[Physics] Entropy and chemical potential of an ideal gas

Question

I am reading Schroeder's book "Thermal Physics". One calculation in the text was not quite clear to me. The entropy of an ideal gas is given by the Sackur-Tetrode equation:
$$
S=Nk\left[\ln\left({V\over N}\left(4\pi m U\over 3Nh^2\right)^{3/2}\right)+{5\over 2}\right]
$$
Also, the chemical potential is given by:
$$
\mu = -T\left(\partial S \over \partial N \right)_{U,V}
$$
So the only thing one needs to do in order to find $\mu$ is to take the derivative of the first equation with respect to $N$. The author goes on and writes:
$$
\mu = -T \left\{k\left[\ln\left(V\left(\frac{4\pi mU}{3Nh^2}\right)^{3/2}\right) -\ln{N^{5/2}} + \frac{5}{2} \right] -Nk \cdot \frac{5}{2}\frac{1}{N} \right\}
$$
So the author used the product rule in this part. Implicitly he assumed that $U$ is constant, but shouldn't we get a term like $\partial U/\partial N$ since $U$ is a function of $N$ $?$ I mean, in the $\ln(\dots)$ term, why isn't there something like:
$$
\frac{\partial \ln(\dots)}{\partial U}\frac{\partial U}{\partial N}
$$

Answer 1

No, there is not. As you wrote down yourself, the definition of the chemical potential is

$$\mu=-T\left(\frac{\partial S}{\partial N}\right)_{U,V} $$ The subscript $U$ and $V$ mean that, in calculating this partial derivative, $U$ and $V$ are held fixed. Thus, you should disregard the fact that $U$ depends on $N$.