# Chain rule: partial derivatives transformation

1. Jan 10, 2009

### 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.

2. Jan 10, 2009

### 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.

3. Jan 10, 2009

### 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

Last edited: Jan 10, 2009
4. Jan 13, 2009

### 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!