Thermodynamics, manipulating partial derivatives

[MexChemE]

Hello PF! It's been a while since I last posted here. I have come across a problem in my textbook, which asks me to find expressions for V as a function of T and P, starting from the coefficients of thermal expansion and compressibility.
$$\alpha = \frac{1}{V} \left(\frac{\partial V}{\partial T} \right)_P$$
$$\beta = -\frac{1}{V} \left(\frac{\partial V}{\partial P} \right)_T$$
I've already solved the problem, I separated the differentials and integrated, then cleared for V. Now, here's where I have trouble. This is how I manipulated the differentials:
$$\alpha \ dT= \frac{1}{V} \ dV$$
$$\beta \ dP= -\frac{1}{V} \ dV$$
Both my professor, and Castellan's PChem text say this is correct, however, neither my professor nor the book explain why the separation of a partial derivative works in this case. Math professors have always said we can't "break" a partial derivative the same way we do for a regular derivative. If anyone could offer some insight about this case, it would be very helpful. Thanks!

 

[sweet springs]

$$\alpha dT=\frac{dV}{V}$$ under the condition p=const or dp=0
$$\beta dP=\frac{-dV}{V}$$ under the condition T=const or dT=0
You need these conditions.

 

[Chestermiller]

If V is a function V(P,T) of P and T, then
$$dV=\left(\frac{\partial V}{\partial P}\right)_TdP+\left(\frac{\partial V}{\partial T}\right)_PdT$$

Chet

 

[MexChemE]

Thank you both! I understand now, I thought they were just magically turning the partial derivative into a regular one. I wasn't aware of how both coefficients relate to the differential of V. We haven't covered partial derivatives or differentials of multivariable functions in Calc III, my professor is going a bit slowly. But we have started to use them in Thermodynamics I, so everything I know about the subject right now is from what I have read on my own.

Oh, and I'm sorry you had to move my thread, I thought it could be posted in the physics section since I wasn't asking for help on solving the problem itself, just a doubt with the math involved, but I'll be more careful next time.