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: 23


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

Work Step by Step

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