Partial Differentiation: What is the Chain Rule and How to Work It Out?

[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.

So could you please help me with a few things?

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: $F_{xx} \quad \text{and} \quad F_{xy}$

Thanks for any help you can give.

 

[dextercioby]

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.

So could you please help me with a few things?

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

What is the chain rule in partial differentiation?

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...

How do you work out the below?

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

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

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

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

Daniel.

Thanks for any help you can give.[/QUOTE]

 

[saltydog]

Prime numbers?

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

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

 

[Zurtex]

Thanks dextercioby, lol.

Just one more question then, what does this mean:

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

 

[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.

 

[dextercioby]

Thanks dextercioby, lol.

Just one more question then, what does this mean:

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

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.

 

[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?

 

[dextercioby]

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

Daniel.

 

[Zurtex]

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

Daniel.

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.

 

[Galileo]

...In general,they commute (Schwarz's theorem)...

*cough*Clairaut's theorem*cough*

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

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?

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

 

[dextercioby]

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

Daniel.

 

[HallsofIvy]

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

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

 

[dextercioby]

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

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

is given...

Daniel.

 

[Zurtex]

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

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.