Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - Review Exercises: 31

Answer

$\frac{d}{dx}\sin^{-1}\left(\frac{1}{x}\right) = \frac{-1}{|x|\sqrt {x^2-1}}$

Work Step by Step

Chain Rule: $\frac{d}{dx}\sin^{-1}\left(\frac{1}{x}\right) = \frac{1}{\sqrt {1-(\frac{1}{x})^2}} \times \frac{-1}{x^2} = \frac{-1}{x^2\sqrt {1-\left(\frac{1}{x}\right)^2}} = \frac{-1}{|x|\sqrt {x^2-1}}$
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