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: 39

Answer

$0$ for $x \gt 1$

Work Step by Step

$y = tan^{-1}\sqrt{x^2-1}+csc^{-1}x $ $= tan^{-1}(x^2-1)^{1/2}+csc^{-1}x $ Derive using the chain rule: $\frac{dy}{dx}=\frac{\frac{(1)}{2}(x^2-1)^{-1/2}(2x)}{1+[(x^2-1)^{1/2}]^2}-\frac{1}{|x|\sqrt{x^2-1}}$ =$\frac{1}{x\sqrt{x^2-1}}-\frac{1}{|x|\sqrt{x^2-1}}=0$ for $x \gt 1$
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