Question

If $f\left(x\right)={\log }_e\left({\log }_{e}x\right)$, then $f\text{'}\left(e\right)$ is equal to

• A. $e^{-1}$
• B. $e$
• C. $1$
• D. $0$

Answer 1

The correct option is A

$e^{-1}$

Explanation for the correct option:

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

Given that,

$f\left(x\right)={\log }_e\left({\log }_{e}x\right)$

Differentiate the above function with respect to $x$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{{\log }_{e}x}\frac{d}{dx}\left({\log }_{e}x\right)\left[\because \frac{d}{\mathrm{dx}}\left(\mathrm{logx}\right)=\frac{1}{\mathrm{logx}}\right]\\ & =& \frac{1}{{\log }_{e}x}\left(\frac{1}{x}\right)\end{array}$

Now, substitute $x=e$ in the above differentiation. we get

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

Hence, the correct option is A.