<s>[INST] How to shorten the derivation of the Laplacian in polar coordinates? Yesterday, I computed the formula for the Laplacian in polar coordinates.
$
\newcommand{p}[2]{\frac {\partial #1} {\partial #2}}
\newcommand{s}[2]{\p {^2 #1} {#2 ^2}}
\newcommand{m}[3]{\p {^2 #1} {#2 \partial #3}}
\newcommand{po}[0]{\p {}}
\newcommand{so}[0]{\s {}}
\newcommand{mo}[0]{\m {}}
\newcommand{t}[0]{\theta}
\newcommand{w}[1]{\left( #1 \right)}
\newcommand{ct}[0]{\cos \t}
\newcommand{st}[0]{\sin \t}
\newcommand{f}[1]{\frac #1 r}
\newcommand{g}[1]{\frac #1 {r^2}}
\newcommand{jx}[2]{\ct #1 - \f \st #2}
\newcommand{jy}[2]{\st #1 + \f \ct #2}
\newcommand{px}[2]{\jx {\po r #1} {\po \t #2}}
\newcommand{py}[2]{\jy {\po r #1} {\po \t #2}}
\newcommand{pox}[0]{\px {} {}}
\newcommand{poy}[0]{\py {} {}}
\newcommand{one}[0]{\sin^2\t + \cos^2\t}
\newcommand{zero}[0]{\st\ct - \st\ct}
$
\begin{align*}
x & = r \ct \\ y & = r \st \\ r^2 & = x^2 + y^2
\end{align*}
First, I used the chain rule to relate $\po r$ and $\po \t$ to $\po x$ and $\po y$:
\begin{align*}
\po x = \p r x \po r + \p \t x \po \t = \pox \\
\po y = \p r y \po r + \p \t y \po \t = \poy \\
\end{align*}
Applying each of $\po x$ and $\po y$ to itself:
\begin{align*}
\so x + \so y
  & = \po x {\w {\po x}} + \po y {\w {\po y}} \\
  & = \px {\w \pox} {\w \pox}  \\
  & \qquad + \py {\w \poy} {\w \poy} \\
  & = \ct \w {\ct \so r + \g \st \po \t - \f \st \mo r \t} \\
  & \qquad+ \f \st \w {\st \po r - \ct \mo \t r + \f \ct \po \t + \f \st \so \t} \\
  & \qquad + \st \w {\st \so r - \g \st \po \t + \f \ct \mo r \t} \\
  & \qquad + \f \ct \w {\ct \po r + \st \mo \t r - \f \st \po \t + \f \ct \so \t} \\
  & = \w \one \so r + \g \zero \po \t \\
  & \qquad - \f \zero \mo r \t + \f \one \po r \\
  & \qquad - \f \zero \mo \t r + \g \zero \po \t \\
  & \qquad + \g \one \so \t \\
  & = \so r + \f 1 \po r + \g 1 \so \t \\
\end{align*}
This was long and tedious. Is there some way to make the calculation shorter.
 [/INST] Apparently, Math.SE does not allow me to “delete for real” my previous wrong answer, so I will try to provide a non-wrong answer.
Let $\newcommand \t \theta F(x,y) = f(r,\t)$, where $(r,\t) = g(x,y)$. Of course, in our case, $g$ is not a globally well-defined function, but who cares? Differential calculus is all about working locally. By the chain rule,
$$F_{xx} = (f \circ g)_{xx} = (\nabla f \cdot g_x)_x = (\nabla f)_x \cdot g_x + \nabla f \cdot g_{xx} = g_x^T \cdot Hf \cdot g_x + \nabla f \cdot g_{xx}$$
Of course, analogously, we have
$$F_{yy} = g_y^T \cdot Hf \cdot g_y + \nabla f \cdot g_{yy}$$
Adding everything, the Laplacian expands to
$$F_{xx} + F_{yy} = g_x^T \cdot Hf \cdot g_y + g_y^T \cdot Hf \cdot g_y + \nabla f \cdot (g_{xx} + g_{yy})$$
Call the sum of the first two terms $A$ and call the last term $B$.

In our particular case, $r^2 = x^2 + y^2$ and $\tan \t = y/x$. Thus,
$$r_x = x/r, \qquad r_y = y/r, \qquad \sec^2 \t \cdot \t_x = -y/x^2, \qquad \sec^2 \t \cdot \t_y = 1/x$$
Rearranging the last two expressions, we have
$$\t_x = -y/r^2, \qquad \qquad \qquad \t_y = x/r^2$$
By the chain rule, $A$ reduces to
$$f_{rr} (r_x^2 + r_y^2) + 2f_{r\t} (r_x \t_x + r_y \t_y) + f_{\t\t} (\t_x^2 + \t_y^2) = f_{rr} + \frac 1 {r^2} f_{\t\t}$$

Differentiating once again, we have
$$r_{xx} = 1/r, \qquad r_{yy} = 1/r, \qquad \t_{xx} = 2yr_x /r^3, \qquad \t_{yy} = -2xr_y / r^3$$
Rearranging the last two expressions, we have
$$\t_{xx} = 2xy / r^4, \qquad \qquad \qquad \t_{yy} = -2xy / r^4$$
By the chain rule, $B$ reduces to
$$f_r \cdot (r_{xx} + r_{yy}) + f_\t \cdot (\t_{xx} + \t_{yy}) = \frac 1r f_r$$

Combining everything, we have
$$F_{xx} + F_{yy} = f_{rr} + \frac 1r f_r + \frac 1 {r^2} f_{\t\t}$$
</s>