If y = \( { \left(\frac{1}{x}\right) }^{x} \), then \(\frac{d^2y}{dx^2}\)
Answer & explanation
Correct answer: option 1
$y=(\frac{1}{x})^x⇒y=x^{-x}$
∴ log y = -x log x
$⇒ \frac{1}{y}\frac{dy}{dx}=\frac{-x}{x}+log x(-1)=-1-logx$
$\frac{dy}{dx}=y-(1+logx)$
$\frac{d^2y}{dx^2}=-y[\frac{1}{x}]-(1+logx)\frac{dy}{dx}$
$⇒-x^{-x-1}+x^{-x}(1+logx)^2=x^{-x}(1+logx)^2-x^{-x-1}$
Option A is correct.