Answer
$e^{-w}\left( \dfrac {1}{w}-\ln w\right) $
Work Step by Step
$\dfrac {d}{dw}\left( e^{-w}\ln w\right) =\left( \dfrac {d}{dv}\left( e^{-w}\right) \right) \ln \omega +\left( \dfrac {d}{dw}\left( \ln w\right) \right) e^{-w}=-e^{-w}\ln w+\dfrac {1}{w}e^{-w}=e^{-w}\left( \dfrac {1}{w}-\ln w\right) $
