Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 51

Answer

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

Work Step by Step

First simplify the given function using the properties of logarithms and then apply the chain rule of differentiation. $y'=\frac{d}{dx}[ln(x^{2}+y^{2})]$ $=\frac{1}{(x^{2}+y^{2})}(2x+2y\frac{dy}{dx})$ $=\frac{2x}{(x^{2}+y^{2})}+\frac{2yy'}{(x^{2}+y^{2})}$ $y'-\frac{2yy'}{(x^{2}+y^{2})}=\frac{2x}{(x^{2}+y^{2})}$ $y'(x^{2}+y^{2}-2y)=2x$ Hence, $y'=\frac{2x}{(x^{2}+y^{2}-2y)}$
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