# Chain rule with multiple variables

• Nov 9th 2009, 06:31 AM
hatsoff
Chain rule with multiple variables
My real analysis textbook asks me to

Let $\displaystyle h(u,v)=f(u+v,u-v)$. Show that $\displaystyle f_{xx}-f_{yy}=h_{uv}$.

It looks like a fairly simple problem, at the end of the chapter on the chain rule. However, when I apply the chain rule, I get

$\displaystyle h_{uv}=f_{xx}-f_{yy}+f_{yx}-f_{xy}$

Now, if I can show that either $\displaystyle f_{xy}$ or $\displaystyle f_{yx}$ is continuous on $\displaystyle \mathbb{R}$, then I can let $\displaystyle f_{xy}=f_{yx}$ and the conclusion follows. But how do I prove that with so little information? Or am I going about this entirely the wrong way?

Thanks!
• Nov 9th 2009, 07:07 AM
HallsofIvy
Quote:

Originally Posted by hatsoff
My real analysis textbook asks me to

Let $\displaystyle h(u,v)=f(u+v,u-v)$. Show that $\displaystyle f_{xx}-f_{yy}=h_{uv}$.

It looks like a fairly simple problem, at the end of the chapter on the chain rule. However, when I apply the chain rule, I get

$\displaystyle h_{uv}=f_{xx}-f_{yy}+f_{yx}-f_{xy}$

Now, if I can show that either $\displaystyle f_{xy}$ or $\displaystyle f_{yx}$ is continuous on $\displaystyle \mathbb{R}$, then I can let $\displaystyle f_{xy}=f_{yx}$ and the conclusion follows. But how do I prove that with so little information? Or am I going about this entirely the wrong way?

Thanks!

What is the exact statement of the problem? What you have written makes no sense. For example, there is nothing said, in the hypothesis, about u and v being functions of x and y! Was there something said about f(x,y)?
• Nov 9th 2009, 07:26 AM
hatsoff
Quote:

Originally Posted by HallsofIvy
What is the exact statement of the problem? What you have written makes no sense. For example, there is nothing said, in the hypothesis, about u and v being functions of x and y! Was there something said about f(x,y)?

What I wrote in my OP is the original wording. I assume that since this is a textbook on real analysis, we have $\displaystyle f:\mathbb{R}^2\to\mathbb{R}$ and that $\displaystyle x=u+v$ and $\displaystyle y=u-v$, where $\displaystyle (u,v)\in\mathbb{R}^2$, but perhaps my assumptions are incomplete (or even wrong).

If you doubt me, the textbook is freely available from the author in pdf format, here.

See exercise 12a from section 5.4, p358---or pdf page 367.