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 the 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
$F_{\beta}=\frac{2\alpha^{2}\beta}{(\alpha^{2}+\beta^{2})}$
