What this notation mean? (I know that is a partial derivative, but I don't understand the meaning of the evaluation bar at the right)
$$\frac{\partial g}{\partial T}\Big|_{SA,p}$$
Is this relation true?
$$\frac{\partial g}{\partial T}\Big|_{SA,p}=\left(\frac{\partial g}{\partial T}\right) _{SA,p}$$
Also used here
$$d\rho=\left(\frac{\partial \rho}{\partial T}\right)_{S,p}+dT\left(\frac{\partial \rho}{\partial S}\right)_{T,p}+dS\left(\frac{\partial \rho}{\partial p}\right)_{T,S}dp$$
And another thing, the $$d\rho$$ must be intended as the differential of the function $$\rho$$?
So if I have understood well, the equality can be rewritten as:
\begin{equation*} \frac{1}{\rho}\left(\frac{\partial \rho}{\partial T}dT+\frac{\partial \rho}{\partial S}dS+\frac{\partial \rho}{\partial p}dp\right)=-\rho\left(\frac{\partial \rho^{-1}}{\partial T}dT+\frac{\partial \rho^{-1}}{\partial S}dS+\frac{\partial \rho^{-1}}{\partial p}dp\right) \end{equation*}
\begin{equation*} \frac{1}{\rho}\sum_{i=1}^3\frac{\partial \rho}{\partial x_i}dx_i=-\rho\sum_{i=1}^3\frac{\partial \rho^{-1}}{\partial x_i}dx_i \end{equation*}
Proof:
$$ \frac{d\rho}{\rho}=-\frac{d\alpha}{\alpha}=-\rho d\rho^{-1}=-\rho\sum_{i=1}^3\frac{\partial \rho^{-1}}{\partial x_i}dx_i $$ Where $$ \alpha=\frac{1}{\rho}\\ \rho=f(S,T,p)\\ $$
$\Box$
