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: 33

Answer

$12$

Work Step by Step

Since, we have $\dfrac{\partial w}{\partial r}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial r}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial r}$ or, $=2(x+y+z) +2(x+y+z)[-\sin (r+s) ]+2(x+y+z))[\cos (r+s) ]$ or, $=2[r-s+\cos (r+s)+\sin (r+s)][1-\sin (r+s)+\cos (r+s)]$ Now, at point $r=1,s=-1$, we get $\dfrac{\partial w}{\partial r}=2[1-(-1)+\cos (1-1)+\sin (1-1)][1-\sin (1-1)+\cos (1-1)]=(2)(3)(2)=12$
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