Differentiating the Free Energy for the Chemical Potential

Question

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.

Answer 1

You missed the $N$ in the logarithm. Since $N_{B}$ is kept constant while changing $N_{A}$, the total number of particles $N=N_{A}+N_{B}$ changes as well. The missing term is

$$\dfrac{\partial N}{\partial N_{A}}\dfrac{\partial}{\partial N}\left[N_{A}\ln\left(\dfrac{N_{A}}{N}\right)+N_{B}\ln\left(\dfrac{N_{B}}{N}\right)\right]=-\dfrac{N_{A}}{N}-\dfrac{N_{B}}{N}$$