Computing second partial derivative with polar coordinates

Solution 1:

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: $$\begin{split}\cos^2\theta \frac{\partial^2}{\partial r^2}-2\cos\theta\sin\theta \frac{1}{r}\frac{\partial^2}{\partial r\partial\theta}+2\cos\theta\sin\theta \frac{1}{r^2}\frac{\partial }{ \partial\theta} \\ +\sin^2\theta \frac1r\frac{\partial}{\partial r}+\sin^2\theta \frac{1}{r^2}\frac{\partial^2 }{ \partial\theta^2}\end{split}$$