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: 18


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

Work Step by Step

Recall that $(x^n)'=nx^{n-1}$ Since $ z=\frac{x}{x-y} $, then by using the quotient rule, we have $$ z_x= \frac{(x-y)(1)-x(1)}{(x-y)^2}=\frac{-y}{(x-y)^2}, \\ z_y= \frac{(x-y)(0)-x(-1)}{(x-y)^2}=\frac{x}{(x-y)^2}.$$
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