Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 41

Answer

$\sin ^{-1}x $

Work Step by Step

$y= x \sin^{-1} x+\sqrt{1-x^2}$ $= x \sin^{-1}x+(1-x^2)^{1/2}$ Derive using the chain rule: $\frac{dy}{dx}=sin^{-1}x+x(\frac{1}{\sqrt{1-x^2}})+(\frac{1}{2})(1-x^2)^{-1/2}(-2x)$ $= sin ^{-1}+ \frac{x}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}$ $=\sin ^{-1}x $
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