I wanted to ask a question about the partial differential of the Free Energy equation.
I learnt to prove the Free Energy equation:
\begin{aligned} \frac{F\left(N_{A}, N_{B}\right)}{k T}=& N_{A} \ln \left(\frac{N_{A}}{N}\right)+N_{B} \ln \left(\frac{N_{B}}{N}\right) \\ &+\left(\frac{z w_{A A}}{2 k T}\right) N_{A}+\left(\frac{z w_{B B}}{2 k T}\right) N_{B}+\chi_{A B} \frac{N_{A} N_{B}}{N} \end{aligned}
and then also learnt about the derivative
$$\mu_{A}=\left(\frac{\partial F} {\partial N_{A}}\right)_{T, N_{B}}$$
where $N$ is the total number of molecules in the mixture, $N_A$ is the number of molecules of type A and $N_B$ is the number of molecules of type B such that
$$N = N_A + N_B$$
so I tried to differentiate the equation above using this idea:
$$\frac{\mu_{A}}{k T}=\left[\frac{\partial}{\partial N_{A}}\left(\frac{F}{k T}\right)\right]_{T, N_{B}}$$
$$=\ln \left(\frac{N_{A}}{N}\right)+1-\frac{N_{A}}{N}-\frac{N_{B}}{N}+\frac{z w_{A A}}{2 k T}+\chi_{A B} \frac{\left(N_{A}+N_{B}\right) N_{B}-N_{A} N_{B}}{\left(N_{A}+N_{B}\right)^{2}}$$
but I couldn't get past the second term.
I could prove the terms
$$\ln \left(\frac{N_{A}}{N}\right) + 1$$
as it was the "product rule" applied on
$$N_{A} \ln \left(\frac{N_{A}}{N}\right)$$
and the rest of the differential apart from
$$-\frac{N_{A}}{N}-\frac{N_{B}}{N}$$
Where do these terms come from?
This derivation taken from here.