University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 41


$\frac{dy}{dx} = \ sin^{-1}{x}$

Work Step by Step

Given that $y$ = $x\sin^{-1}{{x}}+\sqrt{(1-x^2)}$, on applying the product rule and the chain rule of the derivative, we get: $\frac{dy}{dx}$ = ${{1\times\sin^{-1}{x}+x\times{\frac{1}{\sqrt{(1-x^2)}}}}+{\frac{1}{2\sqrt{(1-x^2)}}}}\frac{d{(1-x^2)}}{dx}$ or $\frac{dy}{dx}$ = $ \ sin^{-1}{x}+{ \frac {x}{\sqrt{({1-x^2})}}-{ \frac {x}{\sqrt{({1-x^2})}}}} =\ sin^{-1}{x}$ The final answer is:$\frac{dy}{dx} = \ sin^{-1}{x}$
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