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 782: 59


$$ h_{vv}(u,v)= \frac{32u}{(u+4v)^3}$$

Work Step by Step

Since $ h(u,v)=\frac{u}{u+4v}=u(u+4v)^{-1}$, then by using the chain rule, we have $$ h_v(u,v)=\frac{\partial h}{\partial v}=- u(u+4v)^{-2}(4)=-\frac{4u}{(u+4v)^2}.$$ Hence, again by using the chain rule, we get $$ h_{vv}(u,v)=\frac{\partial h_v}{\partial v}=8 u(u+4v)^{-3}(4)=\frac{32u}{(u+4v)^3}$$
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.