Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Practice Exercises - Page 864: 22


$h_x=2πcos(2πx+y-3z)$ $h_y=cos(2πx+y-3z)$ $h_z=-3cos(2πx+y-3z)$

Work Step by Step

Take the first partial derivatives of the given function. When taking partial derivative with respect to x, treat y and z as constants. Use the same method when taking partial derivative with respect to y and z: $h_x=cos(2πx+y-3z)\times(2π+0-0)=2πcos(2πx+y-3z)$ $h_y=cos(2πx+y-3z)\times(0+1-0)=cos(2πx+y-3z)$ $h_z=cos(2πx+y-3z)\times(0+0-3)=-3cos(2πx+y-3z)$
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