For $z=f(x,y)$ with $(x,y)=(r\cos\theta,r\sin\theta)$, show that $z_{xx}+z_{yy}=z_{rr}+ \frac{1}{r^2}z_{\theta \theta} + \frac{1}{r} z_r$ 
Let $ z = f (x, y) $ with $ x = r \cos \theta $ and $ y = r \sin \theta. $ Show that$$z_{xx}+z_{yy}=z_{rr}+ \frac{1}{r^2}z_{\theta \theta} + \frac{1}{r} z_r$$

Solution: I would like to see an elegant solution. I've solved this problem before by plugging in expressions for all of the variables and expanding everything out ($\phi$ is in place of $z$):
\begin{align*}
\frac{\partial^2 \phi}{\partial r^2} = \frac{\partial}{\partial r} \frac{\partial \phi}{\partial r} = \frac{\partial}{\partial r}\left(\frac{\partial \phi}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial \phi}{\partial y}\frac{\partial y}{\partial r}\right) = \frac{\partial}{\partial r}\left(\frac{\partial \phi}{\partial x}\cos \theta+\frac{\partial \phi}{\partial y}\sin \theta\right) = \\
\cos \theta \frac{\partial}{\partial x}\frac{\partial \phi}{\partial r}+\sin \theta \frac{\partial}{\partial y} \frac{\partial \phi}{\partial r} = \\
\cos \theta \frac{\partial}{\partial x} \left(\frac{\partial \phi}{\partial x}\cos \theta+\frac{\partial \phi}{\partial y}\sin \theta\right)+ \sin \theta \frac{\partial}{\partial y} \left(\frac{\partial \phi}{\partial x}\cos \theta+\frac{\partial \phi}{\partial y}\sin \theta\right) =\\
\cos^2 \theta \frac{\partial^2 \phi}{\partial x^2} + 2\sin\theta\cos\theta\frac{\partial^2 \phi}{\partial x \partial y}+\sin^2\theta\frac{\partial^2 \phi}{\partial y^2}.
\end{align*}By a similar computation,$$\frac{\partial^2 \phi}{\partial \theta^2} = -r\frac{\partial \phi}{\partial r}+r^2\sin^2\theta \frac{\partial^2 \phi}{\partial x^2}-2r^2\sin \theta\cos\theta \frac{\partial^2 \phi}{\partial x\partial y}+r^2\cos^2\theta\frac{\partial^2 \phi}{\partial y^2}.$$Using the identity $\sin^2\theta+\cos^2\theta=1$ and adding, we get$$\frac{1}{r^2}\frac{\partial^2 \phi}{\partial \theta^2}+\frac{\partial^2 \phi}{\partial r^2}=-\frac{1}{r}\frac{\partial \phi}{\partial r}+\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2},$$and the result follows upon rearranging.
 A: A convenient way is to treat partial derivatives as operators. By the Chain Rule, one has $$\frac{\partial}{\partial r}=\frac{\partial}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial}{\partial y}\frac{\partial y}{\partial r}$$ and $$\frac{\partial}{\partial \theta}=\frac{\partial}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial}{\partial y}\frac{\partial y}{\partial \theta}.$$ Using the relations between Cartesian and polar coordinates, one then records this in terms of matrices: $$\left[\begin{array}{c}\frac{\partial}{\partial r}\\ \frac{\partial}{\partial \theta}\end{array}\right]=\left[\begin{array}{cc}\cos\theta&\sin\theta\\ -r\sin\theta&r\cos\theta\end{array}\right]\left[\begin{array}{c}\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\end{array}\right].$$ Taking inverse, one has $$A:=\left[\begin{array}{c}\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\end{array}\right]=\left[\begin{array}{cc}r\cos\theta&-\sin\theta\\ r\sin\theta&\cos\theta\end{array}\right]\left[\begin{array}{c}\frac 1 r\frac{\partial}{\partial r}\\ \frac 1 r \frac{\partial}{\partial \theta }\end{array}\right].$$ Then it is straightforward to compute $$\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}=A^tA$$
$$=\left[\frac 1 r\frac{\partial}{\partial r}\quad \frac 1 r\frac{\partial}{\partial\theta}\right]\left[\begin{array}{cc}r\cos\theta&r\sin\theta\\ -\sin\theta&\cos\theta\end{array}\right]  \left[\begin{array}{cc}r\cos\theta&-\sin\theta\\ r\sin\theta&\cos\theta\end{array}\right] \left[\begin{array}{c}\frac 1 r\frac{\partial}{\partial r}\\ \frac 1 r\frac{\partial}{\partial\theta}\end{array}\right]$$
$$=\left[\frac 1 r\frac{\partial}{\partial r}\quad \frac 1 r\frac{\partial}{\partial\theta}\right]\left[\begin{array}{cc}r^2&0\\ 0&1\end{array}\right]\left[\begin{array}{c}\frac 1 r\frac{\partial}{\partial r}\\ \frac 1 r\frac{\partial}{\partial\theta}\end{array}\right]$$
$$=\left[\frac 1 r\frac{\partial}{\partial r}\quad \frac 1 r\frac{\partial}{\partial\theta}\right]\left[\begin{array}{c}r\frac{\partial}{\partial r}\\ \frac 1 r\frac{\partial}{\partial\theta}\end{array}\right]$$
$$=\frac 1 r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac 1 r\frac{\partial}{\partial\theta}\left(\frac 1 r\frac{\partial}{\partial\theta}\right)$$
$$=\frac 1 r\left(\frac{\partial}{\partial r}+r\frac{\partial^2}{\partial r^2}\right)+\frac 1{r^2}\frac{\partial^2}{\partial\theta^2}$$
$$=\frac 1 r\frac{\partial}{\partial r}+\frac{\partial^2}{\partial r^2}+\frac 1 {r^2}\frac{\partial^2}{\partial\theta^2},$$ as required.
