Internal Energy/Helmholtz Free Energy Proof

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mobwars
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Homework Statement



Show that (∂(βA)/∂β)N,V = E, where A = E - TS is the Helmholtz Free Energy and E is the Internal Energy.

Homework Equations


A = E - TS
dE = TdS - pdV + ΣUidni
β = 1 / (kBT)

The Attempt at a Solution


(∂(βA)/∂β)N,V = (∂/∂β) * (βE - βTS)
(∂(βA)/∂β)N,V = (∂/∂β) * (βE - TS/(kBT))
(∂(βA)/∂β)N,V = (∂/∂β) * (βE - S/(kB))
(∂(βA)/∂β)N,V = (∂(βE)/∂β) [S/(kB) goes away because S is constant for Helmholtz Free Energy]
(∂(βA)/∂β)N,V = E

This solution just feels entirely too easy and simplified. I think you're supposed to do something with the fact that T and S are actual variables in the equation and some chain rule is needed, but that didn't seem to get me anywhere either. Anyone know what's really going on here?
 
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mobwars said:
(∂(βA)/∂β)N,V = (∂(βE)/∂β) [S/(kB) goes away because S is constant for Helmholtz Free Energy]
(∂(βA)/∂β)N,V = E

The two lines above are incorrect. As you suspected, S is not generally constant for a process in which N and V are kept constant.
Also, ∂(βE)/∂β ≠ E because E is not generally constant either.

It might help to make use of the first law as written in your second relevant equation.