Express partial derivatives of second order (and the Laplacian) in polar coordinates $z=f(x,y)$ where $x=rcosθ$ and $y=rsinθ$
Find $ \frac{\partial z}{\partial x}$ and $ \frac{\partial^2 z}{\partial x^2}$
I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have...
$ \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}$. So I'm guessing that $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question. 
Full Question:

 A: One source of confusion in calculations like this is that $z$ is being used to denote two different (but related) functions.  Let's define a function $\hat{z}$ by
\begin{equation}
\hat{z}(r,\theta) = z(r \cos \theta, r \sin \theta).
\end{equation}
Note that $\hat{z}$ is not the same function as $z$, so it should have its own name.
Now it is more clear how to apply the chain rule.  I also think the calculation is clarified by writing inputs to functions explicitly.
\begin{align}
\frac{\partial \hat{z}(r,\theta)}{\partial r}
&= \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial x} \cos \theta
+ \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial y} \sin \theta.\\
\frac{\partial \hat{z}(r,\theta)}{\partial \theta}
&= \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial x} (-r \sin \theta)
+ \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial y} (r \cos \theta).
\end{align}
Higher order partial derivatives can be computed next.  Then we can combine and simplify to get an expression for
\begin{equation}
\frac{\partial^2 z(r \cos \theta, r \sin \theta)}{\partial x^2}
+ \frac{\partial^2 z(r \cos \theta, r \sin \theta)}{\partial y^2}
\end{equation}
in terms of partial derivatives of $\hat{z}$.
