how to differentiate skew symmetric matrix $[\mathbf{v}]_{\times}$ with respect to $\mathbf{v}$ can anyone explain how to differentiate a skew-symmetric matrix $\mathbf{v}_{\times}$ with respect to $\mathbf{v}$ i.e. $\frac{\partial{[\mathbf{v}]_{\times}}}{\partial \mathbf{v}}$ where $\mathbf{v}\in\mathbb{R}^3$?
In addition, how to derive $\frac{\partial [\mathbf{v}]^2_{\times}}{\partial \mathbf{v}}$? 
Thank you! 
 A: Using the Levi-Civita (aka Permutation) tensor $\varepsilon$ the problem has a straightforward approach $$\eqalign{
 [v]_\times &= -\varepsilon\cdot v \cr\cr
d[v]_\times &= -\varepsilon\cdot dv \cr\cr
\frac{\partial [v]_\times}{\partial v} &= -\varepsilon \cr
\cr
}$$
This is the same as what you derived, but the notation is more standard.
Update
Thinking about the second part of your question, your initial expansion is incorrect
$$\eqalign{
[v]^2_\times &= vv^T - (v^Tv)\,I \,\,\,\neq vv^T \cr\cr
}$$
Update #2
Here's a partial result for the question that you linked in your comment.
Given the vector $w$ and the cross-matrix $$W=[w]_\times$$ you generate a rotation matrix $$R=\exp W$$ which you'd like to differentiate.
First, define a scalar $\theta$ representing the length of vector and take its differential
$$\eqalign{
 \theta^2 &= w\cdot w = \frac{1}{2}W:W \cr
 2\theta\,d\theta &= W:dW \cr
 d\theta &= \frac{W:dW}{2\theta} \cr
\cr
}$$
Next expand $R$ via Rodrigues' formula
$$\eqalign{
 R &= I + \frac{\sin\theta}{\theta}W + \frac{1-\cos\theta}{\theta^2}W^2 \cr
  &= I + \alpha W + \beta W^2 \cr
\cr
}$$
Now let me denote the $4^{th}$ order isotropic tensor by ${\mathcal A}$ with components 
$${\mathcal A}_{ijkl}=\delta_{ik}\,\delta_{jl}$$
and the dyadic ($\star$) product $C = A\star B$ with components
$$C_{ijkl} = A_{ij}\,B_{kl}$$
and a colon to  denote the double-dot product, i.e. 
$$A:B = A_{ijkl}\,B_{klmn}$$
Finally we're ready to differentiate the rotation matrix
$$\eqalign{
  dR &= \alpha dW + \beta(W\cdot dW + dW\cdot W) + W\alpha^\prime d\theta + W^2\beta^\prime d\theta \cr
 &= \Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \frac{\alpha^\prime}{2\theta}\,W\star W + \frac{\beta^\prime}{2\theta}\,W^2\star W\Big]:dW \cr
 &= -\Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \Big(\frac{\alpha^\prime W}{2\theta}+\frac{\beta^\prime W^2}{2\theta}\Big)\star W\Big]:\varepsilon\cdot dw \cr\cr
\frac{\partial R}{\partial w} &= -\Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \Big(\frac{\alpha^\prime W}{2\theta}+\frac{\beta^\prime W^2}{2\theta}\Big)\star W\Big]:\varepsilon \cr
}$$where
$$\alpha^\prime=\frac{d\alpha}{d\theta},\,\,\,\,\beta^\prime=\frac{d\beta}{d\theta}$$
You can rearrange this result to better suit your tastes if you keep in mind a few things. 
${\mathcal A}$ is the identity for the double-dot product
$${\mathcal A}:X = X:{\mathcal A} = X$$
and $$W:\varepsilon = \varepsilon:W = -2\,w$$
Update #3
Looking more closely at your linked question, you appear to be interested in functions of the $W$ matrix of the form
$$\eqalign{
 F &= I + \alpha W + \beta W^2 \cr
 \alpha &= \alpha(\theta),\,\,\,\beta = \beta(\theta) \cr
}$$
The case $F=e^W$ was analyzed above. The good news is that all of that analysis carries over to the case $F=\frac{e^W-I}{W}\,\,$ just using different functions for the scalar coefficients 
$$\eqalign{
 \alpha &= \frac{1-\cos(\theta)}{\theta^2} \cr
 \beta &= \frac{\theta-\sin(\theta)}{\theta^3} \cr
}$$
Therefore
$$\eqalign{
\frac{\partial F}{\partial w} &= -\Big[\alpha{\mathcal A} + \beta(W\cdot{\mathcal A} + {\mathcal A}\cdot W^T) + \Big(\frac{\alpha^\prime W}{2\theta}+\frac{\beta^\prime W^2}{2\theta}\Big)\star W\Big]:\varepsilon \cr
\cr
 &= \Big(\frac{\alpha^\prime W}{\theta}+\frac{\beta^\prime W^2}{\theta}\Big)\star w - \alpha\,\varepsilon - \beta\,W\cdot\varepsilon - (\beta{\mathcal A}\cdot W^T):\varepsilon \cr
}$$
A: I derive a solution using the notations of linear algebra:
$$
\frac{\partial [\mathbf{v}]_\times}{\partial \mathbf{v}} = \frac{\partial \begin{bmatrix}
0 & -v_3  & v_2\\ 
v_3 & 0 & -v_1 \\ 
-v_2 & v_1 & 0
\end{bmatrix}}{\partial \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}}
$$
using $v_1$ as an example:
$$
\frac{\partial [\mathbf{v}]_\times}{\partial v_1} = \begin{bmatrix}
0 & 0  & 0\\ 
0 & 0 & -1 \\ 
0 & 1 & 0
\end{bmatrix} = [e_1]_\times
$$
where $e_1$ is the first column of an identity matrix $\mathbf{I} \in \mathbb{R}^{3\times3}$. Then, we have:
$$
\frac{\partial [\mathbf{v}]_\times}{\partial \mathbf{v}} = \begin{bmatrix}
[e_1]_\times\\ 
[e_2]_\times\\ 
[e_3]_\times
\end{bmatrix}
$$
since we have $[\mathbf{v}]_\times^2=\mathbf{v}\mathbf{v}^\top$, we could derive that (thank to greg's help, I made the following correction.)
$$
\frac{\partial [\mathbf{v}]_\times^2}{\partial v_i} = \frac{\partial \mathbf{v}\mathbf{v}^\top-\mathbf{v}^\top\mathbf{v}\mathbf{I}}{\partial v_i} = \frac{\partial \mathbf{v}}{\partial v_i}\mathbf{v}^\top + \mathbf{v} [\frac{\partial \mathbf{v}}{\partial v_i}]^\top-\frac{\partial (\mathbf{v}^\top\mathbf{v})}{\partial v_i}\mathbf{I}=e_i\mathbf{v}^\top+\mathbf{v}e_i^\top-2v_i\mathbf{I}
$$
where $e_i$ is the i-th column of the $\mathbf{I}$. Correct me if anyone see any problem with this.
