Answer
$$\frac{\partial }{\partial u}\ln(u^2+uv)= \frac{2u+v}{u^2+uv}.$$
Work Step by Step
$$\frac{\partial }{\partial u}\ln(u^2+uv)=\frac{1}{u^2+uv}\frac{\partial }{\partial u}(u^2+uv)\\=\frac{2u+v}{u^2+uv}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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