# Partial Differentiation 2

• Dec 19th 2007, 03:23 PM
UbikPkd
Partial Differentiation 2
Sorry, i'm stuck, here goes....

$u=ax+by$

$v=bx-ay$

In the first two parts of the question, you are asked to show that:

$\frac{\partial u}{\partial x})_{y}.\frac{\partial x}{\partial u})_{v}=\frac{a^{2}}{a^{2}+b^{2}}$

and

$\frac{\partial y}{\partial v})_{x}.\frac{\partial v}{\partial y})_{u}=\frac{a^{2}+b^{2}}{a^{2}}$

which i have done, and it's all fine and dandy :)

However, the last part confuzes me.... :(

If $f(x,y)=f(u,v)$, show that:

$\frac{\partial^{2} f}{\partial x \partial y}=ab (\frac{\partial^{2} f}{\partial u^{2}} - \frac{\partial^{2} f}{\partial v^{2}}) + (b^{2}-a^{2})\frac{\partial^{2} f}{\partial u \partial v}$

The problem is i'm not sure what $f(x,y)=f(u,v)$actually means. I understand it to be: f is a function of two variables, x and y, and f is a function of two variables, u and v, and these two functions are equal. But u and v are themselves functions of x and y so what is f?

A push in the right direction would be greatly appreciated! :)
• Dec 20th 2007, 12:49 AM
Opalg
Quote:

Originally Posted by UbikPkd
However, the last part confuzes me.... :(

If $f(x,y)=f(u,v)$, show that:

$\frac{\partial^{2} f}{\partial x \partial y}=ab (\frac{\partial^{2} f}{\partial u^{2}} - \frac{\partial^{2} f}{\partial v^{2}}) + (b^{2}-a^{2})\frac{\partial^{2} f}{\partial u \partial v}$

The problem is i'm not sure what $f(x,y)=f(u,v)$ actually means. I understand it to be: f is a function of two variables, x and y, and f is a function of two variables, u and v, and these two functions are equal. But u and v are themselves functions of x and y so what is f?

This simply means that f is a function of u and v, which becomes a function of x and y when you substitute u=ax+by and v=bx-ay in the definition of f. (It does not mean that the formula for f as a function of x and y is the same as the formula for f as a function of u and v.)

To give a concrete example, if $f(u,v) = u^2+v^2$ then $f(x,y) = (ax+by)^2 + (bx-ay)^2$. You may think that this is a bad and confusing notation (so do I), but that has to be what the question means. Once you interpret it that way, the question becomes an easy exercise in using the chain rule.
• Dec 20th 2007, 03:54 AM
UbikPkd
Thanks Opalg, i got it! :)