Is $C _v=T\left(\frac{\partial S}{\partial T}\right)_{V}$ true irrespective of reversibility of the process? Without specifying if the process is reversible or not my textbook uses the relation

$C_{V}=\left(\frac{\partial Q}{\partial T}\right)_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V}$

Is this equation true irrespective of reversibility of the process?  If that's so then how can one prove the last two expressions in the above equation?
In my view it should apply only on reversible process because only for them we have $d Q=T d S$.
Thank you
 A: The definition of $C_{V}$ is:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_v\tag{1}$$
$C_{V}$ is independent on how the process actually goes. It's the heat capacity "if you had chosen a specific path where volume is constant". When we write $)v$, we are specifying a path, a path were the volume is constant. So there is no other path option. The actual path is irrelevant.
Furthermore, as $C_{V}$ is the ratio between two state functions ($U$ and $T$), it follows that it must also be a state function, path-independent.
We can then use relation you provided
$C_{V}=\left(\frac{\partial Q}{\partial T}\right)_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V}$
which indeed is valid assuming reversibility. But since $C_{V}$ is path-independent, we can calculate using a hypothetical reversible path, which conforms to that equation. And the value will hold for any other irreversible situation. So the equation is correct, even in irreversible processes.
$C$ is path dependant. But $C_{V}$ is not.
A: The relationship between Cv, S, and T refers to thermodynamic equilibrium states of the system.  It involves physical properties of the material comprising the system, and is thus independent of any process path, either reversible or irreversible.
