Deriving the TdS Equation for Thermal Physics

[Elzair]

Homework Statement

Derive the following equation

Homework Equations

TdS = C_{V} \left( \frac{\partial T}{\partial P} \right)_{V}dP + C_{P} \left( \frac{\partial T}{\partial V} \right)_{P}dV

The Attempt at a Solution

dU = \delta Q - \delta W

\delta Q = TdS for a closed system

C_{P} = T \left( \frac{\partial S}{\partial T} \right)_{P}

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

I am not sure where to go from here.

 

[siddharth]

I think the trick here is playing around with the relations.

If you consider S as a function of p and v as independent variables, then
dS = \left(\frac{\partial S}{\partial P}\right)_V dP + \left(\frac{\partial S}{\partial V}\right)_P dV

But, \frac{\partial S}{\partial P}_V = \left(\frac{\partial S}{\partial T}\right)_V \left(\frac{\partial T}{\partial P}\right)_V, and so on.