Computing the Laplacian in Polar Coordinates Similar questions have been asked on this site but none of them seemed to help me. I'm asked to compute the Laplacian $$\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$ in terms of polar coordinates.
I did do it, but I don't understand why what I did is correct, and I don't understand the more "brute force" way to do it at all. 
Here is what I did:

I calculated $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ in terms of $r,$ $\theta,$ $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}.$ This gave me a system of linear equations which I wrote as
  $$\begin{pmatrix}\frac{\partial}{\partial r} \\\frac{\partial}{\partial \theta}\end{pmatrix} = \begin{pmatrix}\cos\theta & \sin\theta \\\ -r\sin\theta & r\cos\theta \end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial x} \\\frac{\partial}{\partial y}\end{pmatrix}.$$
  I inverted to get $$\begin{pmatrix}\frac{\partial}{\partial x} \\\frac{\partial}{\partial y}\end{pmatrix} = \frac{1}{r}\begin{pmatrix}r\cos\theta & -\sin\theta \\\ r\sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial r} \\\frac{\partial}{\partial \theta}\end{pmatrix},$$
  and then I simply wrote \begin{align}\frac{\partial^2}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial}{\partial x}\right)\\ &= \left(\cos\theta \frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\right)\left(\cos\theta \frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\right)\\ &= \cos\theta\frac{\partial}{\partial r}\left(\cos\theta \frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\right) -\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\left(\cos\theta \frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\right)\\ &= \cos^2\theta \frac{\partial^2}{\partial r^2} - \frac{2}{r}\sin\theta\cos\theta\frac{\partial^2}{\partial r
\partial\theta} +\frac{1}{r^2}\sin^2\theta\frac{\partial^2}{\partial \theta^2}.\end{align}
  I similarly got $$\frac{\partial^2}{\partial y^2} = \sin^2\theta \frac{\partial^2}{\partial r^2} + \frac{2}{r}\sin\theta\cos\theta\frac{\partial^2}{\partial r
\partial\theta} +\frac{1}{r^2}\cos^2\theta\frac{\partial^2}{\partial \theta^2}.$$ Adding the two yields $$\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2},$$
  which Spivak says is correct.

Explicitly, here are my questions:


*

*In my solution, when I found $\frac{\partial^2}{\partial x^2},$ I simply "multiplied" the expressions in the second line of the large aligned equation (treating multiplication of the partial operators as composition). Why am I allowed to do this? Why does the expression on the left not act on the thing on the right, forcing me to do the product rule and other nonsense to get the answer?

*My original idea was just to compute the Laplacian using the chain rule. That is, write $\frac{\partial}{\partial x}=\frac{\partial}{\partial r} \frac{\partial r}{\partial x}$ and compute $\frac{\partial^2}{\partial x^2}$ from there. My problem with this is that I keep getting confused about in which variables I should be writing everything, and how the partial derivative operators act on these other expressions. For example, I compute $\frac{\partial}{\partial x}=\frac{\partial}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta} \frac{\partial \theta}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}\frac{\partial}{\partial r} - \frac{y}{x^2+y^2} \frac{\partial}{\partial \theta},$ but then I don't know where to go. Help understanding this method would be greatly appreciated.
If anything is unclear let me know and I'll make the necessary edits.
 A: 
I simply "multiplied" 

Imagine a function $f(x, y) = f(x(r,\theta),y(r,\theta)) = f(r,\theta)$, you just found
$$
\frac{\partial f(x, y)}{\partial x} = \cos\theta \frac{\partial f(r,\theta)}{\partial r} - \frac{1}{r}\sin\theta \frac{\partial f(r,\theta)}{\partial \theta} \equiv h
$$
Then apply it again
\begin{eqnarray}
\frac{\partial^2 f(x, y)}{\partial x^2} &=&\frac{\partial h}{\partial x} \\
&=& \cos\theta \frac{\partial h(r,\theta)}{\partial r} - \frac{1}{r}\sin\theta \frac{\partial h(r,\theta)}{\partial \theta}  \\
&=& \cos\theta \frac{\partial}{\partial r}\left[\cos\theta \frac{\partial f(r,\theta)}{\partial r} - \frac{1}{r}\sin\theta \frac{\partial f(r,\theta)}{\partial \theta}\right] - \frac{1}{r}\sin\theta\left[\cos\theta \frac{\partial f(r,\theta)}{\partial r} - \frac{1}{r}\sin\theta \frac{\partial f(r,\theta)}{\partial \theta} \right]
\end{eqnarray}
That when you think about it is the same thing you did. So actually multiplying is the right way of doing it. In general this is true for any operator, if $T$ is an operator then
$$
T^2f = T(Tf)
$$

My original idea was just to compute the Laplacian using the chain rule

It is completely equivalent to what you did in the first step. Here's the first term
\begin{eqnarray}
\frac{\partial f}{\partial x} &=& \frac{\partial r}{\partial x}\frac{\partial f}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial f}{\partial \theta} \\
&=& \frac{x}{r}\frac{\partial f}{\partial r} - \frac{y}{r^2}\frac{\partial f}{\partial \theta} \\
&=& \cos\theta \frac{\partial f}{\partial r} - \frac{\sin\theta}{r}\frac{\partial f}{\partial \theta}
\end{eqnarray}
and now apply the same logic as before
