Partial derivatives of matrix exponential I am trying to find an expression for
$$\frac{\partial \text{tr}\, Ae^{K}}{\partial v_i}$$
Given that $K$ is skew-symmetric
$$K = \begin{bmatrix}
0 & v_1 & v_2 \\
-v_1 & 0 & v_3 \\
-v_2 & -v_3 & 0
\end{bmatrix}$$
and $A$ is an arbitrary $3 \times 3$ matrix.  It's easy to solve when $A = I_3$, but I'm having difficulties otherwise.
 A: Given an an arbitrary $3\times 3$ skew symmetric matrix, $K$, calculate the quantity 
$$\eqalign{
 \beta &= \frac{\|K\|_F}{\sqrt 2} = \sqrt\frac{K:K}{2} \cr
}$$ 
The differential of this quantity can be expressed in terms of the differential of $K$ like so
$$\eqalign{
 \beta^2 &= \frac{K:K}{2} \cr
 2\,\beta\,d\beta &= K:dK \cr\cr
 d\beta &= \frac{K:dK}{2\,\beta} \cr
}$$ 
Then messing around with Rodrigues' rotation formula you'll find that
$$\eqalign{
 \exp(K) &= I + K\frac{\sin\beta}{\beta} + K^2\frac{1-\cos\beta}{\beta^2} \cr
}$$
The function in question can be written as 
$$\eqalign{
 f &= A^T:\exp(K) \cr
 &= A^T:\Big(I + K\frac{\sin\beta}{\beta} + K^2\frac{1-\cos\beta}{\beta^2}\Big) \cr
}$$
Expand the differential $df$ in terms of $\{dK,d\beta\}$, then substitute the above expression for $d\beta$ to arrive at an expression entirely in terms of $dK$. 
From that differential, the gradient with respect to $K$ is found to be
$$\eqalign{
  \frac{\partial f}{\partial K} &= A^T\frac{\sin\beta}{\beta}+(A^TK+KA^T)\frac{\cos\beta-1}{\beta^2} \cr
 &+ \,K\,\Big[(A^T:K)\frac{\beta\cos\beta-\sin\beta}{2\,\beta^3}+(A^T:K^2)\frac{\beta\sin\beta+2\cos\beta-2}{2\,\beta^4}\Big] \cr
}$$
The derivative with respect to the components of $v$ can be found by taking the Frobenius product with various single-entry matrices $J_{ik}$
$$\eqalign{
\frac{\partial f}{\partial v_1}&=J_{12}:\frac{\partial f}{\partial K}\cr\cr
\frac{\partial f}{\partial v_2}&=J_{13}:\frac{\partial f}{\partial K}\cr\cr
\frac{\partial f}{\partial v_3}&=J_{23}:\frac{\partial f}{\partial K}\cr
}$$ 
A: That follows is the Maple procedure for the calculation of $\frac{\partial \text{tr}\, Ae^{K}}{\partial v_1}$ where $v_1:=x,v_2:=y,v_3:=z$. Time of calculation: $<1"$. The result meets 5 screens...
restart:
 with(LinearAlgebra):
 K := Matrix(3, 3, [[0, x, y], [-x, 0, z], [-y, -z, 0]]):
A := Matrix(3, 3, symbol = a):
 B := MatrixExponential(K):
 u := simplify(Trace(A.B)):
 f := unapply(u, x):
 simplify(diff(f(x), x));
