Derivative of matrix exponential with respect to its symmetric matrix parameter Consider $\textbf{A} = exp(t\textbf{X})$ where $t \in \mathbb{R}$, $\textbf{X}, \textbf{A} \in \mathbb{R}^{n\times n}$ and $exp: \mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n\times n}$ is the matrix exponential. If we restrict $\textbf{X}$ to real-valued symmetric matrices, such that $\textbf{X}$ is diagonalizable and and $\textbf{A}$ is symmetric positive definite, is there a closed form of $\frac{d \textbf{A}}{d \textbf{X}}$?
Note that I am aware of the question here (Derivative of matrix exponential w.r.t. to each element of the matrix) but I believe my more restricted case of symmetric $\textbf{X}$ may yield a simpler and more concise solution.
 A: $
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\vc#1{\op{vec}\LR{#1}}
\def\Diag#1{\op{Diag}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\p{\partial}
\def\grad#1#2{\frac{\p #1}{\p #2}}
$A scalar function
can be applied to a square matrix
$$\eqalign{
f(x) = e^{tx} \qiq
A &= f(X) \\
}$$
and a real symmetric matrix can be diagonalized with orthogonal factors
$$\eqalign{
X &= QBQ^T,\quad B=\Diag b,\quad Q^TQ=I \\
}$$
The Daleckii-Krein theorem followed by vectorization yields
$$\eqalign{
Q^T dA\:Q &= R\odot\LR{Q^T dX\:Q} \\
\LR{Q\otimes Q}^T\,da &= \vc{R}\odot\LR{Q\otimes Q}^Tdx \\
}$$
where $(\odot,\otimes)$ denote the Hadamard and Kronecker products, respectively.
By defining the matrices
$$\eqalign{
D &= \Diag{\vc{R}},\qquad P = \LR{Q\otimes Q}\qiq P^T=P^{-1} \\
}$$
the gradient calculation simplifies to
$$\eqalign{
P^T da &= DP^Tdx \qiq \grad{a}{x} = PDP^T \\
}$$
All that remains is to calculate the symmetric $R$ matrix
which lies at the heart of the theorem
$$\eqalign{
R_{ij} = \begin{cases}
{\Large\frac{f(b_i)-f(b_j)}{b_i-b_j}} \qquad{\rm if}\;b_i\ne b_j \\
\\
\quad f'(b_j) \qquad\qquad {\rm otherwise} \\
\end{cases}
}$$
NB:$\:$ For the current problem $\;f'(x) = te^{tx}$
