Derivative of Tangent Vector Bases in Polar Coordinates In physics using polar coordinates, I have seen $\frac{d e_r}{d \theta}=e_\theta$ and $\frac{d e_\theta}{d \theta}=-e_r$ where $e_r,e_\theta$ is the usual basis of the tangent space at point $(r,\theta)$ in the plane in polar coordinates. 
See e.g. https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf. 
In particular, the vectors $e_r$ and $e_\theta$ actually changes depending on where the point $(r,\theta)$ is. I am struggling to understand these equations mathematically.
First, I am most used to describing the tangent plane as something like $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}$. But then e.g. $\frac{d\frac{\partial}{\partial r}}{d \theta}$ makes no sense to me. Is there any way to recover this point of view or is it too algebraic? 
Now, geometrically if we consider $e_r, e_\theta$ to be just vectors then I can understand how to obtain these equations I think. They basically come from identifying all the tangent spaces as a single $\mathbb{R}^2$ plane in the obvious way so that we can actually compare vectors that belong in different tangent spaces. Then e.g. the first equation $\frac{d e_r}{d \theta}=e_\theta$ comes from looking at $\lim_{h \to 0} \frac{e_r^2-e_r^1}{h}$ where $e_r^1$ is that basis vector at $(r,\theta)$ and $e_r^2$ is that basis vector at $(r,\theta+h)$. I switched to cartesian coordinates to compute the limit then switched back to polar coords.
Is there any way to do this from the point of view of the differential operators $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}$?
 A: To expand a bit on my comment.
Suppose you work on an arbitrary smooth manifold with local coordinates $(x_1, x_2)$. These coordinates define a basis of each tangent space given by the operators $\frac{\partial}{\partial x_1}$ and $\frac{\partial}{\partial x_2}$. For simplicity of the notation, call these operators $\mathbf{b}_{1}$ and $\mathbf{b}_{2}$, respectively. In an arbitrary smooth manifold, this is all that can be done: there is not enough structure to calculate the "second derivatives" $\frac{\partial\mathbf{b}_i}{\partial x_j}$.
The extra data that is needed is supplied by an affine connection or a covariant derivative (each determines the other). The covariant derivative is perhaps simplest to understand in terms of the Christoffel symbols $\Gamma^k_{i,j}$, which specify the "second derivatives" of the standard basis vectors:
$$\frac{\partial \mathbf{b}_i}{\partial x_j}=\sum_{k=1}^n\Gamma^k_{i,j}\mathbf{b}_k$$
The coefficients $\Gamma^k_{i,j}$ just tell how to express the partial derivative of $\mathbf{b}_i$ with respect to $x_j$ as a linear combination of the basis vectors $\mathbf{b}_k$.
This page works out the Christoffel symbols for planar polar coordinates.
A: Simple case the parameter is $\theta$
You have the equation of the vector for the parametric equation, velocity and acceleration, by taking derivatives using "derivatives of unit basis vectors", as if they where a product of two function.
$$\begin{aligned}
\vec r &= r(\theta) \hat e_r\\
\frac {d\vec r} {d\theta} &= \frac {dr} {d\theta}\hat e_r+r\hat e_\theta\\
\frac {d^2\vec r} {d\theta^2} &= \left(\frac {d^2 r} {d\theta^2}-r\right) \hat e_r +\frac {dr} {d\theta} \hat e_\theta
\end{aligned}$$
The tangent unit vector will always be the velocity over the speed.
Case two you can compose the angle as a function of $t$, and that make it a little bet more difficult but it have the same logic but with more chain rule.
