Calculus 10th Edition

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: 64

Answer

$y'=(x^{x-1})(1-\frac{1}{x}+\ln x)$

Work Step by Step

$y=x^{x-1}$ $y’=(x^{x-1})(\frac{d}{dx}(\ln x)(x-1))$ $u=\ln x$ $u’=\frac{1}{x}$ $v=x-1$ $v’=1$ $y’=(x^{x-1})(\frac{x-1}{x}+\ln x)$ $=(x^{x-1})(1-\frac{1}{x}+\ln x)$
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