University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 70



Work Step by Step

$$x^2y^2=1$$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2xy^2+2x^2y\times y'=0$$ $$xy^2+x^2y\times y'=0$$ Now we set all elements with $y'$ to one side and the rest to the other side: $$x^2y\times y'=-xy^2$$ Finally, we calculate $y'$: $$y'=-\frac{xy^2}{x^2y}=-\frac{y}{x}$$
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