differentiating an expression in differential form

Question

I haven't been familiar with what looks like differential forms so hoping to get some help here.

The extensive total enthalpy of a system is $H$. The intensive specific enthalpy of this system is $h=\frac{H}{m}$ where m is the mass of the substance.

So, $H = U + PV$ so its corresponding intensive specific enthalpy is $h = u + pv$ after dividing $H, U, PV$ by $m$ with $H$ the enthalpy, $U$ the internal energy of the system, $P$ the pressure and $V$ the volume.

In a text I am reading, the author worked:

$dh = du + Pdv + vdP$ which appears to be in differential form.

To be clearer:

dV(or dP) - what is $V( or P)$ being differentiated with respect to?

I fail to follow this. Any help is appreciated.

Edit: I realise what was wrong - provided an answer below

Answer 1

I think it follows from the fact that $pv = \frac{PV}{m}$ so if you consider the differential of that, you get, using the product rule that

$ d(pv) = d(\frac{PV}{m}) = dP \cdot \frac{V}{m} + \frac{P}{m}dV = vdP+pdV $

Answer 2

The issue isn't with the mathematics but with the physical implications following from the expression I am working with.

For the benefit of future audiences:

Note that (some included for completion):

$(S, V) \rightarrow U = Q - pdV: internal \space energy$

$(S, p) \rightarrow H = U + pV: enthalpy$

$(T, V) \rightarrow F = U - TS: helmholtz \space free \space energy$

$(T, p) \rightarrow G = U + pV - TS: gibbs \space free \space energy$

Invoking total differentiability on $U = U(S, V):$

$dU = \frac{\partial U}{dS}|_{v}dS + \frac{\partial U}{dV}|_{S}dV$

But $U = Q + pV = TS + pV$

$ \rightarrow \frac{\partial U}{dS}|_{v} = T$

$ \rightarrow \frac{\partial U}{dV}|_{S} = p$

By substitution, this yields:

$dU = TdS + pdV$ as is claimed by the author.