Derivative of $tr(A \exp{\hat{\omega}})$ with respect to $\omega \in \mathbb{R}^3$ I want to calculate the partial derivative of $tr(A \exp{\hat{\omega}})$ with respect to $\omega = [\omega_1, \omega_2, \omega_3 ]^\top \in \mathbb{R}^3$, where $A \in \mathbb{R}^{3\times 3}$ is a constant matrix and $\exp$ is the matrix exponential map. 
Here 
$$
\hat{\omega} = \left[\begin{array}{ccc}
0 & -\omega_{3} & \omega_{2}\\
\omega_{3} & 0 & -\omega_{1}\\
-\omega_{2} & \omega_{1} & 0
\end{array}\right]
$$
Obviously the derivative should be in $\mathbb{R}^3$. $A$ could be any matrix with determinant non-zero and I would like the answer not to depend on the individual terms of $A$. 
Any solution or suggestion of approach or reference is highly appreciated
Attempt so far : let $\delta $ be a real scalar and let $v \in \mathbb{R}^3$ be a variation of $\omega$.
then 
$$
tr(A\exp(\hat{\omega}+\delta\hat{v}))=tr(A\exp(\hat{\omega})\exp(\delta\hat{v}))=tr(A\exp(\hat{\omega})(I+\delta\hat{v}+\frac{1}{2!}\delta^{2}\hat{v}^{2}+\cdots)) 
$$
Now 
$$
tr(A\exp(\hat{\omega}+\delta\hat{v}))-tr(A\exp(\hat{\omega}))= tr(A\exp(\hat{\omega})(\delta\hat{v}+\frac{1}{2!}\delta^{2}\hat{v}^{2}+\cdots))
$$
removing higher order terms of $\delta$ and dividing by $\delta$, we get $tr(A\exp(\hat{\omega})\hat{v})$. However the expression is not of correct order. I am not understanding how to obtain a $3\times 1$ expression from this. 
 A: Note that in general, we don't have $\exp(A + B) = \exp(A)\exp(B)$!
If you want to calculate the derivative of your map $f$ at the origin, we have
$$ df|_{0}(v) = \frac{d}{dt} \operatorname{tr}(A \exp(t\hat{v}))|_{t=0} = \operatorname{tr} \left( \frac{d}{dt} A \exp(t \hat{v})|_{t = 0} \right) = \operatorname{tr}(A \hat{v}).$$
For other points, you'll need to calculate
$$ df|_{\omega}(v) = \frac{d}{dt} \operatorname{tr} (A \exp(\hat{\omega} + t\hat{v}))|_{t = 0} = \operatorname{tr} \left( A \frac{d}{dt} \exp(\hat{\omega} + t\hat{v})|_{t = 0}\right).$$
Set $X = \frac{\omega}{t} + v$. Using Rodrigues' formula, we have
$$ \exp(\hat{\omega} + t\hat{v}) = \exp (t \hat{X}) = I + \sin(t) \hat{X} + (1 - \cos t) \hat{X}^2 = \\
I + \frac{\sin t}{t} \hat{\omega} + \sin(t) \hat{v} + \frac{1 - \cos t}{t^2} \hat{\omega}^2 + \frac{1 - \cos t}{t}(\hat{\omega}\hat{v} + \hat{v}\hat{\omega}) + (1 - \cos t) \hat{v}^2 = \\
I + t \left( \hat{v} + \frac{1}{2}(\hat{\omega}\hat{v} + \hat{v}\hat{\omega})\right) + \dots $$
where the $\dots$ denote higher order terms. Thus,
$$ df|_{\omega}(v) = \operatorname{tr} \left( A \left( \hat{v} + \frac{1}{2}(\hat{\omega}\hat{v} + \hat{v}\hat{\omega})\right) \right). $$
