Why Change Total Derivative to Partial in dU=TdS-PdV?

[asdf1]

for the equation:
dU=TdS-PdV
why can you can the total derivative to a partial derivative, which makes the equation become
T=(partial derivative U)/(particle derivative S) ?

 

[StatusX]

You are taking the partial derivative of U with respect to S with V held fixed. So you take dV=0, and then divide through by dS.

 

[dextercioby]

Because dU is an integrable differential one-form.

Daniel.

 

[asdf1]

thank you very much! :)