Answer
$y'=(x^{x-1})(1-\frac{1}{x}+\ln x)$
Work Step by Step
$y=x^{x-1}$
$y’=(x^{x-1})(\frac{d}{dx}(\ln x)(x-1))$
$u=\ln x$
$u’=\frac{1}{x}$
$v=x-1$
$v’=1$
$y’=(x^{x-1})(\frac{x-1}{x}+\ln x)$
$=(x^{x-1})(1-\frac{1}{x}+\ln x)$
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 - 5.5 Exercises - Page 362: 64
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