Error with mixed derivatives during derivation of Laplacian in polar coordinates

Question

It can be shown that, if $f:\mathbb{R}^2\to\mathbb{R}$ is a smooth function and the Cartesian and polar coordinates are related by $$ x = \rho\cos\phi\\ y = \rho\sin\phi $$ that $\partial f/\partial x$ may be written (using the chain rule) like $$ \dfrac{\partial f}{\partial x} = \cos\phi\dfrac{\partial f}{\partial \rho} - \dfrac{\sin\phi}{\rho}\dfrac{\partial f}{\partial \phi} \tag{1} $$ Because $\partial f/\partial \rho:\mathbb{R}^2\to\mathbb{R}$ is also smooth, we can replace $f$ by $\partial f/\partial \rho$ in equation $(1)$, implying that smooth functions also obey $$ \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial \rho}\right) = \cos\phi\dfrac{\partial^2f}{\partial\rho^2} - \dfrac{\sin\phi}{\rho} \dfrac{\partial^2 f}{\partial\phi\partial\rho} \tag{2} $$ However, when I try to calculate $(2)$ by mixing up the order of differentiation, I get an extra term: $$ \begin{align*} \dfrac{\partial}{\partial\rho}\left(\dfrac{\partial f}{\partial x}\right) &= \dfrac{\partial}{\partial\rho}\left( \cos\phi\dfrac{\partial f}{\partial \rho} - \dfrac{\sin\phi}{\rho} \dfrac{\partial f}{\partial \phi} \right) \\ &= \cos\phi\dfrac{\partial^2f}{\partial\rho^2} - \dfrac{\sin\phi}{\rho}\dfrac{\partial^2f}{\partial\rho\partial\phi} + \dfrac{\sin\phi}{\rho^2}\dfrac{\partial f}{\partial \phi} \tag{3} \end{align*} $$ So could anybody please explain to me why expression $(2)$ and $(3)$ are not the same?

Answer 1

You can't change the order of differentiation with respect to a pair of variables belonging to different coordinate systems! While it's true that $f''_{xy}=f''_{yx}$ and $f''_{\rho\phi}=f''_{\phi\rho}$ (if $f$ is nice enough), there is no reason to believe that $f''_{\rho x} = f''_{x \rho}$, and in fact it's not true in general, as simple examples show.

Remember that taking a partial derivative with respect to $x$ involves keeping $y$ constant, and taking a partial derivative with respect to $\rho$ involves keeping $\phi$ constant, so even if the "partner variables" $y$ and $\phi$ are not "visible" in the expressions $f''_{\rho x}$ and $f''_{x \rho}$, they are very much a part of the game.