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: 88

Answer

$$y'=\sin t+t\cos t$$

Work Step by Step

$$y=t\log_3(e^{(\sin t)(\ln3)})$$ We have $\log_3(e^{(\sin t)(\ln3)})=\log_3(e^{\ln3})^{\sin t}=\log_3(3)^{\sin t}=\sin t$ Therefore, $$y=t\sin t$$ The derivative of $y$ is: $$y'=\sin t+t(\sin t)'$$ $$y'=\sin t+t\cos t$$
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