I recently saw an exercise to derive the Laplacian for Polar Coordinates by using the chain rule.
$\phi: \mathbb{R}^2 \rightarrow \mathbb{R}^2, (r, \varphi) \mapsto (r\cos{\varphi},r\sin{\varphi}) = (x, y)$
$\psi: \mathbb{R}^2 \rightarrow \mathbb{R}^2, (x, y) \mapsto (\sqrt{x^2 + y^2}, arctan\frac{y}{x})= (r,\varphi) = \phi^{-1}$
The exercise reads "Calculate $\Delta (f(\phi(r, \varphi))$ by applying the chain rule to $f(\phi(\psi))$".
This confuses me, since I think $\phi(\psi(x,y))$ isn't doing anything. What am I missing here? How does the $\phi(\psi(x,y))$ help me?
Hint: If you are given $f$ in terms of polar coordinates, you are really being given $f \circ \phi$ in the above notation. So let $F(r, \varphi) = (f \circ \phi)(r,\varphi)$ and apply the chain rule to $f = F \circ \psi$.