Change of variable, second order differentiation

**Blamo_slamo** · Nov6-09, 12:07 AM

Okay, so my problem lies within taking the second derivative of a change of variable equation.

w = f(x,y); x = u + v, y = u - v

so far I have the first derivative:

dw/dx = (dw/dv)(dv/dx) + (dw/du)(du/dx) = (d/dv + d/du)w

Now I'm having problems in finding my second derivative:

d²w/dx² = ?

I don't necessarily want the answer, but more of how to get to the answer. I figure if I get how to do it with dw/dx, dw/dy would follow suit.
Any help would be greatly appreciated!
Thanks.

**Mark44** · Nov6-09, 02:37 AM

All your derivatives are partial derivatives, and your formula for dw/dx is completely wrong.

$LaTeX Code: \\frac{\\partial w}{\\partial x}$ is just itself; you can't do much with it without knowing more about f(x, y). What I think you want is the partial of w with respect to u, and possibly the partial of w with respect to v. From these you can get the second partials.

Here's the formula for the chain rule in Leibniz notation for your problem:

$LaTeX Code: \\frac{\\partial w}{\\partial u}~=~\\frac{\\partial w}{\\partial x} \\frac{\\partial x}{\\partial u}+ \\frac{\\partial w}{\\partial y} \\frac{\\partial y}{\\partial u}$

This formula can be simplified, since
$LaTeX Code: \\frac{\\partial x}{\\partial u}~=~\\frac{\\partial x}{\\partial v}~=~1$

and

$LaTeX Code: \\frac{\\partial y}{\\partial u}~=~1$
$LaTeX Code: \\frac{\\partial y}{\\partial v}~=~-1$

$LaTeX Code: \\frac{\\partial w}{\\partial v}$ is calculated in a similar manner as was used for the partial of w with respect to u, by replacing u everywhere it appears in the formula above with v.

**Blamo_slamo** · Nov6-09, 12:51 PM

First I would like to thank you very much, also I knew they were partials, but I'm a noob to this forum, and didn't know how to put down partials!

The way I started to set it up, was with how the book explained to do it, but your way would probably be my normal approach. I did manage to get the right answer thanks to your help!