Chain rule: partial derivatives transformation

[notReallyHere]

Hello.

Let g(x,y) be a function that has second order partial derivatives. Transform the differential equation

\frac{\delta ^{2}g}{\delta x^{2}}-\frac{\delta ^{2}g}{\delta y^{2}}=xyg

by chaning to the new variables u=x^2-y^2 and v=xy
The equation doesn't have to be solved.

Okay, so this is basically an exercise in using the chain rule.
The first part is pretty easy.

A
\frac{\delta g}{\delta x}=\frac{\delta g}{\delta u}\cdot \frac{\delta u}{\delta x}+\frac{\delta g}{\delta v}\cdot \frac{\delta v}{\delta x}=2x\cdot \frac{\delta g}{\delta u}+y\frac{\delta g}{\delta v}
and
B
\frac{\delta g}{\delta y}=\frac{\delta g}{\delta u}\cdot \frac{\delta u}{\delta y}+\frac{\delta g}{\delta v}\cdot \frac{\delta v}{\delta y}=-2y\cdot \frac{\delta g}{\delta u}+x\frac{\delta g}{\delta v}

Now I try to find \frac{\delta ^{2}g}{\delta x^{2}} using the information from A. I realize I'm supposed to use the chain rule again, but I just can't get the right answer, so I guess I'm missing some fundamental understanding of what's going on here.

Here's what the first part of the problem eventually should boil down to:
\frac{\delta ^{2}g}{\delta x^{2}}=2\frac{\delta g}{\delta u}+4y^{2}\cdot \frac{\delta^{2} g}{\delta u^{2}}+4xy\cdot \frac{\delta^{2} g}{\delta u\delta v}+y^{2}\cdot \frac{\delta^{2} g}{\delta v^{2}}

This part above is what I need help with, if I understand that part, I think I can manage to solve the rest of the problem.

Help is much appreciated. If I find out how to do it I will post here later. Thanks.

 

[AEM]

There is a rather old textbook, "Advanced Calculus" by Angus Taylor that addresses the kind of problem you are experiencing very carefully in Chapter 6 in a section titled "Second derivatives by the chain rule". Your library may have a copy. As I recall, when I first ran into this problem, the issue was as much notation as anything else.

 

[rootX]

I see where you are stumbling:

do you know how to do:

d/dx [ dg/du ] ?

Are you correctly using the product rule as:

it's d/dx[ 2x . dg / du]?

Do you know that dg/du consists of two variables (u,v) so you need to take care of that .. ?

I got the first part.. apparently, there aren't any tough work involved. You just need to understand:
d/dx [df (x,y) /dx] = d^2.f / (dx.dy) + d^2 . f/ (dx^2)

Use trees - they help: write f, and then two children u,v .. both have x,y children

 

[Mmmm]

\frac{\delta ^2 g}{\delta x^2}=\frac{\delta }{\delta x} \left(\frac{\delta g}{\delta x}\right)

= \frac{\delta }{\delta x} \left( \frac{\delta g}{\delta u}\cdot \frac{\delta u}{\delta x} \right) + \frac{\delta }{\delta x} \left( \frac{\delta g}{\delta v}\cdot \frac{\delta v}{\delta x} \right)

Then use the product rule on each. then the chain rule again when differentiating a partial differential of g. It's long winded but it'll do the trick!