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 14.3 - Partial Derivatives - Exercises 14.3 - Page 807: 6

Answer

$f_x=3(2x-3y)^2\times 2$ $f_y=3(2x-3y)^2\times -3$

Work Step by Step

Take the first partial derivatives of the given function. When taking a partial derivative with respect to x, treat y as a constant, and vice versa: $f_x=3(2x-3y)^2\times 2$ $f_y=3(2x-3y)^2\times -3$
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