Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises - Page 221: 2

Answer

Slope of Tangent Line: 1

Work Step by Step

$f(x) = sin^{-1}x$ $f'(x) = \frac{1}{\sqrt {1-x^2}}$ At $x=0$, $f'(0) = \frac{1}{\sqrt {1-0}} = 1$
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