Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 45


$$ y' =\frac{1}{t^2\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}.$$

Work Step by Step

Since $ y=\cos^{-1}t^{-1}-\sec^{-1}t $, then the derivative $ y'$ is given by $$ y'= \frac{-(t^{-1})'}{\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}\\ = \frac{-(-t^{-2})}{\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}\\ =\frac{1}{t^2\sqrt{1-t^{-2}}}- \frac{1}{|t|\sqrt{t^2-1}}.$$
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