Starting from the definition of the function $v$ in terms of $\gamma_{i}$ (cf. (4.11)), we can differentiate with respect to $\eta_{p}$ to get

$$\frac{\partial v}{\partial\eta_{p}}=-2\pi A^{1/2}\int_{S}\frac{\partial\gamma}{\partial\eta_{p}}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}).$$

Using the differential relations in Lemma 4.20,

$$\begin{split}2\frac{\partial v}{\partial\eta_{p}}&=-2\pi A^{1/2}\int_{S}\frac{\partial\gamma_{p3}}{\partial\mu}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})\\ &=-2\pi A^{1/2}\frac{\partial}{\partial\mu}\int_{S}\gamma_{p3}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})\\ &=\frac{\partial\beta_{p3}}{\partial\mu},\end{split}$$

and similarly $2\frac{\partial v}{\partial\eta_{p}}=-\frac{\partial\beta_{p4}}{\partial\mu}.$

Next we study $\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}$ in the complement of $\{f_{S}=0,\mu\leq 0\}$. We have

$$\frac{\partial^{2}\beta_{p3}}{\partial\bar{\eta}_{q}\partial\mu}=2\frac{\partial^{2}v}{\partial\eta_{p}\partial\bar{\eta}_{q}}=-\frac{1}{2}\frac{\partial^{2}w^{p\bar{q}}}{\partial\mu\partial\mu},$$

where the second equality uses the distributional equation (4.6). But by Lemma 4.24, for fixed $\eta_{1},\eta_{2}$,

$$\lim_{\mu\to+\infty}\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}(\eta_{1},\eta_{2},\mu)=0,$$

and the asymptotes we obtained in Section 4.2, 4.3 easily imply

$$\lim_{\mu\to+\infty}\frac{\partial w^{p\bar{q}}}{\partial\mu}(\eta_{1},\eta_{2},\mu)=0.$$

Thus we can integrate from $\mu=+\infty$ to obtain

$$\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}=-\frac{1}{2}\frac{\partial w^{p\bar{q}}}{\partial\mu}.$$

A completely parallel argument shows

$$\frac{\partial\beta_{p4}}{\partial\bar{\eta}_{q}}=\frac{1}{2}\frac{\partial w^{p\bar{q}}}{\partial\mu}.$$

Finally by integrating the second part of Lemma 4.20 we see

$$\frac{\partial\beta_{p3}}{\partial\eta_{q}}=\frac{\partial\beta_{q3}}{\partial\eta_{p}},\quad\frac{\partial\beta_{p4}}{\partial\eta_{q}}=\frac{\partial\beta_{q4}}{\partial\eta_{p}},\quad p,q=1,2.$$

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