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: 58


$$y'=\cot t-1$$

Work Step by Step

$$y=\ln(2 e^{-t}\sin t)$$ We have $$(\ln t)'=\frac{1}{t}$$ $$(e^{t})'=e^t$$ So the derivative of $y$ is $$y'=\frac{1}{2e^{-t}\sin t}(2e^{-t}\sin t)'=\frac{2\Big(e^{-t}\sin t(-t)'+e^{-t}\cos t\Big)}{2e^{-t}\sin t}$$ $$y'=\frac{-e^{-t}\sin t+e^{-t}\cos t}{e^{-t}\sin t}=\frac{e^{-t}(\cos t-\sin t)}{ e^{-t}\sin t}$$ $$y'=\frac{\cos t-\sin t}{\sin t}=\frac{\cos t}{\sin t}-1$$ $$y'=\cot t-1$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.