Higher Partial Derivatives & Chain Rule

[kingwinner]

Higher Partial Derivatives & Chain Rule (urgent)

I'll have a test this evening, and I don't want to fail on a question like this, so please help me out! I will greatly appreciate for any help provided.

The question:
http://www.geocities.com/asdfasdf23135/advcal11.JPG

My attempt:
http://www.geocities.com/asdfasdf23135/advcal10.JPG

From the solutions manual:
http://www.geocities.com/asdfasdf23135/advcal12.JPG


My first-order derivative is correct, but my second-order partial derivative Fjj is wrong. But I don't understand WHY I am wrong. Can someone please show me HOW to calculuate Fjj? I even tried to draw tree diagrams to clarfiy the dependence of variables, but I still got messed up, this is very very frustrating. I spent an hour looking through my work (I'm not kidding), but I can't locate my mistake, nor do I understand the way the solutions manual calculate Fjj.

Thanks again!

 

[Avodyne]

You have $F_j = x_j r^{-1}f'(r)$, but when you differentiate this with respect to x_j, you neglect to differentiate the factor of x_j itself. This is what gives the first term in F_jj in the solutions manual.

 

[kingwinner]

You have $F_j = x_j r^{-1}f'(r)$, but when you differentiate this with respect to x_j, you neglect to differentiate the factor of x_j itself. This is what gives the first term in F_jj in the solutions manual.

[Let D=curly d=partial derivative]


DF_j/Dx_j = (DF_j/Dr)[/color] (Dr/Dx_j) by the tree diagram (or the chain rule). What's wrong with this?


(DF_j/Dr)[/color] = D(f '(r) * x_j * r^-1) / Dr <----here I am differentiating with respect to r, so I just pulled out the constant x_j (it's index is j, not r, so it's constant with repsect to r), should be correct I guess.

 

[HallsofIvy]

[Let D=curly d=partial derivative]


DF_j/Dx_j = (DF_j/Dr)[/color] (Dr/Dx_j) by the tree diagram (or the chain rule). What's wrong with this?


(DF_j/Dr)[/color] = D(f '(r) * x_j * r^-1) / Dr <----here I am differentiating with respect to r, so I just pulled out the constant x_j (it's index is j, not r, so it's constant with repsect to r), should be correct I guess.

?? r is NOT an index to begin with! r depends upons every x_i so every x_i depends upon r. You cannot change r without changing every x_i.

 

[kingwinner]

?? r is NOT an index to begin with! r depends upons every x_i so every x_i depends upon r. You cannot change r without changing every x_i.

I see...
Then how can I calculate (DF_j/Dr)? Can someone please explain this?

 

[Avodyne]

F_j is *not* a function of r alone; it *also* depends on x_j explicitly, through the factor of x_j in front. So, DF_j/Dx_j = DF_j/Dx_j + (DF_j/Dr) (Dr/Dx_j), but DF_j/Dx_j means two different things on each side; on the left, it means differentiate with respect to both explicit and implicit (through r) appearances of x_j; on the right, it means differentiate only with respect to the explicit dependence on x_j. I would prefer to call the left side dF_j/dx_j, that is, the derivative with respect to *all* dependence on x_j, both implicit and explicit. But it's a partial derivative in the sense that all other components of x are being held fixed.

 

[kingwinner]

But, say, if f is a function of x,y,z, then all partial derivatives fx,fy,fz are functions of x,y,z as well. I remember seeing some remarks or theorem like this. So f, fx, fy, fz have the same tree diagram.

So in this case, F and Fx should have the same tree diagram. Following the branches, there is only one way to get from fx to x_j, i.e. from fx to r to x_j, so shouldn't DF_j/Dx_j = (DF_j/Dr) (Dr/Dx_j) be correct?

 

[Avodyne]

But, say, if f is a function of x,y,z, then all partial derivatives fx,fy,fz are functions of x,y,z as well. ... So in this case, F and Fx should have the same tree diagram.

Yes, this is your first tree diagram. But your second tree diagram, the one that goes through r, is correct for Z but not for F_j, because F_j depends explicitly on x_j; F_j cannot be written as a function of r alone, even though Z can. So, for F_j, your second diagram is missing a branch, one that connects F_j directly to x_j without going through r.