## Differential, why three members

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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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: $$(4/3)\sigma T^3dV= (1+ 1/3)\sigma T^3dV= \sigma T^3dV+ (1/3)\sigma T^3dV$$
 Recognitions: Homework Help 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.

## Differential, why three members

HallsofIvy nailed it, thanks!

U.

Mentor
 Quote by 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 differential of a multi variable function is taken.
Fixed that for you

Recognitions:
Homework Help
 Quote by Mark44 Fixed that for you
Thanks.