University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.4 - The Chain Rule - Exercises - Page 711: 34



Work Step by Step

Since, we have $\dfrac{\partial w}{\partial v}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial v}$ or, $=(y)(\dfrac{2v}{u})+(x)(1)+(\dfrac{1}{z})(0)$ or, $=(u+v)(\dfrac{2v}{u})+(\dfrac{v^2}{u})$ Now, at point $u=-1,v=2$, we get $\dfrac{\partial w}{\partial v}=(-1+2)(\dfrac{2(2)}{-1})+(\dfrac{2^2}{(-1)})=-8$
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