Answer
$h_x=(y^2+1)e^x$
$h_y=2ye^x$
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:
$h_x=(y^2+1)e^x$
$h_y=(2y+0)e^x=2ye^x$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 12 - Functions of Several Veriables - 12.4 Partial Derivatives - 12.4 Exercises - Page 904: 18
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