Question

If $f\left(x\right)=x^{\frac{1}{x}}$, then $f\text{'}\text{'}\left(e\right)$ is

• A. $-e^{\frac{1}{e}-3}$
• B. $e^{\frac{1}{e}}$
• C. $e^{\frac{-1}{2}}$
• D. None of these

Answer 1

The correct option is A

$-e^{\frac{1}{e}-3}$

Explanation for the correct option:

Step 1: Find the value of $f\text{'}\left(x\right)$ at $x=e$:

Given that,

$f\left(x\right)=x^{\frac{1}{x}}$

$f\left(e\right)=e^{\frac{1}{e}}...\left(1\right)$

By taking $\log$ on both sides,

$$\begin{gathered} \log f\left(x\right)=\log \left(x^{\frac{1}{x}}\right) \\ \log f\left(x\right)=\frac{1}{x}\log x\left[\because \log a^n=n\log a\right] \end{gathered}$$

Differentiate the above function with respect to $x$,

$\begin{array}{rcl}\frac{1}{f\left(x\right)}f\text{'}\left(x\right)& =& \frac{1}{x}\left(\frac{1}{x}\right)-\log x\left(\frac{1}{x^2}\right)\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{u}\mathbf{v}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{u}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{u}\mathbf{\text{'}}\mathbf{}\mathbf{,}\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{log}\mathbf{x}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\\ & \Rightarrow & f\text{'}\left(x\right)=f\left(x\right)\left(\frac{1}{x^2}-\frac{\log x}{x^2}\right)\end{array}$

At $x=e$,

$\begin{array}{rcl}f\text{'}\left(e\right)& =& e^{\frac{1}{e}}\left(\frac{1}{e^2}-\frac{\log e}{e^2}\right)\\ & =& 0...\left(2\right)\end{array}$

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

Now, differentiate $f\text{'}\left(x\right)$ both sides, we get

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& f\text{'}\left(x\right)\left(\frac{1}{x^2}-\frac{\log x}{x^2}\right)+f\left(x\right)\left(\frac{-2}{x^3}-\left(\frac{-2\log x}{x^3}+\frac{1}{x^2}\left(\frac{1}{x}\right)\right)\right)\left[\because \frac{d\left(uv\right)}{dx}=vu\text{'}+v\text{'}u\right]\\ & =& f\text{'}\left(x\right)\left(\frac{1}{x^2}-\frac{\log x}{x^2}\right)+f\left(x\right)\left(\frac{-2}{x^3}+\frac{2\log x}{x^3}-\frac{1}{x^3}\right)\end{array}$

Substitute $x=e$,

$\begin{array}{rcl}f\text{'}\text{'}\left(e\right)& =& f\text{'}\left(e\right)\left(\frac{1}{e^2}-\frac{\log x}{x^2}\right)+f\left(e\right)\left(\frac{-2}{e^3}+\frac{2\log x}{e^3}-\frac{1}{e^3}\right)\\ & =& 0+e^{\frac{1}{e}}\left(\frac{-2}{e^3}-\frac{2}{e^3}-\frac{1}{e^3}\right)\left[\mathrm{from}\left(1\right)\mathrm{and}\left(2\right)\right]\\ & =& -e^{\frac{1}{e}-3}\end{array}$

Hence, the correct option is A.