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: 48


$w_{xx}=4x^2ye^{x^2-y}+2ye^{x^2-y}$ $w_{yy}=y^2e^{x^2-y}-2e^{x^2-y}$ $w_{xy}=w_{yx}=-2xye^{x^2-y}+2xe^{x^2-y}$

Work Step by Step

Our aim is to take the first partial derivative of the given function $w(x,y)$ with respect to $x$, by treating $y$ as a constant and vice versa: $w_x=2xye^{x^2-y}$ $w_y=-ye^{x^2-y}+e^{x^2-y}$ Now, compute second partial derivatives. $w_{xx}=4x^2ye^{x^2-y}+2ye^{x^2-y}$ $w_{yy}=y^2e^{x^2-y}-e^{x^2-y}-e^{x^2-y}=y^2e^{x^2-y}-2e^{x^2-y}$ and $w_{xy}=w_{yx}=-2xye^{x^2-y}+2xe^{x^2-y}$
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