Within FH theory, for a binary polymer blend the following is commonly found: $$ \mu - \mu_0 = \left(\frac{\partial A_{mix}}{\partial n_A}\right)_{V, T, n_b} =\frac{\partial}{\partial n_a}\left(\chi \phi_a \phi_b + \frac{\phi_a}{N_a}\ln(\phi_a)\ +\frac{\phi_b}{N_b}\ln(\phi_b) \right) = RT\ln(\phi_a) + \left(\frac{N_a}{N_b}-1\right)\phi_b\ + N_a \chi \phi_b^2 $$ I only retrieve the latter when I write $\phi_a = \frac{n_a N_a}{V_{lat}} = \frac{n_a N_a}{n_a N_a + n_b N_b}$. Here my first question arises, because in order to retrieve the chemical potential, we take the deriavtive at constant volume. Shouldn't this mean that one can factor the denominator $\frac{1}{n_a N_a + n_b N_b}$ from the derivative? Doing so will not yield the correct answer however.
Secondly, one would expect that by the chain rule: $$ \frac{\partial A_{mix}}{\partial n_a} = \frac{\partial A_{mix}}{\partial \phi_a}\frac{\partial \phi_a}{\partial n_a} $$ It is easily shown that: $$ \frac{\partial A_{mix}}{\partial \phi_a} = RT\left(\chi(1-2\phi_a) + \frac{\ln\phi_a}{N_b} + \frac{1}{N_a} - \frac{\ln(1-\phi_a)}{N_b} - \frac{1}{N_b} \right) $$ and also: $$ \left(\frac{\partial \phi_a}{\partial n_a}\right)_{V,T,n_b} = \frac{\partial}{\partial n_a}\left(\frac{n_a N_a}{n_a N_a + n_b N_b}\right) = \frac{\phi_a\phi_b}{n_a} \left(\neq \frac{1}{V_{lat}}\frac{\partial}{\partial n_a}\left(n_a N_b\right) \right) $$ (again related to my first question, why does the last equality not hold?). However, I endlessly fail to retrieve the very first equality by using these relations (perhaps it's due to my mathematical incompetence).
Any help on this matter is very much appreciated :) Luuk