Answer
$y' = -\frac{y^2}{x^2}$
Work Step by Step
Applying the rules of derivatives:
$\frac{1}{y}+\frac{1}{x} = 1$
$\frac{d}{dx} [\frac{1}{y}+\frac{1}{x} ] = \frac{d}{dx} 1$
$-y^{-2} \frac{dy}{dx} -x^{-2} = 0$
$\frac{dy}{dx}=-\frac{y^2}{x^2}$
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 2 - The Derivative - Chapter 2 Review Exercises - Page 185: 47
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