Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 15

Answer

$f'(t)=e^{at}(a\sin bt+b\cos bt)$

Work Step by Step

$f(t)=e^{at}\sin bt$ (Here, $a$ and $b$ are constants) Differentiate using the product rule: $f'(t)=(e^{at})(\sin bt)'+(\sin bt)(e^{at})'=...$ Now, use the chain rule to find both $(\sin bt)'$ and $(e^{at})'$: $...=(e^{at})[(\cos bt)(bt)']+(\sin bt)[(e^{at})(at)']=...$ $...=(e^{at})(b\cos bt)+(\sin bt)(ae^{at})=ae^{at}\sin bt+be^{at}\cos bt=...$ Take out common factor $e^{at}$ to present a better looking answer: (Optional) $f'(t)=e^{at}(a\sin bt+b\cos bt)$
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