If y = ( 1 x ) x , then the value of e e ( d 2 y d x 2 ) x = e is:

Question

If $y = {(\frac{1}{x})}^{x},$ then the value of $e^{e} {(\frac{d^{2} y}{d x^{2}})}_{x = e}$ is:

Smart Achievers · Accepted Answer

(2)

Ans. $4 - \frac{1}{e}$

Explanation:

$e^{e} {(\frac{d^{2} y}{d x^{2}})}_{x = e}$

Taking logarithm on both sides,

$\log y = x \log (\frac{1}{x}) = - x \log x$

Differentiating with respect to $x$ ,

$\frac{1}{y} \frac{d y}{d x} = - [x \cdot \frac{1}{x} + \log x]$

$\frac{d y}{d x} = - y (1 + \log x)$

$\frac{d^{2} y}{d x^{2}} = - [y \cdot \frac{1}{x} + (1 + \log x) \frac{d y}{d x}]$

$= - [\frac{y}{x} + (1 + \log x) {- y (1 + \log x)}]$

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

When $x = e$ ,

$e^{e} {[\frac{d^{2} y}{d x^{2}}]}_{x = e} = [- {(\frac{1}{e})}^{e + 1} + 4 {(\frac{1}{e})}^{e}] e^{e}$

$= - e^{- e - 1 + e} + 4 e^{- e + e}$

$= 4 - \frac{1}{e}$

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