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: 38

Answer

$\frac{s|s|-1}{|s|\sqrt{s^2-1}}$

Work Step by Step

$y=\sqrt{s^2-1}-sec^{-1}s $ $=(s^2-1)^{1/2}-sec^{-1}s $ Derive using the chain rule: $\frac{dy}{dx}=(\frac{1}{2}(s^2-1)^{1/2})(2s)-\frac{1}{|s|\sqrt{s^2-1}}$ $=\frac{s}{\sqrt{s^2-1}}-\frac{1}{|s|\sqrt{s^2-1}}$ $=\frac{s|s|-1}{|s|\sqrt{s^2-1}}$
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