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
$$ z_x=y^{-1}=\frac{1}{y}, \quad z_y=-xy^{-2}=-\frac{x}{y^2}.$$
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: 17
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