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: 35



Work Step by Step

Consider $\dfrac{\partial w}{\partial v}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial v}$ and $(2x-\dfrac{y}{x^2})(-2)+(1/x)(1)=(2(u-2v+1)-(-2)\dfrac{(2u+v-2)}{(u-2v+1)^2})+(\dfrac{1}{(u-2v+1)})$ For the point $u=0,v=0$: $\dfrac{\partial w}{\partial v}=(2(u-2v+1)-(-2)\dfrac{(2u+v-2)}{(u-2v+1)^2})+(\dfrac{1}{(u-2v+1)})=[2(0-0+1)-\dfrac{(0+0-2)}{(0-0+1)^2})](-2)+\dfrac{1}{1}=-7$
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