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: 33



Work Step by Step

Since, we have $\dfrac{\partial w}{\partial r}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial r}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial r}$ and, $2(x+y+z) +2(x+y+z)[-\sin (r+s) ]+2(x+y+z))[\cos (r+s) ]=2[r-s+\cos (r+s)+\sin (r+s)][1-\sin (r+s)+\cos (r+s)]$ Now, for the point $r=1,s=-1$: Thus, $\dfrac{\partial w}{\partial r}=2[r-s+\cos (r+s)+\sin (r+s)][1-\sin (r+s)+\cos (r+s)]=2[1-(-1)+\cos (1-1)+\sin (1-1)][1-\sin (1-1)+\cos (1-1)]=12$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.