homework and exercises - Finding solution to this differential equation

In this paper http://arxiv.org/abs/hep-th/9506035 equation (3.11) was written as: $$\frac{\partial L}{\partial u}\frac{\partial L}{\partial v} = -1$$

The author then said p.9 that "approximate solutions to equation (3.11) can be obtained by a power series expansion. It is convenient to define new variables $x$ and $y$ by:"

$x= u+v$ and $y=u-v$

Then he said equation (3.11) becomes $$(\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$$

How come? I didn't understand his argument.

Answer

This is just the chain rule: $\frac{\partial L}{\partial u}=\frac{\partial L}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial L}{\partial y}\frac{\partial y}{\partial u}$. And similar thing for $\frac{\partial L}{\partial v}$. We have the derivative of $x$ and $y$ with respect to $u$ and $v$.