Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 31


$$ G'(s)= \frac{1}{s^2\sqrt{1-(s^{-1})^2}}.$$

Work Step by Step

Since $ G(s)=\cos^{-1}(s^{-1})$, then the derivative, using the chain rule, is given by $$ G'(s)=-\frac{1}{\sqrt{1-(s^{-1})^2}}(s^{-1})'=-\frac{-s^{-2}}{\sqrt{1-(s^{-1})^2}}=\frac{1}{s^2\sqrt{1-(s^{-1})^2}}.$$
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