Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 817: 25



Work Step by Step

Consider $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ Plug the derivatives in equation the $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ This implies that $\dfrac{3x^2+y}{-4y+x} \times (-1)=\dfrac{-3x^2-y}{-4y+x} $ For point $(1,1)$, we have $\dfrac{-3-1}{-4+1}=\dfrac{4}{3}$
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