Expanding partial derivatives Let $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$  Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$. I want to show that $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac\partial{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial \theta^2}.$$

I use the chain rule to compute $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$ and $$\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta}$$

So $$\Delta=\left (\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta} \right )^2 + \left (\frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta} \right )^2$$

Expanding, I get $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1{r^2}\frac{\partial^2}{\partial\theta^2}$$

Where does the term $\frac1r\frac\partial{\partial r}$ disappear?
You have to rewrite everything in terms of $r$ and $\theta$, so where you are leaving $x$ and $y$ behind you have to clear up. Also, these are operators, so you have to have them act on a test function $f$ -- this also makes you be a bit more careful.

What you'll find is that, since $\Delta = \nabla \cdot \nabla$, you need to compute the Jacobian for the transformation, that is
\begin{equation}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial x}
\end{equation}
etc. This gets you how to rewrite the $\nabla$ operator in polar coordinates. Then you need to figure out what $\nabla \cdot (\nabla f)$ is to figure out what $\Delta f$ is.

There is also a metric tensor way to rewrite the Laplacian, see under http://mathworld.wolfram.com/Laplacian.html for example.
You have to rewrite everything in terms of $r$ and $\theta$, so where you are leaving $x$ and $y$ behind you have to clear up. Also, these are operators, so you have to have them act on a test function $f$ -- this also makes you be a bit more careful.

What you'll find is that, since $\Delta = \nabla \cdot \nabla$, you need to compute the Jacobian for the transformation, that is
\begin{equation}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial x}
\end{equation}
etc. This gets you how to rewrite the $\nabla$ operator in polar coordinates. Then you need to figure out what $\nabla \cdot (\nabla f)$ is to figure out what $\Delta f$ is.

There is also a metric tensor way to rewrite the Laplacian, see under http://mathworld.wolfram.com/Laplacian.html for example.