Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 970: 26

Answer

As a result $F_\alpha=-\sqrt{\alpha^3+1}$, $F_\beta=\sqrt{\beta^3+1}$.

Work Step by Step

$F(\alpha,\beta)=\int_\alpha^\beta\sqrt{t^3+1}dt$ In order to find $F_\alpha$ we treat$\beta$ as a constant and differentiate with respect to $\alpha$. $F_\alpha=-\sqrt{\alpha^3+1}$ In order to find $F_\beta$ we treat $\alpha$ as a constant and differentiate with respect to $\beta$. $F_\beta=\sqrt{\beta^3+1}$
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