University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 22


$$y'=\ln\sqrt t+\frac{1}{2}$$

Work Step by Step

$$y=t\ln\sqrt t$$ Recall the following Derivative Rules: $$\frac{d}{dt}(\ln u)=\frac{1}{u}\frac{du}{dt}$$ Therefore, we have $$y'=\Big(t\ln\sqrt t\Big)'=(t)'\ln\sqrt t+t(\ln\sqrt t)'$$ $$y'=\ln\sqrt t+t\times\frac{1}{\sqrt t}\times(\sqrt t)'$$ $$y'=\ln\sqrt t+\sqrt t\times\frac{1}{2\sqrt t}$$ $$y'=\ln\sqrt t+\frac{1}{2}$$
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