derivation of partial differential equations for changes in U


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> Changes in internal energy at 
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> 1. At constant temperature, $$d U = \left({\partial U \over \partial V}\right)\_TdV$$
> 2. At constant volume,
>  $$d U = \left({\partial U \over \partial T}\right)\_VdT$$
> 3. When both temperature and volume is changing,
>  $$d U = \left({\partial U \over \partial V}\right)\_TdV +\left({\partial U \over \partial T}\right)\_VdT$$
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> > I want to know how do we get these equations. 
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I think I can speculate a bit about how do we get the first and second equation. 


For the first equation, since the temperature is kept constant, we can plot a U-V graph. 


**Tell me if I am wrong here** but for that graph the slope at a given point is $\displaystyle \left({\partial U \over \partial V}\right)\_T$.


Now since the change is a infinitesimal change, the slope of tangent at that point is given by $\displaystyle \left({\partial U \over \partial V}\right)\_T$. 


So $\displaystyle {\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)\_T$, since the change is infinitesimal, $\Delta U = dU, \Delta V = dV$ and substituting this we will get out equation 1. Same can be said for equation 2.


I have no idea how do we get the equation 3, I think it is a clever use of product rule or something like that. 




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* Is my understanding for the equations 1 and 2 correct ?
* How do we get equation 3 from 1 and 2 ? because simply adding 1 and 2 does only give the RHS of 3 not the LHS.




One definition of a state function such as internal energy $U$ is that its infinitesimal change is equal to an exact differential. Thus in general


$$\mathrm{d}U=\sum\_{i=1}^n\left(\frac{\partial U}{\partial x\_i}\right)\_{x\_j} \mathrm{d}x\_i.$$


There are various variables with regard to which one can apply the partial derivative. Entropy, volume, temperature to name a few. Usually we assume that our thermodynamic state functions are two-dimensional but this is only if the number of moles is constant.


$$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum\_{i=1}^k\frac{\partial U}{\partial n\_i}\mathrm{d}n\_i \overset{\mathrm{d}n\_j\ =\ 0}{\implies} \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)\_{V,n\_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)\_{T,n\_j}\mathrm{d}V.$$


A good way to think of this assumption is that other functions are also simply functions of $T$ and $V$, *e.g.*, $S = S(T,V)$. From the equation


$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)\_{V,n\_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)\_{T,n\_j}\mathrm{d}V$$


you can reach the other two equations by imposing $\mathrm{d}T = 0$ and $\mathrm{d}V = 0$, respectively.


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