Understanding the use of $d$ and $\partial$ in thermodynamics It seems a hundred variations of this question have been asked, and it's difficult to find which of those questions relates to exactly what I'm asking. My apologies if exactly this question has already been asked.
I am self-studying Elements of Gasdynamics by Liepmann and Roshko. In a section on adiabatic, reversible processes, the authors write

Very important in later applications to fluid flow is the adiabatic, reversible process: i.e., a process in which no heat is transferred to or from the system and in which the work is done reversibly. Then ... depending on whether we choose $v,T$ or $p,T$ as the variables, the adiabatic, reversible process is given by
\begin{align*}
de&=-pdv\\
dh&=vdp
\end{align*}
It follows that
$$
\frac{\partial e}{\partial v}dv +\frac{\partial e}{\partial T}dT=-pdv
$$

By manipulating the latter equation, they go on to derive this equation
$$
\frac{dT}{dv}=-\frac{1}{c_v}\left(\frac{\partial e}{\partial v}+p\right)
$$
My question is why the authors choose to write the partial derivative of $e=e(v,T)$ with respect to $v$ as $\frac{\partial e}{\partial v}$ but don't write $\frac{dT}{dv}$ with similar partial derivative notation. Obviously, $T$ depends on more than just $v$, so I would think that a partial derivative symbol is appropriate. My only thought as to why is that in this case, the authors have specifically identified $e$ as a function of $v,T$, whereas they have not made that identification for the other variables.
In a similar equation on the same page, the authors derive
$$
\frac{dp}{dv}=-\frac{p}{v}\frac{dh}{de}
$$
Again, I'm confused why the partial derivative symbol is not used.
 A: The partial and total derivatives are different things, but they are related via the Chain Rule:
For $f(x,y,z)$, the differential of $f$ is:
$$df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz
$$
If we then divide by $dx$, we can obtain the Total Derivative of $f$ with respect to $x$:
$$\frac {df}{dx} = \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac {dy}{dx}+\frac{\partial f}{\partial z}\frac {dz}{dx}
$$
So let us apply this to the case at hand.  Your book takes $e$ as a function $e(v,T)$, and then applies the same differential rule we used above, setting that equal to:
$$de = \frac{\partial e}{\partial v}dv +\frac{\partial e}{\partial T}dT=-pdv $$
Now, technically we can be even more clear by giving subscripts:
$$de = \left(\frac{\partial e}{\partial v}\right)_T dv +\left(\frac{\partial e}{\partial T}\right)_vdT=-pdv $$
Where "subscript $v$" means "at constant $v$."
The authors then divide thru by $dv$, obtaining:
$$\left(\frac{\partial e}{\partial v}\right)_T +\left(\frac{\partial e}{\partial T}\right)_v \frac{dT}{dv}=-p $$
Finally, they use the definition of specific heat capacity at constant volume:
$$c_v \equiv \left(\frac{\partial e}{\partial T}\right)_v $$
And rearrange algebraically to produce your equation:
$$\frac{dT}{dv}=-\frac{1}{c_v}\left(\left(\frac{\partial e}{\partial v}\right)_T+p\right)
$$
except that they omit the subscript which is understood.
