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: 41

Answer

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

Work Step by Step

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