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: 8

Answer

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

Work Step by Step

$f(x) = x\sin^{-1}x$ Product Rule: $f'(x) = x \times \frac{1}{\sqrt {1-x^2}} + (1)\sin^{-1}x = \sin^{-1}x + \frac{x}{\sqrt {1-x^2}}$
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