Answer
$$\frac{dy}{dx} = \frac{-y}{x(ln(x)+2y)}$$
Work Step by Step
$yln(x)+y^2=0$
differentiating both sides w.r.t x gives,
$y'ln(x)+(1/x)y +2yy'=0 $
Or, $$y'(ln(x)+2y)=-\frac{y}{x}$$
i.e. $$\frac{dy}{dx} = \frac{-y}{x(ln(x)+2y)}$$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - Review Exercises - Page 393: 45
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