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: 60


$$ h_{xy}(x,y) =-\frac{9x^2y^2}{(x^3+y^3)^{2}}.$$

Work Step by Step

Since $ h(x,y)=\ln (x^3+y^3)$, then by using the chain rule, we have $$ h_x(x,y)=\frac{\partial h}{\partial x}=\frac{3x^2}{x^3+y^3}=3x^2(x^3+y^3)^{-1}.$$ Again using the chain rule, we get $$ h_{xy}(x,y)=\frac{\partial^2 h}{\partial x\partial y}= -3x^2(x^3+y^3)^{-2}(3y^2)=-\frac{9x^2y^2}{(x^3+y^3)^{2}}.$$
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.