Question

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

• A. $e^{1/(e-3)}$
• B. $e^{1/e}$
• C. $e^{1/(e-2)}$
• D. $-e^{\left(\right(1/e)-3)}$

Answer 1

Answer: (D)

$-e^{\left(\right(1/e)-3)}$

Given that:

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

which can also bhi written as:

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

Differentiating both sides, we get
$f^{\prime }\left(x\right)=(x{)}^{\frac{1}{x}}(\frac{1}{x^2}-\frac{\log x}{x^2})$

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

Differentiating $f^{\prime }\left(x\right)$ both sides, we get

$f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)(\frac{1}{x^2}-\frac{\log x}{x^2})+f\left(x\right)(\frac{-2}{x^3}+\frac{2}{x^3}\log x-\frac{1}{x^3})$

Put $x=e$, we get
$f^{\prime \prime }\left(e\right)=f\left(e\right)\left(\frac{-1}{e^3}\right)$
$=-(e{)}^{\frac{1}{e}-3}$