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: 29


$\dfrac{1}{4}$ ; $\dfrac{-3}{4}$

Work Step by Step

Here,we have $F(x,y,z)= z^3-xy+yz+y^3-2=0$ and $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ and $\dfrac{\partial z}{ \partial x}=-(\dfrac{-y}{3z^2+y})=\dfrac{y}{y+3z^2}$ For the point $(1,1,1)$, we have $\dfrac{\partial z}{ \partial x}=\dfrac{1}{4}$ $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ $\implies \dfrac{\partial z}{ \partial y}=\dfrac{(x-z-3y^2)}{y+3z^2}$ Now, $\dfrac{\partial z}{ \partial y}(1,1,1)=\dfrac{1-1-3}{1+3}=\dfrac{-3}{4}$
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