Q:

How to find the rotation vector by deriving the final vector with respect to the displacement?

My understanding of a rotation of a vector can be done by using a 2D rotation matrix as shown below,
$R(\theta )=\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \\\end{bmatrix}$.
This rotates column vectors by means of the following matrix multiplication,
$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \\\end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix}$
For example, if you rotate the vector 
x=$\begin{bmatrix}1\\1 \end{bmatrix}$ by 45 degrees (clockwise), then the new vector is  $\begin{bmatrix} \sqrt2 \\ 0 \end{bmatrix}$.
Other Method:
If I have only initial and final coordinates of the vectors

The initial vector is, V = $\begin{bmatrix}1\\1 \end{bmatrix}$ and the final vector is, v = V+d = $\begin{bmatrix} \sqrt2 \\ 0 \end{bmatrix}$.
The displacement between these vectors is d = $\begin{bmatrix} \sqrt2-1 \\ -1 \end{bmatrix}$.
Can I derive the final vector v with respect to displacement $\frac{\partial{v}}{\partial{d}}$ to get the rotation vector? 
[but returns a identity matrix]
If so, does $\frac{\partial{v}}{\partial{d}} * d $ can be used to cross-check?

A:

The rotation vector transforms the initial vector to the final vector. These operations are carried out in polar space, making it non feasible to conduct using differentiation.

