Answer
$y'=e^x(ln(x)+\frac{1}{x})$
Work Step by Step
$y=e^xln(x)$
$u=e^x$
$u’=e^x$
$v=ln(x)$
$v’=\frac{1}{x}$
$y’=(e^x ln(x))+(e^x)(\frac{1}{x}$
$=e^x(ln(x)+\frac{1}{x})$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 39
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