Partial derivative with respect to a function? I have $f(x, y)$ is a class $C^2$ function where $x = u + v$, and $y = u - v$
How do I get $$\frac{\partial^2 f}{\partial x^2}$$
Well before we even talk about that one, I don't even know how to get the first partial. I'm confused about how to go about solving it when its with respect to x. Any hints?
Edit: I don't have a specific equation for $f(x, y)$. The full question was to show 
$$\frac{\partial^2f}{\partial{u}\ \partial{v}} = \frac{\partial^2f}{\partial{x^2}} - \frac{\partial^2f}{\partial{y^2}}$$
 A: You can show this by using the chain rule twice. First,
$$\frac{\partial{f}}{\partial{v}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{v}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{v}} \\
= \frac{\partial{f}}{\partial{x}} - \frac{\partial{f}}{\partial{y}}$$
and then
$$\frac{\partial^2 f}{\partial{u}\ \partial{v}}  = \frac{\partial}{\partial{u}}\frac{\partial{f}}{\partial{x}} - \frac{\partial}{\partial{u}}\frac{\partial{f}}{\partial{y}} \\
= \left(\frac{\partial^2 f}{\partial{x}^2}\frac{\partial{x}}{\partial{u}} + \frac{\partial^2 f}{\partial{y}\ \partial{x}}\frac{\partial{y}}{\partial{u}}\right) - \left(\frac{\partial^2 f}{\partial{x}\ \partial{y}}\frac{\partial{x}}{\partial{u}} + \frac{\partial^2 f}{\partial{y}^2}\frac{\partial{y}}{\partial{u}}\right) \\
= \left(\frac{\partial^2 f}{\partial{x}^2} + \frac{\partial^2 f}{\partial{y}\ \partial{x}}\right) - \left(\frac{\partial^2 f}{\partial{x}\ \partial{y}} + \frac{\partial^2 f}{\partial{y}^2}\right) \\
= \frac{\partial^2 f}{\partial{x}^2} - \frac{\partial^2 f}{\partial{y}^2}$$
The last equality is valid because $$\frac{\partial^2 f}{\partial{y}\ \partial{x}} = \frac{\partial^2 f}{\partial{x}\ \partial{y}}$$ when $f$ is of class $C^2$.
