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: 20


$$ z_x= \frac{y^2}{( x^2+y^2)^{3/2}}, $$ $$ z_y= \frac{ -xy}{( x^2+y^2)^{3/2}}. $$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Since $ z= \frac{x}{\sqrt{x^2+y^2}} $, then by using the quotient rule, we have $$ z_x= \frac{\sqrt{x^2+y^2} (1)-x\frac{2x}{2\sqrt{x^2+y^2}}}{ x^2+y^2}= \frac{x^2+y^2-x^2}{( x^2+y^2)^{3/2}}, \\ =\frac{y^2}{( x^2+y^2)^{3/2}}, $$ $$ z_y= \frac{\sqrt{x^2+y^2} (0)-x\frac{2y}{2\sqrt{x^2+y^2}}}{ x^2+y^2}= \frac{ -xy}{( x^2+y^2)^{3/2}}. $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.