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 780: 4


$$\frac{\partial }{\partial u}\ln(u^2+uv)= \frac{2u+v}{u^2+uv}.$$

Work Step by Step

$$\frac{\partial }{\partial u}\ln(u^2+uv)=\frac{1}{u^2+uv}\frac{\partial }{\partial u}(u^2+uv)\\=\frac{2u+v}{u^2+uv}.$$
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.