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: 53

Answer

$w_{xy}=w_{yx}=2y+6xy^2+12x^2y^3$

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: $w_x=y^2+2xy^3+3x^2y^4$ $w_y=2xy+3x^2y^2+4x^3y^3$ Then take the derivative of the first order partial derivatives to find second partial derivatives: $w_{xy}=2y+6xy^2+12x^2y^3$ $w_{yx}=2y+6xy^2+12x^2y^3$
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