University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 13 - Section 13.3 - Partial Derivatives - Exercises - Page 702: 33


$$sech^2 (x+2y+3z) \\ 2sech^2 (x+2y+3z) \\ 3 \space sech^2 (x+2y+3z)$$

Work Step by Step

$f_x=sech^2 (x+2y+3z) \times \dfrac{\partial }{\partial x} (x+2y-3z)=sech^2 (x+2y+3z)$ $f_y=sech^2 (x+2y+3z) \times \dfrac{\partial}{\partial y} (x+2y-3z)=2sech^2 (x+2y+3z)$ $f_z= sech^2 (x+2y+3z) \times \dfrac{\partial }{\partial z} (x+2y-3z) =3 \space sech^2 (x+2y+3z)$
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