Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 51

Answer

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

Work Step by Step

Apply the Chain Rule: $y=sin^{-1}(2x+1)\\ y'=\frac{1}{\sqrt{1-(2x+1)^2}}.2\\ y'=\frac{2}{\sqrt{1-(2x+1)^2}}$
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