Derivative of the Frobenius norm of an exponential function using chain rule Let $G=\text{exp}(\sum_{k=1}^{6}a_{k}E_{k})$,
where $E_{k}$ is a constant square matrix, scalars  $\{a_{k}\}_{k=1,...6}$ is the coefficients of matrix $E_{k}$.
Let $\phi= \parallel G \parallel^{2}_{F}$.
My question is, how to calculate the derivative $\frac{\partial \phi}{\partial a_{k}}$?
I use chain rule to calculate it, and have $\frac{\partial \phi}{\partial a_{k}}=2\text{trace}(G^{T}E_{k})$. I don't know if it is correct?
Please help.
 A: For typing convenience let $\,X=\sum_k a_kE_k$
The block triangular method of Kenney & Laub says that
$${\rm G}\Bigg(\begin{bmatrix}X&E_k\\0&X\end{bmatrix}\Bigg) =  \begin{bmatrix}G&L_k\\0&G\end{bmatrix} $$
where $G = {\rm G}(X)\,\,\,$ 
and $$L_k = \lim_{h\rightarrow\,0}\, \frac{{\rm G}(X+hE_k)-{\rm G}(X)}{h}$$
For the current problem the function is: $\,\,{\rm G}(X) = \exp(X)$ 
Incrementing coefficient $a_k$ by $da_k\,$ will increment $(X,G)$ by the following amounts
$$\eqalign{
dX&=E_k\,da_k\cr dG &= L_k\,da_k
}$$ [NB: Your question assumes that $L_k=E_k$, but that assumption is false.]
Now consider how the Frobenius norm will be affected. 
$$\eqalign{
 \phi &= \|G\|^2_F = G:G \cr
d\phi &= 2G:dG = 2G:L_k\,da_k \cr
\frac{d\phi}{da_k} &= 2G:L_k = 2\,{\rm tr\,}(G^TL_k) \cr
}$$
where a colon denotes the trace/Frobenius product; an alternate notation for the trace function $$A:B = {\rm tr\,}(A^TB)$$
While this is not a closed-form solution, as long as you are able to evaluate the function of interest on a block triangular argument, you can calculate the required derivative.
