Chemical potential from a given entropy

[rayman123]

Homework Statement

Knowing some gas entropy
$$S= Nk_{B}[ln(\frac{V}{N})+\frac{3}{2}ln(\frac{mU}{3\pi\hbar^2N})+\frac{5}{2}]$$ calculate chemical potential.

Homework Equations

$$\mu = -T(\frac{\partial S}{\partial N})$$

The Attempt at a Solution

I multiply all the terms by themselvs and get
$$S= Nk_{B}ln(\frac{V}{N})+\frac{3}{2} Nk_{B}ln(\frac{mU}{3\pi\hbar^2N})+\frac{5}{2}Nk_{B}$$

Now following the formula
$$\mu = -T(\frac{\partial S}{\partial N})$$
I differentiate each term separately
$$(\frac{\partial S}{\partial N})= Nk_{B}ln(\frac{V}{N})=k_{B}ln(\frac{V}{N})+Nk_{B}\frac{-NV}{VN^2}=k_{B}ln(\frac{V}{N})-k_{B}$$

second therm
$$(\frac{\partial S}{\partial N})= \frac{3}{2} Nk_{B}ln(\frac{mU}{3\pi\hbar^2N})=\frac{3}{2}k_{B}ln(\frac{mU}{3\pi\hbar^2N})+\frac{3}{2}Nk_{B}\frac{3\pi\hbar^2N}{mU}\frac{-mU}{3\pi\hbar^2N^2}=\frac{3}{2}k_{B}ln(\frac{mU}{3\pi\hbar^2N})-\frac{3}{2}k_{B}$$

the last term
$$(\frac{\partial S}{\partial N})= \frac{5}{2}Nk_{B}=\frac{5}{2}k_{B}$$



could someone show me where I made mistake?

I found the mistake. I forgot the inner function derivative

 

[NateD]

in the second therm.(\frac{\partial S}{\partial N})= \frac{3}{2} Nk_{B}ln(\frac{mU}{3\pi\hbar^2N})=\frac{3}{2}k_{B}ln(\frac{mU}{3\pi\hbar^2N})+\frac{3}{2}Nk_{B}(\frac{1}{3\pi\hbar^2N}\frac{-mU}{3\pi\hbar^2N^2})=\frac{3}{2}k_{B}ln(\frac{mU}{3\pi\hbar^2N})-\frac{3}{2}k_{B}