# Differential, why three members

1. Jun 28, 2012

### Uku

Hello!

I have this thermodynamical expression:
$dS=\sigma T^3 dV+4V\sigma T^2 dT+\frac{1}{3}\sigma T^3 dV=d(\frac{4}{3}\sigma T^3 V)$
Basically saying:
$\frac{4}{3}\sigma T^3 V=S$

Now, I do not get this.. d(expr) part, why are there three members to d(expr), with 2x dV and 1x dT.. nope..
I might add that $\sigma$ is a constant.

Last edited: Jun 28, 2012
2. Jun 28, 2012

### HallsofIvy

Where did you find that? With $\sigma$ constant, there should be two parts:
$$d((4/3)\sigma T^3V)= 4\sigma T^2V dT+ (4/3)\sigma T^3 dV$$

Oh, wait, what they have done is just separate that last term:
$$(4/3)\sigma T^3dV= (1+ 1/3)\sigma T^3dV= \sigma T^3dV+ (1/3)\sigma T^3dV$$

3. Jun 29, 2012

### I like Serena

Hi Uku!

Can you clarify what you do not get?

What you have is similar to:
$$df(x,y)={\partial f \over \partial x}dx + {\partial f \over \partial y}dy$$
This is how the derivative of a multi variable function is taken.

4. Jun 29, 2012

### Uku

HallsofIvy nailed it, thanks!

U.

5. Jun 29, 2012

### Staff: Mentor

Fixed that for you

6. Jun 29, 2012

Thanks.