If $z=f(x,y)$, where $x=r \cos\theta$ and $y=r\sin\theta$, how can I use Mathematica to prove that:
$$\frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}=\frac{\partial^2z}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2z}{\partial \theta^2}+\frac1r\frac{\partial z}{\partial r}$$

**Understanding the chain rule in Mathematica:**

Jen's and Ian's effort on my behalf is helping me understand Mathematica notation involving the chain rule. If $z=f(x,y)$, where $x=r \cos t$ and $y=r\sin t$, here is how we start to find $\partial z/\partial r$ by hand in class. We draw this image as a reminder of the chain rule.

![enter image description here](https://i.stack.imgur.com/0EfwW.jpg)

Then, because $x=r\cos t$ and $y=r\sin t$, we get:
$$\begin{align\*}
\frac{\partial z}{\partial r}
&=\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}\\
\frac{\partial z}{\partial r}
&=\cos t\ \frac{\partial z}{\partial x}+\sin t\ \frac{\partial z}{\partial y}
\end{align\*}$$

Now we enter just part of Ian's code, concentrating only on $\partial z/\partial r$ in order to understand Mathematica notation.

```
Clear[z, f, x, y, r, t];
z[r, t] = f[x[r, t], y[r, t]];
x[r_, t_] = r Cos[t];
y[r_, t_] = r Sin[t];
D[z[r, t], r]

```

The output is:

![enter image description here](https://i.stack.imgur.com/l0HBY.png)

It's in a different order from my answer, but it is the same. 

I am interpreting **Cos[t] $f^{(1,0)}$[r Cos[t], r Sin[t]] + Sin[t] $f^{(0,1)}$ [r Cos[t], r Sin[t]]** as meaning 

$$\cos t\ \frac{\partial z}{\partial x}+\sin t\ \frac{\partial z}{\partial y},$$ 

or equivalently:

$$\cos t\ \frac{\partial}{\partial x}\ f(r \cos t, r\sin t)+\sin t\ \frac{\partial}{\partial y}\ f(r \cos t, r\sin t)$$

I also computed $\frac{\partial^2z}{\partial r^2}$ by hand and got the same answer using Ian's Mathematica notation.

**Was Able to Understand Another Answer:**

The question was find $\partial^2z/\partial t^2$, where $z=f(x,y)$ and $x=g(s,t)$ and $y=h(s,t)$. I entered:

```
Clear[z, f, x, y, g, s, t];
D[f[g[s, t], h[s, t]], {t, 2}] // Simplify

```

Which gave this answer:

![enter image description here](https://i.stack.imgur.com/JEFCb.png)

And I interpreted that as meaning:

$$
\left(\frac{\partial y}{\partial t}\right)^2\frac{\partial^2f}{\partial y^2}
+\frac{\partial f}{\partial y}\frac{\partial^2y}{\partial t^2}
+\frac{\partial^2 x}{\partial t^2}\frac{\partial f}{\partial x}
+2\frac{\partial x}{\partial t}\frac{\partial y}{\partial t}\frac{\partial^2f}{\partial x\partial y}
+\left(\frac{\partial x}{\partial t}\right)^2\frac{\partial^2f}{\partial x^2}
$$

Which agrees with the answer in the textbook.

Thanks to Jens and Ian, I have learned quite a bit.