Is it valid to divide by $\mathrm{d}T$? I'm trying to prove the following relation 
$$c_p -c_v = \Biggl[p+\biggl(\frac{\partial E}{\partial V}\biggr)_T\Biggr]\biggl(\frac{\partial V}{\partial T}\biggr)_p$$
Using the first law of thermodynamics as a start, I have gotten to the stage of 
$$c_p\,\mathrm{d}T = c_v\,\mathrm{d}T + \Biggl[\biggr(\frac{\partial E}{\partial V}\biggr)_T + p\Biggr]\mathrm{d}V$$
Can I simply say in my next line "Dividing by $\mathrm{d}T$ and keeping pressure constant yields"?
Then I would get the original expression but I'm not sure this is mathematically sound.
 A: When there is only one independent variable, procedures such as dividing throughout by infinitesimals like $dT$ are unambiguous (perhaps even rigorous; only someone acquainted with infinitesimal calculus can tell). However in multivariable calculus, in particular thermodynamics which always contains at least two independent variables, such a procedure is potentially confusing. For example, reasoning as "Dividing by $dT$ and keeping (say) enthalpy constant yields..." would result in incorrect conclusion. So why are we permitted to hold only pressure constant while transforming from the infinitesimal version (containing $dT,dV$) to partial-derivative version (containing $\frac{\partial V}{\partial T}|_p$)?
To see the reason let us go through the derivation. First law states:
\begin{align}
T~dS=dE+p~dV
\end{align}
which is true of arbitrary infinitesimal process. If we take entropy $S$ as a function of $T,p,$ and internal energy $E$ as a function of $T,V,$ then:
\begin{align}
dS&=\frac{\partial S}{\partial T}\Biggr|_pdT+\frac{\partial S}{\partial p}\Biggr|_pdp\\
dE&=\frac{\partial E}{\partial T}\Biggr|_VdT+\frac{\partial E}{\partial V}\Biggr|_TdV
\end{align}
Substituting into first law, and using definitions $T\frac{\partial S}{\partial T}|_p=c_p,~\frac{\partial E}{\partial T}|_V=c_V$, gives:
\begin{align}
c_pdT+T\frac{\partial S}{\partial p}\Biggr|_pdp=c_VdT+\left(p+\frac{\partial E}{\partial V}\Biggr|_T \right)dV
\end{align}
This is again valid for arbitrary infinitesimal process. 
Now we apply the equation above to a constant pressure process, to obtain:
\begin{align}
c_pdT=c_VdT+\left(p+\frac{\partial E}{\partial V}\Biggr|_T \right)dV
\end{align}
Above equation is valid only for constant pressure process. That is why when you divide by $dT$ throughout any partial derivatives that result must be taken to be at constant pressure:
\begin{align}
c_p=c_V+\left(p+\frac{\partial E}{\partial V}\Biggr|_T \right)\frac{\partial V}{\partial T}\Biggr|_p
\end{align}
