Differential, why three members

[Uku]

Hello!

I have this thermodynamical expression:
[itex]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)[/itex]
Basically saying:
[itex]\frac{4}{3}\sigma T^3 V=S[/itex]

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 [itex]\sigma[/itex] is a constant.

 

[HallsofIvy]

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

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

 

[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.

 

[Uku]

HallsofIvy nailed it, thanks!

U.

 

[Mark44]

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 [STRIKE]derivative[/STRIKE] differential of a multi variable function is taken.

Fixed that for you

 

[I like Serena]

Fixed that for you

Thanks.