Derivative of matrix with respected to vector (matrix in se(3)) We know that the matrix $A \in SE(3)$ can be written in exponential formula, i.e.
$$e^{M} = \begin{bmatrix} R_{3\times3} & \vec{t} \\\vec{0}^T & 1
\end{bmatrix}$$
$$M = \begin{bmatrix} [\vec{\omega}]_{\times} & V^{-1}\vec{t} \\\vec{0}^T & 0
\end{bmatrix}$$
,where $\omega = [\omega_x, \omega_y, \omega_z]^T$ can be retrieved from rodrigues' formula. and 
$$V = I_3 + \frac{1-cos{\theta}}{\theta^2}[\vec{\omega}]_{\times} + \frac{\theta - \sin{\theta}}{\theta^3}[\vec{\omega}]_{\times}^2$$
We know that matrix $V$ is a function of $[\omega_x, \omega_y, \omega_z]$. I wish to know how to compute the derivative of $V$ with respected to each element. My attempt is to use the definition of derivative.
$$f^\prime = \lim_{h\rightarrow0}\frac{f(x+h) - f(x)}{h}$$
However, even with the help from wolframalpha, I couldn't simplify the term. The image below shows my attempt with wolframalpha for the second term. Does anyone here know more convenient method of doing this? Thank you very much. 

Not enough reputation point to comment, so I'll post this as an "answer".
The phi-function is 
$$\phi(x) = \frac {e^{x}-1} {x}$$
The function you're attempting to take the derivative of is a matrix version of the phi-function
$$V=\phi \big([\vec{\omega}]_{\times}\big)$$
That's going to be difficult.  Even more difficult is that fact that you want a matrix-by-vector derivative.
