Formula for heat in thermodynamics. happy to join this community. :)
I have a simple problem. We have a well known formula
$C_v:=(\frac{\partial U}{\partial T})_V=T(\frac{\partial S}{\partial T})_V $. It comes from (we assume $N=const.$) $dU=TdS-pdV$ and taking $(\frac{\partial }{\partial T})_V$ of both sides. Of course we have 2nd term vanish, but why don't we differentiate T in a 1st term? I mean, why is rather $$(\frac{\partial U}{\partial T})_V=T(\frac{\partial S}{\partial T})_V$$
instead of something like
$$(\frac{\partial U}{\partial T})_V=(\frac{\partial T}{\partial T})_VdS+T(\frac{\partial S}{\partial T})_V=dS+T(\frac{\partial S}{\partial T})_V$$ (which is surely wrong, but I don't have better idea and just like to underline my doubts)?
Similarly, in a proof of a formula $$(\frac{\partial z}{\partial x})_y=-\frac{(\frac{\partial y}{\partial x})_z}{(\frac{\partial y}{\partial z})_x}$$ we conclude $$0=(\frac{\partial z}{\partial x})_y\cdot 1+(\frac{\partial z}{\partial y})_x\cdot (\frac{\partial y}{\partial x})_z $$ by taking partial derivative $(\frac{\partial}{\partial x})_z$ of a form $dz$ as a function of $(x,y)$ $$dz=(\frac{\partial z}{\partial x})_y dx +(\frac{\partial z}{\partial y})_xdy.$$
Why don't we have some 2nd order derivatives or sth like that? Why in those both examples we don't differentiate coefficients? 
Best regards
 A: "Formally", you divide the entire thing by this funny differential $(dT)_V$ (a differential change in $T$ along the surface $V=V_0$). So $\frac{dU}{(dT)_V}=\frac{T dS}{(dT)_V}-\frac{p dV}{(dT)_V}$. The first term combines together the way you expect while the second term vanishes, since $V$ doesn't change on the surface $V=V_0$. 
Note that what I've written above is not in standard notation. To my knowledge there is actually no standard notation for these constrained differentials.
There are two basic tricks required to make the manipulations we like to do in thermodynamics rigorous. These are the Legendre transform and the implicit function theorem. In this case all we need is the Legendre transform.
The Legendre transform is how we get away with making things like $T$ and $p$, which are really derivatives of things, into independent variables in their own right. It does this in a somewhat opaque way, so I will try to explain. In this context we take the Legendre transform of $U$ with respect to $S$. What we get is the Helmholtz free energy:
$$A(T,V)=\min_S U(S,V)-TS.$$
Why does this formula even make sense? Because by a second law argument, $U$ is convex with respect to $S$, so for each $V$ and $T$ there is exactly one point where $\left ( \frac{\partial U}{\partial S} \right )_V=T$. By knowing this value of $S$, we get new functions $S(T,V)$ and $U(T,V)$ for the entropy and internal energy depending on temperature. The latter is precisely the function we want to work with. The only remaining step is to differentiate this last function with respect to $T$. This turns out to be simple: the derivative of $U$ with respect to $S$ is $T$, so the derivative of the $S$-Legendre transform of $U$ with respect to $T$ is $-S$. In other words, when you switch from one to the other, you get the other conjugate variable but with a minus sign. You probably already knew this from the less rigorous procedure $dA=dU-TdS-SdT=TdS-pdV-TdS-SdT=-SdT-pdV$.
So now we can finish:
$$\left ( \frac{\partial U}{\partial T} \right )_V = \left ( \frac{\partial A}{\partial T} \right )_V + \left ( \frac{\partial (TS)}{\partial T} \right )_V = -S + S + T \left ( \frac{\partial S}{\partial T} \right )_V = T \left ( \frac{\partial S}{\partial T} \right )_V.$$
