# Thread: Chain rule with multiple variables

1. ## Chain rule with multiple variables

My real analysis textbook asks me to

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

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

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

Now, if I can show that either $f_{xy}$ or $f_{yx}$ is continuous on $\mathbb{R}$, then I can let $f_{xy}=f_{yx}$ 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!

2. Originally Posted by hatsoff
My real analysis textbook asks me to

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

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

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

Now, if I can show that either $f_{xy}$ or $f_{yx}$ is continuous on $\mathbb{R}$, then I can let $f_{xy}=f_{yx}$ 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)?

3. 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 $f:\mathbb{R}^2\to\mathbb{R}$ and that $x=u+v$ and $y=u-v$, where $(u,v)\in\mathbb{R}^2$, 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.