Answer
$$ f_{uuv}= 2v\sin(u+v^2).$$
Work Step by Step
Since $ f(u,v)=\cos(u+v^2)$, then using the chain rule, we have
$$ f_u=-\sin(u+v^2), \quad f_{uu}=-\cos(u+v^2).$$
Then, we have
$$ f_{uuv}=\sin(u+v^2)(2v)=2v\sin(u+v^2).$$
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 782: 65
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