Transforming the Laplace operator from Polar to Cartesian coordinates I'm trying to find the error in my logic here. 
Let's say we are given the Laplace operator in polar coordinates:
$$ \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial}{\partial \theta^2} \tag{1} $$
and we're interested in transforming back to Cartesian coordinates.  We could first make use of the usual coordinate transformation, namely
$$ u = r \sin(\theta), ~~v = r \cos(\theta), \tag{2}$$
to write the partial derivatives in polar coordinates as follows,
$$ \frac{\partial}{\partial r} = \sin(\theta) \frac{\partial}{\partial u} + \cos(\theta)\frac{\partial}{\partial v}, \tag{3}$$
$$ \frac{\partial}{\partial \theta} = r \cos(\theta) \frac{\partial}{\partial u} -  r \sin(\theta)\frac{\partial}{\partial v}. \tag{4}$$
I think we should then rewrite all $\theta$ and $r$ in terms of $u$ and $v$, making use of $r^2 = u^2 + v^2$, so that in particular 
$$ \sin(\theta) = \frac{u}{\sqrt{u^2 + v^2}}, \tag{5}$$
$$ \cos(\theta) = \frac{v}{\sqrt{u^2 + v^2}} \tag{6}.$$
With this, (3) and (4) then become
$$ \frac{\partial}{\partial r} = \frac{u}{\sqrt{u^2 + v^2}} \frac{\partial}{\partial u} + \frac{v}{\sqrt{u^2 + v^2}} \frac{\partial}{\partial v}, \tag{3*}$$
$$ \frac{\partial}{\partial \theta} = v \frac{\partial}{\partial u} -  u \frac{\partial}{\partial v}. \tag{4*}$$
Our next task is to then compute $\frac{\partial^2}{\partial r^2}$ and $\frac{\partial^2}{\partial \theta^2}$, which we can carefully do by multiplying the right-hand sides of (3*) and (4*), keeping in mind that the operators will act on the coefficients. 
However, executing this and adding the terms together yields nothing that looks remotely near the known form, namely $$\frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2}.$$  So I'm wondering, what am I missing here?  
In addition to this particular example, I'm interested in any general information you have about transforming partial derivatives from polar to Cartesian coordinates. 
 A: Here is my proof:
$$ \frac{\partial}{\partial x}=\frac{\partial}{\partial r} \frac{\partial r}{\partial x} +\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}$$ so by applying the product rule
$$ \frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}\right)=
\frac{\partial^2}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2+
\frac{\partial}{\partial r}\frac{\partial^2 r}{\partial x^2} +
\frac{\partial^2}{\partial \theta^2}\left(\frac{\partial \theta}{\partial x}\right)^2+
\frac{\partial}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2}$$
$$ +\frac{\partial^2 }{\partial r \partial \theta}\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}$$
You get the same relation with $y$. Now you just have to calculate the derivatives of $r,\theta$ with respect to $x,y$. You have 
$$ r=\sqrt{x^2+y^2},\ \theta=\arctan \frac{y}{x}$$.
What follows is a simple calculus exercise on derivatives. You just need to prove that
$$\left(\frac{\partial r}{\partial x}\right)^2+\left(\frac{\partial r}{\partial y}\right)^2=1, $$
etc.(the relations you need so that when you sum $\frac{\partial^2}{\partial x^2}$ and $\frac{\partial^2}{\partial y^2}$ you'll get the polar form of the laplacian).

Going on the lines you started, you shouldn't change $\sin \theta,\cos \theta$ in terms of $x,y$. Just calculate the next derivative with respect to $r,\theta$, using appropriately the formula of the partial derivative of composition of functions.
A: Using a bit of differential geometry. 
The key is the writing $\Delta f = \text{div grad}f$: if we manage to express div and grad in a coordinate-independent manner, we can write $\Delta$ easily in any coordinate system, be it cartesian or polar.
The metric and the inverse metric are
$$
g_{ab}=\left(\begin{array}[cc]
\ 1&0\\
0&1
\end{array}
\right)=g_{ab}^{-1}
$$
in cartesian coordinates and
$$
g'_{ab}=\left(\begin{array}[cc]
\ 1&0\\
0&r^2
\end{array}
\right),\ \ \
g'^{-1}_{ab}=\left(\begin{array}[cc]
\ 1&0\\
0&r^{-2}
\end{array}
\right)
$$
in polar coordinates. The volume form is $\omega=dx\wedge dy=r\, dr\wedge d\varphi$ and the differential of a function is
$$
df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy,\ \ \ 
dF= \frac{\partial F}{\partial r}dr + \frac{\partial F}{\partial \varphi}d\varphi.
$$
Now, div is given by 
$
d[\omega(E)]=\text{div}(E) \omega,
$
for any vector field $E=E^x \partial_x + E^y\partial_y$ or $B=B^r \partial_r + B^\varphi\partial_\varphi$;
so
$$
d[\omega(E)] = d[E^x dy - E^y dx]=\left(\frac{\partial E^x}{\partial x}+\frac{\partial E^y}{\partial y}\right) dx\wedge dy\implies\text{div}(E)=\frac{\partial E^x}{\partial x}+\frac{\partial E^y}{\partial y}.
$$
Or
$$
d[\omega(B)] = d[B^r r\,d\varphi - B^\varphi r\,dr]
=\left(\frac{1}{r}\frac{\partial (rB^r)}{\partial r}+\frac{\partial B^\varphi}{\partial \varphi}\right)r\, dr\wedge d\varphi\implies\text{div}(B)=
\frac{1}{r}\frac{\partial (rB^r)}{\partial r}+\frac{\partial B^\varphi}{\partial \varphi}.
$$
And grad$f=g^{-1}df$, so
$$
\text{grad}f = \frac{\partial f}{\partial x}\partial_x + \frac{\partial f}{\partial y}\partial_y
\ \text{ and } \ 
\text{grad}F = \frac{\partial F}{\partial r}\partial_r + \frac{1}{r^2}\frac{\partial F}{\partial \varphi}\partial_\varphi.
$$
Finally, $\Delta=$div grad:
$$
\Delta f = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}
\ \text{ and } \ 
\Delta F = \frac{1}{r}\left( r \frac{\partial F}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 F}{\partial \varphi^2}.
$$
