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: 51



Work Step by Step

Take the first partial derivatives of the given function. When taking partial derivative with respect to x, treat y as a constant, and vice versa: $w_x=\frac{1}{2x+3y}\times2=\frac{2}{2x+3y}$ $w_y=\frac{1}{2x+3y}\times3=\frac{3}{2x+3y}$ Then take the derivative of the first order partial derivatives to find second partial derivatives: $w_{xy}=\frac{(2x+3y)(0)-(2)(3)}{(2x+3y)^2}=\frac{-6}{(2x+3y)^2}$ $w_{yx}=\frac{(2x+3y)(0)-(3)(2)}{(2x+3y)^2}=\frac{-6}{(2x+3y)^2}$
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