Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 35

Answer

$\frac{-1}{\sqrt{e^{2t}-1}}$

Work Step by Step

$y=csc^{-1}(e^t)$ Derive using the chain rule: $\frac{dy}{dt}=\frac{-e^t}{|e^t|\sqrt{(e^t)^2-1}}$ =$\frac{-1}{\sqrt{e^{2t}-1}}$
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