Question

if $f\left(x\right)=x^{\frac{1}{x}}$ then $f^{\prime }\left(e\right)$ is equal to

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

Answer 1

Answer: (D)

None of these

To find $f{}^{\prime }\left(e\right)$

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

Let $y=x^{\frac{1}{x}}$

Taking $\log$ on both side, we get

$\log y=\log x^{\frac{1}{x}}$

$\log y=\frac{1}{x}\log x$

On differentiating w.r.t. $x$, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{-1}{x^2}\log x+\frac{1}{x^2}$

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}(1-\log x)$

$\frac{dy}{dx}=\frac{y}{x^2}(1-\log x)$

$\frac{dy}{dx}=x^{(\frac{1}{x}-2)}(1-\log x)$

This can be written as ,

$f{}^{\prime }\left(x\right)=x^{(\frac{1}{x}-2)}(1-\log x)$

Now, at $x=e$, we have

$f{}^{\prime }\left(e\right)=e^{(\frac{1}{e}-2)}(1-\log e)$

$f{}^{\prime }\left(e\right)=0$