Answer
$-x^{-x} (1+\ln x) $
Work Step by Step
We have: $y=x^{-x}$
We need to take natural logarithm of both sides.
$\ln y=-x \ln x$
We differentiate both sides with respect to $x$.
$ \dfrac{1}{y}\dfrac{dy}{dx}=-x \times \dfrac{1}{x}-\ln x \\ \dfrac{dy}{dx}=-y(1+\ln x) \\ \dfrac{dy}{dx}=-x^{-x} (1+\ln x) $
