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: 32


$$ P_y= \frac{y}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2},\\ P_z = \frac{z}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2}. $$

Work Step by Step

Recall that $(e^x)'=e^x$ Recall that $(x^n)'=nx^{n-1}$ Since $ P=e^\sqrt{ y^2+z^2}$, then by using the chain rule, we have $$ P_y=e^\sqrt{ y^2+z^2} \frac{2y}{2\sqrt{ y^2+z^2}}= \frac{y}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2},\\ P_z=e^\sqrt{ y^2+z^2} \frac{2z}{2\sqrt{ y^2+z^2}}= \frac{z}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2}. $$
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.