Answer
$f'(x) = \sin^{-1}x + \frac{x}{\sqrt {1-x^2}}$
Work Step by Step
$f(x) = x\sin^{-1}x$
Product Rule:
$f'(x) = x \times \frac{1}{\sqrt {1-x^2}} + (1)\sin^{-1}x = \sin^{-1}x + \frac{x}{\sqrt {1-x^2}}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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: 8
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