Calculus (3rd Edition)

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


$$ h_z(3,0) =3 .$$

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 $$ h_z(3,0)=(3-0)e^{0}=3 .$$
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