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: 6


$f_x=6(2x-3y)^2$ and $f_y=-9(2x-3y)^2$

Work Step by Step

Our aim is to take the first partial derivatives of the given function $F(x,y)$ with respect to $x$, by keeping $y$ as a constant, and vice versa: $f_x=3(2x-3y)^2\times 2 \\f_y=3(2x-3y)^2\times -3$ Our results are: $f_x=6(2x-3y)^2$ and $f_y=-9(2x-3y)^2$
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