Complicated application of chain rule If we are given two functions $g$ and $f$ such that $g(u,v) = f(Au+Bv,Cu+Dv)$ for some constants $A,B,C,D$, how do we use the chain rule to show that
$$\nabla^2 g = (A^2+B^2)\frac{\partial^2 f}{\partial x^2}+2(AC+BD)\frac{\partial^2 f}{\partial x \partial y}+(C^2+D^2)\frac{\partial^2 f}{\partial y^2}$$?
I fixed the typo and am only typing this so that my edit will be accepted - Jared
 A: So I'm pretty sure based on the answer, you are actually trying to find $\nabla^2 g$.  First, let's go over the chain rule:
$$
f(x,y) \rightarrow g(u, v) = f(\alpha(u, v), \beta(u, v)) \\
\frac{\partial g}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial \alpha}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial \beta}{\partial u}
$$
The way I reason through this is that you need hit everything that may or may not have a $u$ in it (or $v$ if it had been $\frac{\partial g}{\partial v}$).  It' the same thing as when you take a full derivative: you take all of the partials and multiply by the thing you took the partial with (it's the chain rule, e.g $\frac{df(x, y, z)}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt}$).
So then this is easy to start doing (it's going to be a pain to write out, but you just have to be very careful):
$$
\vec{\nabla} g(u, v) = \left\langle\frac{\partial g}{\partial u}, \frac{\partial g}{\partial v}\right\rangle = \left\langle \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}, \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} \right\rangle \\
\vec{\nabla} \circ \vec{\nabla}g = \frac{\partial }{\partial u}g_u + \frac{\partial }{\partial v}g_v
$$
Let's just do the first one (the other is the same, just substitute $v$ everywhere you see $u$:
\begin{align}
\frac{\partial }{\partial u}g_u =& \frac{\partial }{\partial u}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\right) \\
\frac{\partial }{\partial u}g_u =& \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\right)\cdot\frac{\partial x}{\partial u} + \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\right)\cdot \frac{\partial y}{\partial u} \\
\frac{\partial}{\partial u}g_u =& \frac{\partial^2f}{\partial x^2}\left(\frac{\partial x }{\partial u}\right)^2 + 2\frac{\partial^2 f}{\partial x\partial y}\frac{\partial x}{\partial u}\frac{\partial y}{\partial u} + \frac{\partial^2 f}{\partial y^2}\left(\frac{\partial y}{\partial u}\right)^2
\end{align}
Now we could just plug and chug using the above formula, but let's just go through one step at a time how we would do this particular problem.  I think some notation will help clarify things.  Here is the setup:
$$
g(u, v) = f(x(u, v), y(u, v))\\ x(u, v) = Au + Bv\\ y(u, v) = Cu + Dv
$$
Now take the first partials:
$$
\frac{\partial g}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u} = A\frac{\partial f}{\partial x} + C\frac{\partial f}{\partial y} \\
\frac{\partial g}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} = B\frac{\partial f}{\partial x} + D\frac{\partial f}{\partial y}
$$
Now you need to find the second partials, I'm only going to do the first one and I'll leave the second to you:
\begin{align}
\frac{\partial^2g }{\partial u^2} =& \frac{\partial }{\partial u}\left(A\frac{\partial f}{\partial x} + C\frac{\partial f}{\partial y}\right)\\
\frac{\partial^2g }{\partial u^2} =& A\left(\frac{\partial^2f }{\partial x^2}\frac{\partial x}{\partial u} + \frac{\partial^2 f}{\partial y\partial x}\frac{\partial y}{\partial u}\right) + C\left(\frac{\partial^2f }{\partial x\partial y}\frac{\partial x}{\partial u} + \frac{\partial^2f}{\partial y^2}\frac{\partial y}{\partial u}\right) \\
\frac{\partial^2g }{\partial u^2} =& A^2\frac{\partial^2f }{\partial x^2} + 2AC\frac{\partial^2f }{\partial x\partial y} + C^2\frac{\partial^2f }{\partial y^2}
\end{align}
