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: 19


$$ z_x= \frac{-x}{\sqrt{9-x^2-y^2}}, \quad z_y= \frac{-y}{\sqrt{9-x^2-y^2}}.$$

Work Step by Step

Recall that $(x^n)'=nx^{n-1}$ Since $ z=\sqrt{9-x^2-y^2} $, then by using the chain rule, we have $$ z_x= \frac{-2x}{2\sqrt{9-x^2-y^2}}=\frac{-x}{\sqrt{9-x^2-y^2}}, \\ z_y= \frac{-2y}{2\sqrt{9-x^2-y^2}}=\frac{-y}{\sqrt{9-x^2-y^2}}.$$
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