Computing second partial derivative with polar coordinates Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$.
I can show using the chain rule that $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$ 
What is the method to compute $\dfrac{\partial^2}{\partial x^2}$? I don't know how to do it.
 A: Rewrite your formula as
$$\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\tag{1}$$
(I like writing in this way because $\frac{1}{r}\frac{\partial}{\partial\theta}$ is more natural than $\frac{\partial}{\partial\theta}$: it means the rate of change in the tangential direction). Now square (1):
$$\frac{\partial^2}{\partial x^2} = \left(\cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\right)\left(\cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\right)\tag{2}$$
The computation of (2) involves no thinking, just some product rule:
$$\begin{split}\cos^2\theta \frac{\partial^2}{\partial r^2}-2\cos\theta\sin\theta \frac{1}{r}\frac{\partial^2}{\partial r\partial\theta}+2\cos\theta\sin\theta \frac{1}{r^2}\frac{\partial }{ \partial\theta} \\ +\sin^2\theta \frac1r\frac{\partial}{\partial r}+\sin^2\theta \frac{1}{r^2}\frac{\partial^2 }{ \partial\theta^2}\end{split}$$ 
