Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises - Page 984: 28

Answer

$$-\dfrac{y \sin xy}{x \sin xy+\cos y}$$

Work Step by Step

We have: $$ \cos x=1+\sin y$$ and $F(x,y)=\cos (xy)=1-\sin y=0$ Now, $$F_x=-y \sin xy \\ F_y= -x \sin (xy) -\cos y$$ Apply equation: $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ Thus, we have: $$\dfrac{dy}{dx}=-\dfrac{-y \sin xy}{-x \sin (xy) -\cos y} \\ =-\dfrac{y \sin xy}{x \sin xy+\cos y}$$
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