Derivative of matrix exponential $\exp(A+xB)$ at $x=0$ Consider two (Hermitian) matrices $A$ and $B$. Is there a nice expression for  the following?
$$ \boxed{ \frac{\mathrm d}{\mathrm d x}  \exp\left( A + x B \right)\big|_{x=0} = \; ? }$$
Of course, if $A$ and $B$ commute, this is simply $B \exp{(A)}$.
One thing I tried was the Suzuki-Trotter formula:
\begin{align}
\boxed{\frac{\mathrm d}{\mathrm d x}  \exp\left( A + x B \right)\big|_{x=0}} 
&= \frac{\mathrm d}{\mathrm d x} \left. \left( \lim_{N \to \infty} \left[ \exp\left( \frac{A}{N} \right) \exp \left( x \frac{B}{N} \right) \right]^N \right) \right|_{x=0} \\
&= \lim_{N\to \infty}  \sum_{n=1}^N \exp\left( \frac{n}{N} A \right) \frac{B}{N}  \exp\left( \frac{N-n}{N} A \right) \\
&= \left( \lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N e^{\frac{n}{N}A }B\; e^{-\frac{n}{N}A } \right)  e^A \\\
&= \boxed{ \int_0^1 e^{t A} B \;e^{(1-t)A} \; \mathrm d t } \; .
\end{align}
Is this as close as it gets to a closed form?
One thing we can do is go to the eigenbasis of $A$, such that we can explicitly perform the integration over $t$. If we index the eigenvectors of $A$ by $i$, with corresponding eigenvalues $\lambda_i$, then we can express the answer in this basis:
\begin{equation}
\boxed{ \left( \frac{\mathrm d}{\mathrm d x}  \exp\left( A + x B \right)\big|_{x=0} \right)_{ij} = \frac{e^{\lambda_i}-e^{\lambda_j}}{\lambda_i-\lambda_j} B_{ij}} \;,
\end{equation}
where $(\cdot)_{ij}$ are the entries of a matrix in the eigenbasis of $A$. (Note that if $\lambda_i = \lambda_j$, we replace $\frac{e^{\lambda_i}-e^{\lambda_j}}{\lambda_i-\lambda_j} \to e^{\lambda_i}$, which is also consistent with l'Hopital's rule.)
 A: Given the $\lambda$-parameterized matrix definitions
$$\eqalign{
C &= C(\lambda) &= A+\lambda B \\
E &= E(\lambda) &= \exp(C) \\
E'&= E'(\lambda)&= \frac{dE}{d\lambda} \\
}$$
an effective method to calculate the value of $\,E'(0)$ is the block-triangular method:
$$\eqalign{
&F = \exp\Bigg(\begin{bmatrix}A&B\\0&A\end{bmatrix}\Bigg)
   =  \begin{bmatrix}C(0)&E'(0) \\0&C(0)\end{bmatrix} \\
&E'(0) = \big[\matrix{I&0}\big]\,F \left[\matrix{0\\I}\right] \\
}$$
This method is quite general and works for any function, not just the exponential.
A: Not an answer, but here is an alternative, algebraic, elementary approach to the final formula you have.  It rests on the observation that if $T:V\to W$ is a linear transformation, $(v_{\alpha})$ a basis for $V$, $(w_{\beta})$ a basis for $W$, and $(w^*_{\gamma})$ the corresponding dual basis, then if $[T]$ is the matrix of $T$ with respect to our bases, $[T]_{ij}=w_i^*Tv_j$.
By using the product rule for matrices that $\frac{d}{dx}(M(x)N(x))=M'(x)N(x)+M(x)N'(x)$ and induction, we get the formula 
$$\frac{d}{dx}(M(x)^n)=\sum_{\substack{0\leq i,j \\i+j=n-1}}M(x)^iM'(x)M(x)^j.$$
Therefore 
$$\left.\frac{d}{dx}(A+Bx)^n\right\rvert_{x=0}=\sum_{i+j=n-1}A^iBA^j,$$
and so
$$C:=\left.\frac{d}{dx}e^{A+Bx}\right\rvert_{x=0}=\sum_n\sum_{i+j=n-1}\frac{A^iBA^j}{n!}.$$
Let $u,v$ be left and right eigenvectors of $A$ so that $uA=\mu u$, $Av=\lambda v$. (Sidedness isn't necessary when working with symmetric matrices over $\mathbb R$, but I want to leave open the possibility of working with not-necessarily symmetric matrices).  Then 
$$uCv=(uBv)\sum_n\sum_{i+j=n-1}\frac{\mu^i\lambda^j}{n!}.$$
By the identity $a^n-b^n=(a-b)\displaystyle \sum_{i+j=n-1}a^ib^j$, we have 
$$(\mu-\lambda)uCv=(uBv)\sum_n\frac{\mu^n-\lambda^n}{n!}=(uBv)(e^{\mu}-e^{\lambda}).$$

Actually, here is an extension of this idea which sort of gives a formula.  Given $A\in \operatorname{GL}(V)$, define $L_A,R_A:\operatorname{End}(\operatorname{GL}(V))$ by $L_A(B)=AB, R_A(B)=BA$.  Then $L_A$ and $R_A$ commute.  We can then write
$$C=\sum_n\sum_{i+j=n-1}\frac{A^iBA^j}{n!}=\left(\sum_n\sum_{i+j=n-1}\frac{L_A^iR_A^j}{n!}\right)B.$$
If we multiply this on the left by $ad_A=(L_A-R_A)$, identical algebra as above yields
$$[A,C]=e^{L_A}(B)-e^{R_A}(B)=e^AB-Be^A=[e^A,B].$$
This formula only determines $C$ up to a map commuting with $A$, but there be another way to to make use of this to find a formula for $C$ itself.  
