Can you please help me in finding out the mistakes I am doing during the calculation of derivative of a vector. I am briefing the problem I am trying to solve as follows.
There is a line joining two points (1) and (3) as shown in the image below. The possible degrees of freedom at each of the points are shown as $d_{1,1}$, $d_{1,2}$ of point (1) in x- and y- directions. It is similar for point (3) as well.
The vector at the original configuration is calculated as,
$V_{130}$ = $X_{30}$ - $X_{10}$
The new orientation of the vector $v_{13}$ due to these displacement is $v_{13} = V_{130}+d_{3}-d_{1}$
And the unit vector $c_{13} = \dfrac{v_{13}}{\det{v_{13}}}$
Now, I am trying to calculate the derivative of the unit vector as follows
$\dfrac{\partial{c_{13}}}{\partial{d_{i,n}}}$ = $\dfrac{\partial}{\partial{d_{i,n}}}$ $(\frac{v_{13} }{\det{v_{13} } }) $
where, i is the point number (1 or 3) and n is the component number (1 or 2).
$\dfrac{\partial{c_{13}}}{\partial{d_{i,n}}}$ = $\dfrac{ \det{v_{13} \dfrac{\partial{v_{13}}}{\partial{d_{i,n}}}} - v_{13} \dfrac{v_{13}^{T}}{\det{v_{13}}} \dfrac{\partial{v_{13}}}{\partial{d_{i,n}}} } {\det{v_{13}}^2}$
= $\dfrac{ \left[ I - \dfrac{v_{13}}{\det{v_{13}}} \dfrac{v_{13}^{T}}{\det{v_{13}}} \right] } {\det{v_{13}}^2} \det{v_{13}} \dfrac{\partial{v_{13}}}{\partial{d_{i,n}}} $
= $\dfrac{ \left[ I - c_{13} c_{13}^{T} \right] } {\det{v_{13}}} \dfrac{\partial{v_{13}}}{\partial{d_{i,n}}}$
= $\dfrac{ \left[ I - c_{13} c_{13}^{T} \right] } {\det{v_{13}}} (\delta_{3i}-\delta_{1i}) \left[ \begin{array}{1} \delta_{1n} \\ \delta_{2n} \end{array} \right] $
The above delta is Kronecker delta.
Can you please correct the above outcome of derivative?
Cross-Checking: When I am trying to cross check if the above derivative is correct, I am finding a discrepancy as shown below.
There is a line with coordinates $X_{10} = (-1,-1)$ and $X_{30} = (1,1)$ at its original configuration as shown in the image
The vector at the original configuration is calculated as,
$V_{130}$ = $X_{30}$ - $X_{10}$ = $\left[\begin{array}{l}2\\2\end{array}\right]$
The displacement of the points 1 and 3 are (new coordinate - old coordinate),
$d_1$ = $\left[\begin{array}{l}1\\ -\sqrt(2)+1 \end{array}\right]$ $d_3$ = $\left[\begin{array}{l}-1\\ \sqrt(2)-1 \end{array}\right]$
$v_{13} = V_{130}+d_3-d_1 = \left[\begin{array}{l}0\\ 2\sqrt(2) \end{array}\right]$
And the unit vector $c_{13} = \dfrac{v_{13}}{\det{v_{13}}}$ = $\left[\begin{array}{l}0\\ 1 \end{array}\right]$
Upon substitution in the above derivation formula,
$\dfrac{\partial{c_{13}}}{\partial{d_{1,1}}} = \dfrac{1}{2\sqrt{2}} \left[ I - \left[ \begin{array}{3} 0 & 0 \\ 0 & 1 \\ \end{array} \right] \right] (-1) \left[ \begin{array}{1} 1 \\ 0 \\ \end{array} \right] $
= $ \dfrac{-1}{2\sqrt{2}} \left[ \begin{array}{1} 1 \\ 0 \\ \end{array} \right]$; $\dfrac{\partial{c_{13}}}{\partial{d_{1,2}}} = 0$
$\dfrac{\partial{c_{13}}}{\partial{d_{3,1}}} = \dfrac{1}{2\sqrt{2}} \left[ \begin{array}{1} 1\\ 0 \\ \end{array} \right]$; $\dfrac{\partial{c_{13}}}{\partial{d_{3,2}}} = 0$
Upon rearranging,
$\dfrac{\partial{c_{13}^{1}}}{\partial{d_{1,n}}} = \dfrac{-1}{2\sqrt{2}} \left[\begin{array}{1}1\\0 \end{array}\right]$; $\dfrac{\partial{c_{13}^{1}}}{\partial{d_{3,n}}} = \dfrac{1}{2\sqrt{2}} \left[\begin{array}{1}1\\0 \end{array}\right]$
By using chain rule of differentiation, $c^{1}_{13} = \dfrac{\partial{c^{1}_{13}}}{\partial{d_{1,n}}} \ d_{1,n} + \dfrac{\partial{c^{1}_{13}}}{\partial{d_{3,n}}} \ d_{3,n} $
$c_{13}^{1}$ = $ \frac{-1}{2\sqrt{2}} \left[\begin{array}{1} 1 & 0 \end{array} \right] * \left[\begin{array}{1} 1 \\ -\sqrt(2)+1 \end{array} \right] + \dfrac{1}{2\sqrt{2}} \left[\begin{array}{1} 1 & 0 \end{array} \right] * \left[\begin{array}{1} -1 \\ \sqrt(2)-1 \end{array} \right] = \frac{-1}{\sqrt{2}}$;
$c_{13}^{2}$ =0
(WHICH IS NOT CORRECT). Can you please help me in which part of the calculation went wrong. Please let me know if you need any clarifications in the whole, I can re-explain to make you understand.
If y = m*x+c is derived by x, which results in m. The derivative, (dy/dx)*x is not equal to y.
Thus lets understanding that this this question is not feasible to get solved.