}{{\log }_{e}x}\left[\because {\log }_{\mathit{a}}\mathit{b}=\frac{\log \mathit{b}}{\log \mathit{a}}\right]\\ & =& \frac{\ln \left(\ln x\right)}{\ln x}\end{array}$

Differentiate the above equation with respect to $x$,

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

Now, substitute $x=e$ in the above differentiation,

$\begin{array}{rcl}f\text{'}\left(e\right)& =& \frac{{\displaystyle \frac{1}{e}}-\ln \left(1\right){\displaystyle \frac{1}{e}}}{1}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{ln}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{lne}\mathbf{=}\mathbf{1}\mathbf{]}\\ & =& \frac{1}{e}\end{array}$

Hence, the correct option is A.