Differential equations without derivation In my introductory physics book there is this equation:
$$U=\frac{PV}{(\gamma -1)}$$
that is transformed to
$$dU=\frac{Pdv+Vdp}{(\gamma -1)}$$
It seems like a derivative operation, but where are the differentials at the denominators?
What kind of operation is this? And under which conditions can be applied?
Thanks.
 A: We starts with
$$U=\frac{PV}{(\gamma -1)}$$
I'll just differentiate this w.r.t a dummy variable $x$, assuming $\gamma$ is a constant.
$$\frac{dU}{dx}=\frac{P\frac{dV}{dx}+V\frac{dP}{dx}}{(\gamma -1)}$$
Multiplying $dx$ throughout,
$$dU=\frac{PdV+VdP}{(\gamma -1)}$$
as shown.
A: If $\gamma$ is a constantand $U,P,V$ are function for the same variable, say $x$, then you get:
$$U=\frac{PV}{(\gamma -1)}\\\dfrac{dU}{dx}=\dfrac{d}{dx}\frac{PV}{(\gamma -1)}=\frac{P\dfrac{dV}{dx}+V\dfrac{dP}{dx}}{(\gamma -1)}$$hence $$dU=\frac{PdV+VdP}{(\gamma -1)}$$
A: It seems as though $U$ is regarding potential (internal) energy. This is always regarded as a state function, which means it can be written as an exact differential (or equivalently that the function is path independent, or conservative). More precisely, for $U = U(P, V)$,
$$dU = \frac{\partial U}{\partial P}dP + \frac{\partial U}{\partial V} dV$$
More generally, if $F(x_1,..., x_n)$ is a state function, then:
$$dF = \frac{\partial F}{\partial x_1}dx_1 +...+ \frac{\partial F}{\partial x_n} dx_n$$
In your case, we have $U = \frac{PV}{\gamma - 1}$, and if we treat $\gamma$ as a constant,
$$\frac{\partial U}{\partial P} = \frac{V}{\gamma - 1} \\ \frac{\partial U}{\partial V} = \frac{P}{\gamma - 1}$$
So that
$$dU = \frac{PdV + VdP}{\gamma - 1}$$
As desired.
Edit:
Basically, you can parametrize $P, V$ as you like, say with respect to some arbitrary variable $t$. The reason this works here is that $U$ is path independent, which means if you start at $P_0, V_0$ and move to $P_1, V_1$, the energy will change from $U_0$ to $U_1$ regardless of which path you moved along in the $PV$-plane. 
Parametrizing $P,V$ as $P = P(t), V = V(t)$, and taking a derivative of your equation with respect to $t$ you get
$$\frac{dU}{dt} = \frac{V \frac{dP}{dt} + P \frac{dV}{dt}}{\gamma - 1}$$
By exploiting the chain rule.
Multiplying through by $dt$ and using the fact that a function parametrized as $F = F(t)$, has small change:
$$dF = \frac{dF}{dt}dt$$
Plugging this in with $F = U, V, P$,  you again get the result you were looking for.
