Why is $\partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1r\partial_r+\frac1{r^2}\partial_{\theta}^2\right)$? I have to show the identity I wrote in the title: it should be $\partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1r\partial_r+\frac1{r^2}\partial_{\theta}^2\right)$ but some computation doesn't match with my book with these operators.
We know that $\partial_z\partial_{\bar z}=\frac14(\partial_x^2+\partial_y^2)$.
Then we pass to polar coordinates writing $x=r\cos\theta$ and $y=r\sin\theta$, from which, obviously we have $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\frac yx$.
We use than the chain rule to write
$$
\partial_x=\frac{\partial r}{\partial x}\partial_r+\frac{\partial\theta}{\partial x}\partial_{\theta}=\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta}
$$
and
$$
\partial_y=\frac{\partial r}{\partial y}\partial_r+\frac{\partial\theta}{\partial y}\partial_{\theta}=\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_{\theta}
$$
and till here all matches with my book. My problem comes now: I used the above cited $\partial_z\partial_{\bar z}=\frac14(\partial_x^2+\partial_y^2)$ but this gives me 
$$
\partial_z\partial_{\bar z}=\frac14\left(\partial_r^2+\frac1{r^2}\partial_{\theta}^2\right)
$$
because the "crossed" terms are equal and opposite: squaring $\partial_x$ the "crossed" term is $-\frac2r\cos\theta\sin\theta\partial_r\partial_{\theta}$, squaring $\partial_y$ the "crossed" term is $\frac2r\cos\theta\sin\theta\partial_r\partial_{\theta}$ hence in the sum they should vanish.
Where is the error?
EDIT: following Andrea's suggest, I wrote
\begin{align*}
\partial_x^2+\partial_y^2
&=\left(\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta}\right)\left(\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_{\theta}\right)+\\
&\;\;\;\;\;\,\left(\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_{\theta}\right)
\left(\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_{\theta}\right)\\
&=\cos^2\theta\partial_r^2+\frac{\sin\theta}{r}\cos\theta\partial_{\theta}
+\frac{\sin^2}{r}\partial_r+\cos\theta\frac{\sin\theta}{r^2}\partial_\theta+\\
&\;\;\;\;\;\;\sin^2\theta\partial_r^2-\frac{\cos\theta}{r}\sin\theta\partial_{\theta}
+\frac{\cos^2}{r}\partial_r-\sin\theta\frac{\cos\theta}{r^2}\partial_\theta\\
=&\partial_r^2+\frac1r\partial_r
\end{align*}
which is false again! I checked several times but I cannot find where my mistake is!
Many thanks!
 A: You have correctly derived
$$
   \frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} \\
   \frac{\partial}{\partial y} = \sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial\theta}
$$
Therefore,
$$
\begin{align}
    \frac{\partial^{2}}{\partial x^{2}} & =\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right) \\
   & = \cos\theta\frac{\partial}{\partial r}\cos\theta\frac{\partial}{\partial r} \\
   & -\cos\theta\frac{\partial}{\partial r}\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} \\
   &  -\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\cos\theta\frac{\partial}{\partial r} \\
   & + \frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}
\end{align}
$$
It is true that
$$
\cos\theta\frac{\partial}{\partial r}\cos\theta\frac{\partial}{\partial r}
      = \cos^{2}\theta\frac{\partial^{2}}{\partial r^{2}}
$$
However, the product rule gives
$$
\begin{align}
   \cos\theta\frac{\partial}{\partial r}\frac{\sin\theta}{r}\frac{\partial}{\partial\theta} & = 
   \cos\theta\sin\theta\frac{\partial}{\partial r}\frac{1}{r}\frac{\partial}{\partial\theta} \\
    & = \cos\theta\sin\theta\frac{1}{r}\frac{\partial^{2}}{\partial r\partial\theta}
   - \cos\theta\sin\theta\frac{1}{r^{2}}\frac{\partial}{\partial\theta}
\end{align}
$$
Therefore, keeping things on separate lines for you to inspect,
$$
\begin{align}
\frac{\partial^{2}}{\partial x^{2}}
    & = \cos^{2}\frac{\partial^{2}}{\partial r^{2}} \\
    & -\frac{\cos\theta\sin\theta}{r}\frac{\partial^{2}}{\partial r\partial \theta}
      +\frac{\cos\theta\sin\theta}{r^{2}}\frac{\partial}{\partial \theta} \\
    & -\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}}{\partial\theta\partial r}
      +\frac{\sin^{2}\theta}{r}\frac{\partial}{\partial r} \\
    & + \frac{\sin^{2}\theta}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
      +\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial}{\partial\theta}
\end{align}
$$
Every first order derivative term came from a product rule. Similarly,
$$
\begin{align}
\frac{\partial^{2}}{\partial y^{2}}
  & = \sin^{2}\frac{\partial^{2}}{\partial r^{2}} \\
  & +\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}}{\partial r\partial\theta}-\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial}{\partial\theta} \\
  & +\frac{\cos\theta\sin\theta}{r}\frac{\partial^{2}}{\partial\theta\partial r}
    + \frac{\cos^{2}\theta}{r}\frac{\partial}{\partial r} \\
  & +\frac{\cos^{2}\theta}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
    -\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial}{\partial\theta}
\end{align}
$$
Now when you add these together, you get
$$
   \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}
    = \frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
$$
A: When you "multiply" two differential operators, like $\cos\theta\,\partial_r$ and $\frac{\sin\theta}r\,\partial_\theta$, the derivative operator in the first factor acts on the coefficient in the second factor.  
A: One has $\theta=\arctan{y\over x}$ (not $=\arctan{x\over y}$) when $x>0$, but
$$\theta_x={1\over1+ {y^2\over x^2}}\left(-{y\over x^2}\right)={-y\over x^2+y^2}=-{\sin\theta\over r}$$
and
$$\theta_y={1\over1+ {y^2\over x^2}}{1\over x}={x\over x^2+y^2}={\cos\theta\over r}$$are valid in all of $\dot{\Bbb R}^2$.
You have the sign wrong in your displayed formula for $\partial_y$.
