What is the significance of the temperature derivative of the isentropic bulk modulus? I've been investigating various properties of the Gruneisen parameter and in my calculations the temperature derivative of the isentropic bulk modulus keeps coming up, i.e.
$$\left( \frac{\partial K_S}{\partial T}\right)_V.$$
Is there an intuitive way to think about this derivative or a way to write it in a different form? So far this is my thinking:
$$
\left( \frac{\partial K_S}{\partial T}\right)_V = \frac{\partial}{\partial T}\left( -V\left(\frac{\partial P}{\partial V}\right)_S\right)_V \\
= V\frac{\partial}{\partial V}\left( \left(\frac{\partial P}{\partial T}\right)_V\right)_S \\
= -V\frac{\partial}{\partial V}\left( \alpha K_T\right)_S \\
= -V\alpha \left( \frac{\partial K_T}{\partial V}\right)_S + VK_T \left( \frac{\partial \alpha}{\partial V}\right)_S
$$
where $\alpha$ is the thermal expansion coefficient and $K_T$ is the isothermal bulk modulus. But I think
$$
\left( \frac{\partial \alpha}{\partial V}\right)_S = \frac{\partial}{\partial V}\left( \left(\frac{\partial V}{\partial T} \right)_P \right)_S = \frac{\partial}{\partial T}\left( \left(\frac{\partial V}{\partial V} \right)_S \right)_P = 0
$$
therefore
$$
\left( \frac{\partial K_S}{\partial T}\right)_V = -V\alpha \left( \frac{\partial K_T}{\partial V}\right)_S.
$$
But I'm not sure this helps me. If the derivative was of the isentropic bulk modulus at constant entropy or the isothermal bulk modulus at constant temperature, I feel like this would be easier to interpret as the third derivative of either the internal energy or Helmholtz free energy respectively, but I'm not sure how to deal with transforming the derivative properly without the cyclic rule.
EDIT - I was missing a factor of V. Thanks to Chester Miller for pointing it out.
Any ideas are appreciated :-)
 A: You have $$C_p=C_v+T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_P$$
Also, $$\left(\frac{\partial V}{\partial T}\right)_P=-\frac{\left(\frac{\partial P}{\partial T}\right)_V}{\left(\frac{\partial P}{\partial V}\right)_T}$$
So, $$C_p=C_v-T\frac{\left[\left(\frac{\partial P}{\partial T}\right)_V\right]^2}{\left(\frac{\partial P}{\partial V}\right)_T}$$where P and $C_v$ are now regarded as functions of V and T.   So, $$K_S=\frac{C_p}{C_v}K_T=K_T+V\frac{T}{C_v}\left[\left(\frac{\partial P}{\partial T}\right)_V\right]^2$$
The partial derivative of this with respect to T at constant V can readily be determined.
A: This is not an answer per se, but it's too long for a comment.
Your maneuver in your first step
$$
\frac{\partial}{\partial T}\left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V 
= \frac{\partial}{\partial V}\left( \left(\frac{\partial P}{\partial T}\right)_V\right)_S 
$$
is not necessarily valid.  Partial derivatives only necessarily commute if you're holding variable #1 constant when you're taking the derivative with respect to variable #2, and vice versa;  so it is true that
$$
\frac{\partial}{\partial S}\left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V 
= \frac{\partial}{\partial V}\left( \left(\frac{\partial P}{\partial S}\right)_V\right)_S 
$$
or the same thing with $T$ replaced with $S$ everywhere.  But the "mixed" version is not necessarily true.
To show that this can in fact fail to be true, let $f = x^2 + x + y/x$, and let $z = xy$.  We wish to take the derivatives
$$
\frac{\partial}{\partial x}\left( \left(\frac{\partial f}{\partial y}\right)_z\right)_y 
\stackrel{?}{=} \frac{\partial}{\partial y}\left( \left(\frac{\partial f}{\partial x}\right)_y\right)_z 
$$
(Your first step is precisely this maneuver with $f = P$, $x = T$, $y = V$, and $z = S$.)
For the left-hand side, we have
$$
\left(\frac{\partial f}{\partial y}\right)_z = \frac{\partial}{\partial y} \left( \frac{z^2}{y^2} + \frac{z}{y} + \frac{y^2}{z} \right)_z \\
= -2\frac{z^2}{y^3} - \frac{z}{y^2} + \frac{2y}{z} \\
= -2 \frac{x^2}{y} - \frac{x}{y} + \frac{2}{x}
$$
and so
$$
\mathrm{LHS} = \frac{\partial}{\partial x} \left( -2 \frac{x^2}{y} - \frac{x}{y} + \frac{2}{x} \right)_y = - \frac{4x}{y} - \frac{1}{y} - \frac{2}{x^2}.
$$
Meanwhile, for the right-hand side, we have
$$
\left(\frac{\partial f}{\partial x}\right)_y = 2x + 1 - \frac{y}{x^2} \\
= \frac{2z}{y} + 1 - \frac{y^3}{z^2}
$$
and so the right-hand side is
$$
\mathrm{RHS} = \frac{\partial}{\partial y} \left( \frac{2z}{y} + 1 - 2 \frac{y^3}{z^2} \right)_z = - \frac{z}{y^2} + 0 - 3\frac{y^2}{z^2} = - \frac{2x}{y} + 0 - \frac{3}{x^2}.
$$
Obviously, we have $\mathrm{LHS} \neq \mathrm{RHS}$;  in fact, none of the three terms in this expression agree.

EDIT: To fix this, we can use the chain rule which holds so long as we're holding the same things constant in all the derivatives:
$$
\frac{\partial}{\partial T}\left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V = \left(\frac{\partial S}{\partial T} \right)_V \frac{\partial}{\partial S} \left( \left(\frac{\partial P}{\partial V}\right)_S\right)_V \\
= \frac{C_V}{T} \frac{\partial}{\partial V} \left( \left(\frac{\partial P}{\partial S}\right)_V\right)_S \\
= \frac{C_V}{T} \frac{\partial}{\partial V} \left(  \left( \frac{\partial T}{\partial S}\right)_V \left(\frac{\partial P}{\partial T}\right)_V \right)_S \\
= \frac{C_V}{T} \frac{\partial}{\partial V} \left( \frac{T}{C_V} \left(\frac{\partial P}{\partial T}\right)_V \right)_S \\
= \frac{\partial}{\partial V} \left( \left(\frac{\partial P}{\partial T}\right)_V \right)_S + \left(\frac{\partial P}{\partial T}\right)_V  \frac{\partial}{\partial V} \left( \ln \left( \frac{T}{C_V} \right) \right)_S.
$$
