# Partial derivative in thermodynamics

1. May 21, 2012

### orgohell

So I have a proof and I can't follow the process, I think its because I haven't learned how to do partial derivatives or I've forgotten, anyways can someone tell me if this is a rule in calculus

(∂Cv/∂V)T=0

I've gotten to
[(∂/∂V)(∂U/∂T)V]T

and the proof I have goes to
[(∂/∂T)(∂U/∂V)T]V
-is this a rule? they were able to switch the denominators of the derivatives which is can see but then they switched the constants aswell If so what is the name so I can read up on it, I can't seem to find it with google

2. May 21, 2012

### Tosh5457

That's allowed because U is a function that is in the conditions of Schwarz's theorem (which can be derived from stokes' theorem)