# Partial Differntiation

1. Feb 2, 2005

### Zurtex

Hi hi, due to no fault of my own I have ended up missing 2 of my lectures on partial differentiation. I have a friend who takes amazing note but unfortunately my lecturer only seems to go through examples rather than standard methods and formulas.

What is the chain rule in partial differentiation?

How do you work out the below?

$$\frac{ \partial g}{\partial x} \quad \text{for} \quad x^2 + y^2 = r^2$$

And what do these mean please: $F_{xx} \quad \text{and} \quad F_{xy}$

2. Feb 2, 2005

### dextercioby

If they don't involve prime numbers :uhh: ,i think i can help... :tongue2:

Basically this one:
$$z=z(x(u,v),y(u,v)) \Rightarrow \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$$

and similar for "v"...I hope u can see the generalization for an arbitrary no.of functions & variables...

Give it a try using the formula i gave u...

That's the Lagrange notation for partial derivatives...In the Jacobi notation:
$$F_{xy}=\frac{\partial^{2} F}{\partial x \partial y}$$

Daniel.

3. Feb 2, 2005

### saltydog

Prime numbers?

Prime numbers? I'm hesitant to challenge you but I don't think prime numbers have anything to do with partial derivatives.

4. Feb 2, 2005

### Zurtex

Thanks dextercioby, lol.

Just one more question then, what does this mean:

$$\frac{\partial^{2} F}{\partial x \partial y}$$

5. Feb 2, 2005

### dextercioby

:rofl: :rofl: It's a famous thread in the HS homework section... :tongue2:

So yes,u didn't know what i was referring to...

Daniel.

6. Feb 2, 2005

### dextercioby

That's the notation for the second order partial derivative of the function F wrt the variables "y" and "x",viz.

$$\frac{\partial^{2} F}{\partial x \partial y}=:\frac{\partial}{\partial x}\frac{\partial}{\partial y} F$$

,where i designated specifically the order of the PD operators...In general,they commute (Schwarz's theorem)...

Daniel.

7. Feb 2, 2005

### Zurtex

I'm going to take a guess that:

$$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y} \right) = \frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x} \right)$$

Right?

8. Feb 2, 2005

### dextercioby

Not always.Any respectable mathematician knows at least one example of function for which the PD operators do not commute....

Daniel.

9. Feb 2, 2005

### Zurtex

Yeah thanks, the first time I saw them in my life was on monday and I've missed all subsequent lectures on them since and due to personal problems haven't even give them the time of day to think about them.

10. Feb 2, 2005

### Galileo

*cough*Clairaut's theorem*cough*

Ofcourse, the same theorem can have different names...

If $F_{xy}$ and $F_{yx}$ are both continuous on a disk D containing the point (a,b), then $F_{xy}(a,b)=F_{yx}(a,b)$.

11. Feb 2, 2005

### dextercioby

That's because your were taught mathematics in one place & me in another...
Anyway,i'll check up on Clairaut...

Daniel.

12. Feb 2, 2005

### HallsofIvy

Staff Emeritus
That really makes no sense at all because there is no "g" in your equation!

13. Feb 2, 2005

### dextercioby

It doesn't unless:
$$g=g(r(x,y))$$

where the functional dependence
$$g(r)$$

is given...

Daniel.

14. Feb 2, 2005

### Zurtex

That's what my friends notes have and I'm told that's the problem our lecturer gave, he's also asked us to partially differentiate a whole bunch of stuff without giving what variable with respect to what variable. I rather imagine a lot of the time they really have no idea what they are going on about (at least some of them), you can usually tell which ones don’t because they don’t offer support like the rest.

Edit: Oh and we were given no g(r)=... we were just asked:

$$\frac{ \partial g}{\partial x} \quad \text{for} \quad x^2 + y^2 = r^2$$

and that alone. Oh well, I'll go back and ask him to go over it again to see if he does know what he is writting.

Last edited: Feb 2, 2005