I'm studying Mathematical Analysis and trying to solve example problems as I go. Specifically, this problem comes after an introduction to $C$ parameterizations of $r:I\subseteq\mathbb R \to \mathbb R^m$.
We have a $C^1$ function $f = f(x,y): \mathbb R^2 \to \mathbb R$ where we switch variables $(x,y) = (s+t,s-t)$.
Prove that $\left( \dfrac{\partial f}{\partial x}\right )^2 - \left( \dfrac{\partial f}{\partial y}\right )^2 = \dfrac{\partial f}{\partial s}\dfrac{\partial f}{\partial t}$
I understand that it's asking me to calculate partial derivatives $f_x$ and $f_y$, then substract them, but I did that and yielded nothing useful.
Plus, shouldn't I be doing something with $C^1$, which gives us that $f$ is continuously derivable in $\mathbb R^2$, and that $f'$ is also continuous? How could I put that into play?
I'm completely stalled, and would appreciate any help.
Call the map $\bar\gamma(s,t) = (s+t,s-t)$.
When you make the change of variables, you are obtaining a new function $f(\bar\gamma(s,t))$; the equation you are asked to prove combines these 2 functions. You need to be able to compute the derivates $f(\bar\gamma(s,t))_s$ and $f(\bar\gamma(s,t))_t$ in terms of $f_x$ and $f_y$.
Here is where $f$ being $C^1$ comes to play; it lets us write:
$$\frac{\partial f}{\partial s} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial s} = f_x + f_y,$$
and similarly for $\frac{\partial f}{\partial t} = f_x - f_y$.