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: 49

Answer

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

Work Step by Step

Take the derivative and apply the chain rule using the derivative for inverse tangent ($\frac{1}{1+x^2}$): $y'=2(tan^{-1}(x))(\frac{1}{1+x^2})$
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