change in unit radial vector and unit polar angle due to change in polar angle In cylindrical coordinates, the set B of basis vectors is
$B=\left \{ \vec{e}_{r},\vec{e}_{\theta} \right \}$.
It is clear, geometrically, that 
$\frac{\partial e_{r}}{\partial r}=0=\frac{\partial e_{\theta}}{\partial r}$
However, I am unable to convince myself that
$\frac{\partial e_{r}}{\partial \theta}=e_{\theta}$ and $\frac{\partial e_{\theta}}{\partial \theta}=-e_{r}$
Any help to shed light on this is appreciated.
Thanks in advance

This is for polar coordinates, cylindrical would have a third basis vector $e_z$, but regardless, as the angle $\Delta\theta$ goes to $0$ in the ccw direction, the differential angle makes an appearance resulting in a differential change in the direction of the radial vector
So wherever $e_r$ was pointing, it now has to rotate over an amount of $d\theta$ (ccw) to point in the new correct direction, and the direction of where it should go is already accounted for by $e_\theta$
Likewise, $e_\theta$ is perpendicular to $e_r$, so when going counterclockwise, the new $e_\theta$ is more flattened horizontally than the original $e_\theta$, so the direction $e_\theta$ must go when rotated by $d\theta$ is in the direction of $-e_r$
This is to say
$$\underline{e_r}=\cos\theta\underline{i}+\sin\theta\underline{j}$$
By definition, they are orthogonal
$$\underline{e_r}\cdot\underline{e_\theta}=0$$
And we want $e_\theta$ to have positive orientation sense, so
$$\underline{e_\theta}=-\sin\theta\underline{i}+\cos\theta\underline{j}$$
And we can now clearly see
$$\frac{d\underline{e_r}}{d\theta}=-\sin\theta\underline{i}+\cos\theta\underline{j}=\underline{e_\theta}$$
$$\frac{d\underline{e_\theta}}{d\theta}=-\cos\theta\underline{i}-\sin\theta\underline{j}=-\underline{e_r}$$
