# Partial derivatives

1. Apr 1, 2008

### ehrenfest

[SOLVED] partial derivatives

1. The problem statement, all variables and given/known data
Can the partial derivative of a function depend depend on the form it is in?

Say, z = f(x,y), and y=g(x,w). If I take

$$\frac{\partial z}{\partial y}$$

then I get

$$\frac{\partial f(x,y)}{\partial y}$$

which is not necessarily 0. But $\frac{\partial z}{\partial y}$ is also equal to

$$\frac{\partial f(x,g(x,w))}{\partial y}$$

which is identically 0. This is DRIVING ME OUT OF MY MIND.

Also, say we have z = f(x,y) = x^2+y^2+y and we also know that x=y. Then z also equals g(x,y) = x^2+y^2+x.

Thus, we get

$$2 y +1 = \frac{\partial f(x,y)}{\partial y} = \frac{\partial z}{\partial y} = \frac{\partial g(x,y)}{\partial y} = 2y$$

which is absurd. What is wrong with my logic?
All of these examples come from thermodynamics.
2. Relevant equations

3. The attempt at a solution

Last edited: Apr 1, 2008
2. Apr 1, 2008

### Dick

Of course the partial derivative depends on the form of the function. When you write a partial derivative you are implicitly assuming that some combination of variables is held constant. When you juggle the form around like that you are changing what you are thinking of as 'constant'. Don't do that. Use the chain rule for partial derivatives and everything will take care of itself.

3. Apr 1, 2008

### genneth

To expound on what Dick just said --- make sure you say which variables are being held constant, *explicitly*.

4. Apr 1, 2008

### ehrenfest

So, you are saying that partial derivatives do not make sense unless you hold enough variables constant to make the partial derivative unambiguous? That is, whenever I write down I partial derivative, I should always make sure that I have specified enough variables SO THAT THE PARTIAL DERIVATIVE IS INDEPENDENT OF THE FORM OF THE FUNCTION, right?

How do you know when you have specified enough variables to make the partial derivative unambiguous?

Last edited: Apr 1, 2008
5. Apr 1, 2008

### Dick

Uh, when you've specified enough that the function is only a function of one unfixed variable.

6. Apr 2, 2008

### ehrenfest

What about the first two questions in my last post?

7. Apr 2, 2008

### HallsofIvy

Staff Emeritus
The answer to both of those questions is "yes".