Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 58



Work Step by Step

Given $$y=x\ln x,\ \ \ (e,e)$$ Since $$y'=x\frac{1}{x}+\ln x=1+\ln x$$ Then $m=y'\bigg|_{(e,e)}=2$, hence the tangent line given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-e}{x-e}&=2\\ y-e&=2x-2e\\ y&=2x-e \end{align*}
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