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: 31


$$ z_x=-2xe^{-x^2-y^2}, \quad z_y=-2ye^{-x^2-y^2}. $$

Work Step by Step

Recall that $(e^x)'=e^x $ Recall that $(x^n)'=nx^{n-1}$ Since $ z=e^{-x^2-y^2}$, then by using the chain rule, we have $$ z_x=-2xe^{-x^2-y^2}, \quad z_y=-2ye^{-x^2-y^2}. $$
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