Thermodynamics: Combining the 1st and 2nd Law I'm confused how combining the first and second laws can derive

$$dS = \left(\frac{C_p}{T}\right)dT+ \left[ \left( \frac{1}{T} \right) \left\{ \left( \frac{\partial H}{\partial P} \right)_T -V \right\} \right] dP$$

The 1st Law states $dU = dq + dw$, and the 2nd Law states $dS = \frac{dq_{rev}} {T}$

I factored out $\frac{1}{T}$ in the given equation to make

$$dS = \frac{1}{T} \left[ \left( C_p\, dT \right) + { \left( \frac{\partial H}{\partial P} \right)_TdP - V  dP } \right]$$

$C_p = \frac{dq}{dT}$

Therefore, 
$$dS = \frac{1}{T} \left[  dq + { \left( \frac{\partial H}{\partial P} \right)_TdP - V  dP } \right] $$

Without analyzing $\left\{ \left( \frac{\partial H}{\partial P} \right)_T - V  dP \right\}$, I already have $dS = \frac{dq}{T}$.

Can anyone explain this for me?

Thanks,
Starting with $$dH=TdS+VdP\tag{1}$$H is a function of T and P, so$$dH=\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP\tag{2}$$Substituting Eqn. 2 into Eqn. 1,
$$\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{3}$$But, from the definition of $C_p$, $$C_p\equiv \left(\frac{\partial H}{\partial T}\right)_P \tag{4}$$Substituting Eqn. 4 into Eqn. 3 gives:$$C_pdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{5}$$Solving for dS then gives the desired relationship.