Solve Clairaut's Theorem: Proving (dT/dV)S & (dS/dV)T

[Miriverite]

Homework Statement

Given dU = TdS - PdV, prove that:
1. (dT/dV)S = -(dP/dS)V
2. (dS/dV)T = (dP/dT)V

Homework Equations

Clairaut's states that d^2U/dXdY = d^2U/dYdX is true, or in my case:
[d/dX (dU/dY)X ]Y = [d/dY (dU/dX)Y ]X

The Attempt at a Solution

1. (dU/dS)V = T(S,V) and (dU/dV)S = -P(T,V)
so [d/dS (dU/dV)S]V = [d/dV (dU/dS)V]S
and that means [-P(S,V)/dS]V = [T(S,V)/dV]S
and thus, if we differentiate again (-dP/dS)V = (dT/dV)S

2. [d/dT (dU/dV)T]V = [d/dV (dU/dT)V]T
(dU/dV)T = -P(S,V) and (dU/dT)V = dS(T) (unsure of this)
so that means [-P(S,V)/dT]V = [dS(t)/dV)]T
This is where I get stuck, because there's a dS on the right hand side, but it's only P on the left hand, not dP. Am I missing an important step here?

Obviously we need to get the equation for dU into a form that depends not on S and V, but something that depends on T and V. I don't know how to do this, though..

 

[mighty2000]

First, let's rewrite the given equation as:
dU = TdS - PdV = T(S,V)dS - P(S,V)dV

1. From the given equation, we have:
(dU/dS)V = T(S,V)
(dU/dV)S = -P(S,V)

Using Clairaut's theorem, we get:
[d/dS (dU/dV)S]V = [d/dV (dU/dS)V]S
Simplifying, we get:
[-P(S,V)/dS]V = [T(S,V)/dV]S

Now, we differentiate both sides with respect to V:
[-dP/dS]V = (dT/dV)S

2. From the given equation, we have:
(dU/dT)V = S(T,V) (Note: this should be S(T,V) instead of dS(T))
(dU/dV)T = -P(T,V)

Using Clairaut's theorem, we get:
[d/dT (dU/dV)T]V = [d/dV (dU/dT)V]T
Simplifying, we get:
[-P(T,V)/dT]V = [S(T,V)/dV]T

Now, we differentiate both sides with respect to V:
[-dP/dT]V = (dS/dV)T

Hence, we have proved the given equations.