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: 32


$-\dfrac{4}{3 \ln 2}$ and $-\dfrac{5}{3 \ln 2}$

Work Step by Step

Consider $F(x,y,z)=xe^y+ye^x+2 \ln x-2-3 \ln 2=0$ Now, $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ $\implies \dfrac{\partial z}{ \partial x}=-(\dfrac{e^y+(2/x)}{ye^z})$ For the point $(1,\ln 2,\ln 3)$, we have : $\dfrac{\partial z}{ \partial x}(1,\ln 2,\ln 3)=-\dfrac{e^{\ln 2}+\dfrac{2}{1}}{(\ln 2)e^{\ln 3}}=-\dfrac{4}{3 \ln 2}$ Now, consider $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ $\implies \dfrac{\partial z}{ \partial y}=-\dfrac{xe^y+e^z}{ye^z}$ For the point $(1,\ln 2,\ln 3)$, we have : $-[\dfrac{e^{\ln 2}+e^{\ln 3}}{(\ln 2)e^{\ln 3}}]=-\dfrac{5}{3 \ln 2}$
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