In introductory calculus, we learn to take derivatives of vectors. For instance, in Cartesian, we have the basis vectors $\vec{e}_x, \vec{e}_y$ and in polars we have $\vec{e}_\rho, \vec{e}_\theta$ given by $$ \vec{e}_\rho=\cos\theta \: \vec{e}_x + \sin\theta \: \vec{e}_y \\ \vec{e}_\theta=\cos\theta \: \vec{e}_y - \sin\theta \: \vec{e}_x $$
Taking derivatives, we get $\partial_x \vec{e}_x = \partial_x \vec{e}_y = \partial_y \vec{e}_x = \partial_y \vec{e}_y = \vec{0}$ and we can differentiate polars as follows:
\begin{align} \partial_\theta \vec{e}_\rho &=\partial_\theta \left( \cos\theta \:\vec{e}_x + \sin\theta \:\vec{e}_y \right) \\ &= -\sin\theta \:\vec{e}_x + \cos\theta \:\partial_\theta (\vec{e}_x)+ \cos\theta \:\vec{e}_y +\sin\theta \:\partial_\theta (\vec{e}_y) \\ &=\vec{e}_\theta \end{align}
Fair enough. This all seems reasonable and the idea that the derivative of a basis vector with respect to the coordinates can be written as some linear combination of the basis vectors seems obvious.
My question is, how can I arrive at the same conclusions having identified basis vectors with the set of partial derivatives. If $\vec{e}_x \to \partial_x$, why should $\partial_x \vec{e}_x \to \partial_x \partial_x = \frac{\partial^2}{\partial x^2}=0$? I'm confused. More worrying still, using the polar example, I am getting
\begin{align} \partial_\theta \vec{e}_\rho \to\partial_\theta \partial_\rho = \frac{\partial^2}{\partial \theta \partial \rho} \neq \partial_ \theta \end{align}
I can't seem to reconcile any of this. It is also not at all obvious to me why the derivatives of basis vectors (partial derivatives) should be vectors - aren't they second derivatives?
Grateful for any help...