Answer
$$ h_z(3,0) =3 .$$
Work Step by Step
Since $ h(x,z))=e^{xz-x^2z^3}$, then by the chain rule, we have
$$ h_z=e^{xz-x^2z^3}(x-3x^2z^2)=(x-3x^2z^2)e^{xz-x^2z^3}.$$
Hence, we get
$$ h_z(3,0)=(3-0)e^{0}=3 .$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 781: 44
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