Answer
$f_{xx}=0$
$f_{yy}=\frac{2x}{y^3}$
$ f_{xy}=f_{yx}=-\frac{1}{y^2}$
Work Step by Step
$f(x,y)=y+\frac{x}{y}$
$f_{x}=\frac{1}{y}, f_{xx}=0, f_{yx}=-\frac{1}{y^2}$
$f_{y}=1-\frac{x}{y^2}, f_{yy}=\frac{2x}{y^3}, f_{xy}= f_{yx}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 14: Partial Derivatives - Practice Exercises - Page 864: 25
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