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: 28


$-(2+\ln 2)$

Work Step by Step

Consider $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ Plug the derivatives in equation $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ This implies that $\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1} (-1)=-\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1}$ For the point $(0,\ln 2)$ As we know that $e^{\ln 2}=2$ So, $-\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1}=-\dfrac{e^{\ln 2} +(\ln 2) \cos (0) (\ln 2)}{1} =-(2+\ln 2)$
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