Computing partial derivatives. I am having trouble solving this problem: If $f$ is $C^{2}$ and $f(x,y)=g(r,\theta)$ where $(r,\theta)$ are polar coordinates in $\mathbb{R}^{2}$, then
$$
\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)f(x,y)=\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}\right)g(r,\theta)
$$
for all $(x,y)\ne(0,0)$.
I computed
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial x}=\cos\theta\frac{\partial f}{\partial r}-\frac{\sin\theta}{r}\frac{\partial f}{\partial\theta}
$$
and
$$
\frac{\partial f}{\partial y}=\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial y}=\sin\theta\frac{\partial f}{\partial r}+\frac{\cos\theta}{r}\frac{\partial f}{\partial\theta}
$$
but I do not know how to compute the second derivatives.
Any help is appreciated.
 A: I assigned this as homework and when no one could do it was wrote out a solution. Maybe it is useful.  The formatting didn't transfer perfectly.
We are given $z=f(x,y)$ where $x=r \cos \theta$ and $y= r \sin \theta$.  We want to show
$$\frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}= \frac{\partial^2z}{\partial r^2} +\frac{1}{r^2} \frac{\partial^2z}{\partial \theta^2} +\frac 1 r \frac{\partial z}{\partial r}$$
Using the chain rule (case 2), we have
$\begin{align}
\frac{\partial z}{\partial r}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}\\
&=\frac{\partial f}{\partial x} \cos \theta +\frac{\partial f}{\partial y} \sin \theta \\&= \frac{\partial f}{\partial x}\frac{x}{\sqrt{x^2+y^2}} +\frac{\partial f}{\partial y} \frac{y}{\sqrt{x^2+y^2}}
\end{align} $
and
$\begin{align}
\frac{\partial z}{\partial \theta}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}\\
&=\frac{\partial f}{\partial x}  (-r \sin \theta) +\frac{\partial f}{\partial y} (r\cos \theta)\\
&=\frac{\partial f}{\partial x} (-y) +\frac{\partial f}{\partial y} x.
\end{align}$
Using the chain rule, the above formula for $\frac{\partial z}{\partial r}$  and the fact that $\cos \theta$ and $\sin \theta$ are constants with respect to $r$, we have
$\begin{align}
\frac{\partial^2 z}{\partial r^2}&=\cos \theta  \frac{\partial}{\partial r}\left(\frac{\partial f}{\partial x}\right )+ \sin \theta \frac{\partial}{\partial r}\left(\frac{\partial f}{\partial y}\right)\\
&= \cos \theta \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial r}+\frac{\partial^2 f}{\partial y \partial x}\frac{\partial y}{\partial r}\right)+ \sin \theta \left(\frac{\partial^2 f}{\partial x \partial y}\frac{\partial x}{\partial r}+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial r}\right)\\
&=\cos \theta \left(\frac{\partial^2 f}{\partial x^2}\cos \theta+\frac{\partial^2 f}{\partial y \partial x}\sin \theta \right)+ \sin \theta \left(\frac{\partial^2 f}{\partial x \partial y}\cos \theta+\frac{\partial^2 f}{\partial y^2}\sin \theta \right)\\
&= \cos^2 \theta \frac{\partial^2 f}{\partial x^2} + \sin^2 \theta \frac{\partial^2 f}{\partial y^2}+2 \cos \theta \sin \theta \frac{\partial^2 f}{\partial x \partial y}\\
&= \frac{x^2}{x^2+y^2} \frac{\partial^2 f}{\partial x^2} +  \frac{y^2}{x^2+y^2}\frac{\partial^2 f}{\partial y^2}+ \frac{2xy}{x^2+y^2} \frac{\partial^2 f}{\partial x \partial y}
\end{align}$
Using the chain rule, the above formula for $\frac{\partial z}{\partial \theta}$ and the product rule, we have
$\begin{align}
\frac{\partial^2 z}{\partial \theta^2}&= \frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x} (-y) +\frac{\partial f}{\partial y} x\right) \\
&= -y \frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)+\frac{\partial f}{\partial x} \frac{\partial}{\partial \theta} (-y) +  x \frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)+\frac{\partial f}{\partial y} \frac{\partial}{\partial \theta} (x) \\
&=-y \left(\frac{\partial^2 f}{\partial x^2}(-y)+\frac{\partial^2 f}{\partial y \partial x} x\right)+\frac{\partial f}{\partial x} (-r \cos \theta) + x \left(\frac{\partial^2 f}{\partial x \partial y}(-y)+\frac{\partial^2 f}{\partial y^2}  x\right)+ \frac{\partial f}{\partial y} (-r \sin \theta)\\
&=y^2\frac{\partial^2 f}{\partial x^2} + x^2 \frac{\partial^2 f}{\partial y^2}-2xy\frac{\partial^2 f}{\partial x \partial y}-x \frac{\partial f}{\partial x}-y\frac{\partial f}{\partial y}
\end{align}$
Putting this all together, we have
$\begin{align}
 \frac{\partial^2z}{\partial r^2} +\frac{1}{r^2} \frac{\partial^2z}{\partial \theta^2} +\frac 1 r \frac{\partial z}{\partial r}&= \frac{\partial^2z}{\partial r^2} +\frac{1}{x^2+y^2} \frac{\partial^2z}{\partial \theta^2} +\frac {1}{\sqrt{x^2+y^2}} \frac{\partial z}{\partial r}\\
&=  \frac{x^2}{x^2+y^2} \frac{\partial^2 f}{\partial x^2} +  \frac{y^2}{x^2+y^2}\frac{\partial^2 f}{\partial y^2}+ \frac{2xy}{x^2+y^2} \frac{\partial^2 f}{\partial x \partial y} \\
&  +   \frac{1}{x^2+y^2} \left(y^2\frac{\partial^2 f}{\partial x^2} + x^2 \frac{\partial^2 f}{\partial y^2}-2xy\frac{\partial^2 f}{\partial x \partial y}-x \frac{\partial f}{\partial x}-y\frac{\partial f}{\partial y}\right)\\
&  +\frac {1}{\sqrt{x^2+y^2}}\left(\frac{\partial f}{\partial x}\frac{x}{\sqrt{x^2+y^2}} +\frac{\partial f}{\partial y} \frac{y}{\sqrt{x^2+y^2}}\right)\\
&=\frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}
\end{align}$
A: An other way ?
$\dfrac{\partial^{2}f}{\partial x^{2}}=\dfrac{\partial}{\partial x}\left(\cos\theta\dfrac{\partial f}{\partial r}-\dfrac{\sin\theta}{r}\dfrac{\partial f}{\partial\theta}\right)=\cos\theta\left(\dfrac{\partial^{2}f}{\partial r^{2}}\dfrac{\partial r}{\partial x}+\dfrac{\partial^{2}f}{\partial r\partial\theta}\dfrac{\partial\theta}{\partial x}\right)+\left(\dfrac{\partial \cos\theta}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial \cos\theta}{\partial\theta}\dfrac{\partial\theta}{\partial x}\right)\dfrac{\partial f}{\partial r}-\dfrac{\sin\theta}{r}\left(\dfrac{\partial^{2}f}{\partial\theta\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial^{2}f}{\partial\theta^{2}}\dfrac{\partial\theta}{\partial x}\right)-\dfrac{\partial f}{\partial\theta}\left(\dfrac{\partial\dfrac{\sin\theta}{r}}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial\dfrac{\sin\theta}{r}}{\partial\theta}\dfrac{\partial\theta}{\partial x}\right)$
$\dfrac{\partial^{2}f}{\partial x^{2}}=\cos^{2}\theta\dfrac{\partial^{2}f}{\partial r^{2}}-2\cos\theta\dfrac{\sin\theta}{r}\dfrac{\partial^{2}f}{\partial r\partial\theta}+\dfrac{\sin^{2}\theta}{r^{2}}\dfrac{\partial^{2}f}{\partial\theta^{2}}+\dfrac{\sin^{2}\theta}{r}\dfrac{\partial f}{\partial r}$
And $\dfrac{\partial^{2}f}{\partial y^{2}}=\dfrac{\partial}{\partial y}\left(\sin\theta\dfrac{\partial f}{\partial r}+\dfrac{\cos\theta}{r}\dfrac{\partial f}{\partial\theta}\right)=\sin\theta\left(\dfrac{\partial^{2}f}{\partial r^{2}}\dfrac{\partial r}{\partial x}+\dfrac{\partial^{2}f}{\partial r\partial\theta}\dfrac{\partial\theta}{\partial x}\right)+\left(\dfrac{\partial \sin\theta}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial \sin\theta}{\partial\theta}\dfrac{\partial\theta}{\partial x}\right)\dfrac{\partial f}{\partial r}+\dfrac{\cos\theta}{r}\left(\dfrac{\partial^{2}f}{\partial\theta\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial^{2}f}{\partial\theta^{2}}\dfrac{\partial\theta}{\partial x}\right)+\dfrac{\partial f}{\partial\theta}\left(\dfrac{\partial\dfrac{\cos\theta}{r}}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial\dfrac{\cos\theta}{r}}{\partial\theta}\dfrac{\partial\theta}{\partial x}\right)$
$\dfrac{\partial^{2}f}{\partial y^{2}}=\sin^{2}\theta\dfrac{\partial^{2}f}{\partial r^{2}}+2\cos\theta\dfrac{\sin\theta}{r}\dfrac{\partial^{2}f}{\partial r\partial\theta}+\dfrac{\cos^{2}\theta}{r^{2}}\dfrac{\partial^{2}f}{\partial\theta^{2}}+\dfrac{\cos^{2}\theta}{r}\dfrac{\partial f}{\partial r}$
So $\nabla f=\left(\dfrac{\partial^{2}}{\partial x^{2}}+\dfrac{\partial^{2}}{\partial y^{2}}\right)f=\left(\dfrac{\partial^{2}}{\partial r^{2}}+\dfrac{1}{r}\dfrac{\partial}{\partial r}+\dfrac{1}{r^{2}}\dfrac{\partial^{2}}{\partial\theta^{2}}\right)f$
