How do I differentiate this? I am given $u=\frac{x+y}{\sqrt 2}$ and $v=\frac{x-y}{\sqrt2}$, how would I find $\frac{d^2}{du^2},\frac{d^2}{dv^2}$?
I rearranged $u$ and $v$ in terms of $x$ and $y$, and I get $x = \frac{u + v}{\sqrt{2}}$ and $y = \frac{u - v}{\sqrt{2}}$. But how do I find $\frac{d^2}{du^2}$ and $\frac{d^2}{dv^2}$? 
 A: if $f$ is a function of $x,y$ then
$$
\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}
$$
and
$$
\frac{\partial^2 f}{\partial u^2}=\frac{\partial}{\partial u}\Big(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\Big)
$$
$$
=\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial x}{\partial u}\Big(\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}\Big)

+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial u^2}+\frac{\partial y}{\partial u}\Big(\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}\Big)
$$
$$
=\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial u^2}+2\frac{\partial^2 f}{\partial x\partial y}\frac{\partial x}{\partial u}\frac{\partial y}{\partial u}+\Big(\frac{\partial x}{\partial u}\Big)^2\frac{\partial^2 f}{\partial x^2}+\Big(\frac{\partial y}{\partial u}\Big)^2\frac{\partial^2 f}{\partial y^2}
$$
note
$$x=(u+v)\sqrt{2}, y=(u-v)\sqrt{2},\frac{\partial x}{\partial u}=\frac{\partial y}{\partial u}=\sqrt{2}$$
so the above simplifies to
$$
4\frac{\partial^2 f}{\partial x\partial y}+2\Big(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\Big)
$$
you can do the same thing with respect to $v$ (i assumed equality of mixed partials above)
A: When you apply $\partial_u$ you take the directional  derivative with respect to the vector $\frac 1{\sqrt 2}(1,1)$, so $\partial u =\frac 1{\sqrt 2}(\partial x+\partial y)$. Now take the directional  derivative with respect to the vector $\frac 1{\sqrt 2}(1,1)$ of $\partial u$. We get $\partial^2_u=\frac 12(\partial^2_x+\partial_x\partial_y+\partial_y\partial_x+\partial_y^2)$. Now, do the same for $\partial_v$ and $\partial_v^2$. 
