If y = \( { \left(\frac{1}{x}\right) }^{x} \), then \(\frac{d^2y}{dx^2}\)

x-x (1 + log x)2 – x-(x+1) x-x (1 + log x)2 – x-(x-1) x-x (1 + log x)-2 – x-(x+1) x-x (1 + log x)-2 – x-(x-1)

x-x (1 + log x)2 – x-(x+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.