Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - Review Exercises: 28

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) $
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