Partial derivative of thermodynamics properties I know that in finding the partial derivative of certain thermodynamics property such as $H=H(P,V)$, we can hold the other variable as constant. But what will happen if the relation have more than two variables? For example, if a certain thermodynamics property of pure substance is given by $\Gamma$ where,
$$ \Gamma=S-\frac{U}{T}-\frac{PV}{T}$$
If I need to to find the value of $\left(\frac{\partial\Gamma}{\partial P}\right)_T$, when I perform a partial derivative can I also consider other thermodynamics properties (S and U) as constant and the equation becomes
$$\left(\frac{\partial\Gamma}{\partial P}\right)_T=0-0-\frac{V}{T}$$
 A: When you evaluate that partial derivative, you do not need to consider other properties, U and S, as constant.  You have $$d(T\ \Gamma)=TdS+SdT-dU+VdP+PdV$$But, $$dU=TdS-PdV$$Therefore, $$d(T\ \Gamma)=\Gamma dT + Td\Gamma=SdT-VdP$$So, $$\left(\frac{\partial \Gamma}{\partial P}\right)_T=-\frac{V}{T}$$
A: My conclusion is the same as in Chet Miller's answer. Still, in this case, I would have better start from the guideline of identifying the two independent variables an expression like $\Gamma$ is a function of. It makes explicit a step that has implicitly been used in Chet Miller's second equation.
In quantity like
$$ \Gamma=S-\frac{U}{T}-\frac{PV}{T}$$
we have the entropy, which is a fundamental thermodynamic function if it is seen as a function of $U$ and $V$, minus the product of $U$ (the first variable) times $\frac1T=\left.\frac{\partial{S}}{\partial{U}}\right|_V$, minus the product of $V$ (second variable of $S$) times  $\frac{P}{T}=\left.\frac{\partial{S}}{\partial{V}}\right|_V$. Therefore, we have an expression which can be used to perform a double Legendre transform of $S(U,V)$ into another fundamental equation $\Gamma(1/T,P/T)$.
It is a property of the Legendre transform that the derivative of $\Gamma$ with respect to one of its variables (in this case $P/T$) gives the minus the conjugate variable $V$, as a function of the same variables $\Gamma$ depends on. Then,
$$
\left.\frac{\partial \Gamma}{\partial{(P/T)}}\right|_T=T\left.\frac{\partial \Gamma}{\partial{P}}\right|_T= -V
$$
Where now $V$ should be intended as a function of $1/T$,$P/T)$ or equivalently $P$ and $T$.
