Question

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

• A. $e$
• B. $\frac{1}{e}$
• C. $1$
• D. None of these.

Answer 1

The correct option is B

$\frac{1}{e}$

Explanation for the correct option

Step 1: Solve for the first derivative of the given function

The given function is $f\left(x\right)={\log }_x\left(\log x\right)$.

$\Rightarrow$ $f\left(x\right)=\frac{\log \left(\log x\right)}{\log x}\left[\because {\mathbf{log}}_{\mathbf{a}}\mathit{b}\mathbf{=}\frac{\mathbf{log}\mathbf{b}}{\mathbf{log}\mathbf{a}}\right]$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[\frac{\log \left(\log x\right)}{\log x}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{\log x·{\displaystyle \frac{d}{dx}}\left[\log \left(\log x\right)\right]-\log \left(\log x\right)·{\displaystyle \frac{d}{dx}}\left(\log x\right)}{{\left(\log x\right)}^2}\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathbf{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}}{{\mathbf{v}}^{\mathbf{2}}}\right] \\ =\frac{\log x·{\displaystyle \frac{1}{\log x}·\frac{d}{dx}}\left(\log x\right)-\log \left(\log x\right)·{\displaystyle \frac{1}{x}}}{{\left(\log x\right)}^2}\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left[}\mathbf{f}\mathbf{\left(}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right)}\mathbf{\right]}\mathbf{=}\mathit{f}\mathbf{\text{'}}\mathbf{\left[}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{·}\mathit{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\right] \\ =\frac{{\displaystyle \frac{1}{x}}-{\displaystyle \frac{\log \left(\log x\right)}{x}}}{{\left(\log x\right)}^2}\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{logx}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\right] \\ =\frac{{\displaystyle \frac{1-\log \left(\log x\right)}{x}}}{{\left(\log x\right)}^2} \\ =\frac{{\displaystyle 1-\log \left(\log x\right)}}{x{\left(\log x\right)}^2} \end{gathered}$$

Step 2: Solve for the required value

Substitute $x=e$ in the above equation.

$$\begin{gathered} \Rightarrow f\text{'}\left(e\right)=\frac{1-\log \left(\log e\right)}{e{\left(\log e\right)}^2} \\ =\frac{1-\log \left(1\right)}{e{\left(1\right)}^2} \\ =\frac{1-0}{e} \\ =\frac{1}{e} \end{gathered}$$

Therefore, the value of $f\text{'}\left(x\right)$ at $x=e$ is $\frac{1}{e}$.

Hence, option(B) is the correct option.