Question

Answer Please .......​

Answer 1

Given,

$\longrightarrow\sf{f(x)=x^{\frac{1}{x}}}$

Since $\sf{a^b=e^{b\log a},}$

$\longrightarrow\sf{f(x)=e^{\frac{1}{x}\,\log x}}$

Differentiating it with respect to $\sf{x,}$

$\longrightarrow\sf{f'(x)=\dfrac{d}{dx}\,\left[e^{\frac{1}{x}\,\log x}\right]}$

By chain rule,

$\longrightarrow\sf{f'(x)=\dfrac{d\left[e^{\frac{1}{x}\,\log x}\right]}{d\left[\dfrac{1}{x}\,\log x\right]}\cdot\dfrac{d\left[\dfrac{1}{x}\,\log x\right]}{dx}}$

Since $\sf{\dfrac{d}{dx}\,\left[e^x\right]=e^x}$ and by product rule,

$\longrightarrow\sf{f'(x)=e^{\frac{1}{x}\,\log x}\left[\dfrac{d}{dx}\left[\dfrac{1}{x}\right]\cdot\log x+\dfrac{1}{x}\cdot\dfrac{d}{dx}\left[\ln x\right]\right]}$

$\longrightarrow\sf{f'(x)=e^{\frac{1}{x}\,\log x}\left[-\dfrac{1}{x^2}\,\log x+\dfrac{1}{x^2}\right]}$

$\longrightarrow\sf{f'(x)=\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}\left[\,1-\log x\,\right]}$

$\longrightarrow\sf{f'(x)=\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}-\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}\cdot\log x}$

Again differentiating,

$\longrightarrow\sf{f''(x)=\dfrac{d}{dx}\,\left[\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}-\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}\cdot\log x\right]}$

$\longrightarrow\sf{f''(x)=\dfrac{d}{dx}\,\left[\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}\right]-\dfrac{d}{dx}\,\left[\dfrac{1}{x^2}\cdot e^{\frac{1}{x}\,\log x}\cdot\log x\right]}$

Applying product rule,

$\begin{aligned}\longrightarrow\sf{f''(x)}=&\sf{-\dfrac{2}{x^3}\cdot e^{\frac{1}{x}\,\log x}+\dfrac{1}{x^4}\cdot e^{\frac{1}{x}\,\log x}\left[\,1-\log x\,\right]+\dfrac{2}{x^3}\cdot e^{\frac{1}{x}\,\log x}\cdot\log x}\\\\&\sf{-\dfrac{1}{x^4}\,e^{\frac{1}{x}\,\log x}\cdot\log x\left[\,1-\log x\,\right]-\dfrac{1}{x^3}\,e^{\frac{1}{x}\,\log x}}\end{aligned}$

$\longrightarrow\sf{f''(x)=\dfrac{1}{x^4}\cdot e^{\frac{1}{x}\,\log x}\cdot\log x\left[\,1-\log x\,\right]^2-\dfrac{1}{x^3}\,e^{\frac{1}{x}\,\log x}\left[\,3-2\log x\,\right]}$

Finding $\sf{f''(e),}$

$\longrightarrow\sf{f''(e)=-\dfrac{e^{\frac{1}{e}}}{e^3}}$

$\longrightarrow\sf{\underline{\underline{f''(e)=-e^{\frac{1}{e}-3}}}}$

Hence (d) is the answer.