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: 17

Answer

$$ z_x= \frac{1}{y}, \quad z_y= -\frac{x}{y^2}.$$

Work Step by Step

Recall that $(x^n)'=nx^{n-1}$ Since $ z=\frac{x}{y}=xy^{-1}$, then we have $$ z_x=y^{-1}=\frac{1}{y}, \quad z_y=-xy^{-2}=-\frac{x}{y^2}.$$
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