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: 36


$g_u=2ve^{2u/v}$ $g_v=-2ue^{2u/v}+2ve^{2u/v}$

Work Step by Step

Take the first partial derivatives of the given function. When taking partial derivative with respect to u, treat v as a constant, and vice versa: $g_u=2/v\times v^2e^{2u/v}=2ve^{2u/v}$ $g_v=v^2e^{2u/v}\times -2u/v^2+2ve^{2u/v}=-2ue^{2u/v}+2ve^{2u/v}$
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