Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - Review - Exercises: 15

Answer

$F_{\alpha}=2 \alpha ln(\alpha^{2}+\beta^{2})+\frac{2\alpha^{3}}{(\alpha^{2}+\beta^{2})}$ and $F_{\beta}=\frac{2\alpha^{2}\beta}{(\alpha^{2}+\beta^{2})}$

Work Step by Step

Given: $F(\alpha, \beta)=\alpha^{2}ln(\alpha^{2}+\beta^{2})$ Need to find first partial derivatives $F_{\alpha}$ and $F_{\beta}$. Differentiate the function with respect to $\alpha$ keeping $\beta$ constant. $F_{\alpha}=2 \alpha ln(\alpha^{2}+\beta^{2})+\frac{\alpha^{2}\times 2\alpha}{(\alpha^{2}+\beta^{2})}$ $F_{\alpha}=2 \alpha ln(\alpha^{2}+\beta^{2})+\frac{2\alpha^{3}}{(\alpha^{2}+\beta^{2})}$ Differentiate the function with respect to $\beta$ keeping $\alpha$ constant. $F_{\beta}=\frac{\alpha^{2}\times 2\beta}{(\alpha^{2}+\beta^{2})}$ $F_{\beta}=\frac{2\alpha^{2}\beta}{(\alpha^{2}+\beta^{2})}$ Hence, $F_{\alpha}=2 \alpha ln(\alpha^{2}+\beta^{2})+\frac{2\alpha^{3}}{(\alpha^{2}+\beta^{2})}$ and $F_{\beta}=\frac{2\alpha^{2}\beta}{(\alpha^{2}+\beta^{2})}$
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