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: 34



Work Step by Step

Consider $\dfrac{\partial w}{\partial v}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial v}$ and, $(y)(\dfrac{2v}{u})+(x)(1)+(\dfrac{1}{z})(0)=(\dfrac{2v}{u})(u+v)+(\dfrac{v^2}{u})$ For the point $u=-1,v=2$: Thus, $\dfrac{\partial w}{\partial v}=(\dfrac{4}{-1}) \times (-1+2)+\dfrac{4}{(-1)}=-8$
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