What is the derivation of the equation for (d/dv)(ds) in lecture notes?

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In the lecture notes
http://webraft.its.unimelb.edu.au/640322/pub/notes/lectures_common/lecture3.pdf

How did they derive the equation on the line above "L.H.S bracket is not a function of V thus"?

I can see that (d/dv)(ds)= "that equation except with the 'equals sign' replaced by '+' "

(dU/dV)=0 because it is an ideal gas.
 
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The expression for dS has three terms so we can write it as

dS = AdT + BdV + CdV

The condition for dS being an exact differential is

d(A)/dV = d(B+C)/dT ( partial differentiation)

But dB/dT is zero because B is held at fixed T, so

d(A)/dV = d(C)/dT

which is the equation you asked about.
 
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I don't understand why the condition on dS in order for it to be an exact differential. The reasoning for the exact differential on the notes I don't follow either.
 
I'm not clear what's troubling you.

dS has to be an exact differential for physical reasons. It just means there are no other sources of variation in S but these.

The little equation in brackets ( 'dz = ...') is a theorem, which says that if a total differential is the sum of two sources, then the sources have a special relationship. Just accept it for now. It's used to get the next line.
 
Mentz114 said:
I'm not clear what's troubling you.

dS has to be an exact differential for physical reasons. It just means there are no other sources of variation in S but these.

The little equation in brackets ( 'dz = ...') is a theorem, which says that if a total differential is the sum of two sources, then the sources have a special relationship. Just accept it for now. It's used to get the next line.


I see now but I seem to be more interested to see why the exact differential demands this form. You want dS to be an exact differential because S is a function of two variables and is a state function which behaves just like a perfect mathematical function of two variables.
 
That's how I see it, but talk to other people about this, it might be important.