how to find matrix from its exponential form I know about the relation $$\frac{d}{dt}e^{At}=Ae^{At}$$
Is the only way to use it is to find the inverse of $e^{At}$ and then post-multiply on both sides? 
Is there a better approach? 
 A: Okay, we are going to use the fact that if $A = P^{-1}DP$ we also have $e^A = P^{-1}e^DP$
And $e^D$ is simply the diagonal matrix $e^{d_{ i,i}}$.
So if you find the $P$ matrix by finding the eigenvectors of $e^{Dt}$ you can use the associated $D$ matrix of eigenvalues (need to take log of each value and divide by $t$, if the result is still a function of $t$ than something is wrong) and plug it back into $A = P^{-1}DP$
I am not sure if that is "better" but it is a nice extension of the theory, I actually like your idea of taking the derivative a lot.
Does that get you where you want to be?
A: Given any exponential matrix $e^{A}$, first find a matrix $P$ such that $P e^A P^{-1}$ is in Jordan normal form.
$$P A P^{-1} = J \stackrel{def}{=}
\begin{bmatrix}
J_{m_1}(\lambda_1) & 0 & \ldots & 0 \\
0 & J_{m_2}(\lambda_2) & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \ldots & J_{m_n}(\lambda_n)
\end{bmatrix}$$
where $J_{m_k}(\lambda)$ is the Jordan block associated with the $k^{th}$ eigenvalue $\lambda_k$ of $e^{A}$ whose multiplicity is $m_k$. 
Each Jordan block $J_m(\lambda)$ is a $m \times m$ matrix of the form 
$$J_m(\lambda) = \lambda I_m + \eta_m = \begin{bmatrix}
\lambda & 0  & 0 & \ldots & 0 & 0 & 0\\
1  & \lambda & 0 & \ldots & 0 & 0 & 0\\
0  & 1 & \lambda & \ldots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & \ldots  & 1 & \lambda & 0\\
0 & 0 & 0 & \ldots  & 0 & 1 & \lambda
\end{bmatrix}
$$
where $\eta_m$ is the $m \times m$ matrix with $1$ on its sub-diagonal and $0$ everywhere else.
Since $e^{A}$ is an exponential matrix for $A$, all its eigenvalues $\lambda_k$ are non-zero. Now $\eta_m$ is a nilpotent matrix and $\eta_m^{m} = 0$, any power series of it truncates after the $m\!-\!1$-term. As a result, we can define a matrix logarithm on $J_m(\lambda)$ as a finite series:
$$\log(J_m(\lambda)) = \log\left[\lambda \left(I_m + \frac{1}{\lambda} \eta_m \right)\right]
= ( \log\lambda ) I_m + \sum_{k=1}^{m-1} \frac{(-1)^{k-1}}{k\lambda^k} \eta_m^k
$$
and it will satisfy
$$e^{\log(J_m(\lambda))} = J_m(\lambda)$$
If we assemble all the $\log(J_{m_k}(\lambda_k))$ blocks into as a single matrix and call it $\log J$, we will have
$$e^{PA P^{-1}} = P e^{A} P^{-1} = J = e^{\log J}
\quad\implies\quad
A = P^{-1}(\log J) P.$$
A: If you have $e^{At}$ and you want to find $A$, note that having $e^{At}$, in whatever form you've got it, and there are many different formulas, sometimes depending on $A$, such as
$e^{Jt} = (\cos t) I + (\sin t) J \tag{1}$
for matrices satisfying
$J^2 = -I; \tag{2}$
the point is, you might have $e^{At}$ in a form which is easier to evaluate/differentiate than the standard power series
$e^{At} = \sum_0^\infty \dfrac{(At)^n}{n!}; \tag{3}$
anyway, as I was saying, note that having $e^{At}$ also gives you $(e^{At})^{-1} = e^{-At}$; you just set $t = -1$ in your expression for $e^{At}$.  So yeah, you just find $d(e^{At})/dt$ from your formula, and then you have, for any $t$,
$A = \dfrac{d(e^{At})}{dt} (e^{-At}). \tag{4}$
Doing this with my example (1) yields
$\dfrac{d(e^{Jt})}{dt} = -(\sin t)I + (\cos t)J \tag{5}$
and
$e^{-Jt} = (\cos t)I - (\sin t)J; \tag{6}$
multiplying them out validates:
$\dfrac{d(e^{Jt})}{dt}e^{-Jt} = -(\sin t)(\cos t) I^2 - (\sin t)(\cos t)J^2 + (\cos^2 t + \sin^2 t)J = J, \tag{7}$
where we used $J^2 = -I$.  Of course, to actually get at $J$ (or $A$) entry-by-entry, we'd have to use the explicit forms of the matrices; for example, if
$J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag{8}$
(this is only one possible $J$; there are many) then
$e^{-Jt} = \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}, \tag{9}$
and
$\dfrac{d(e^{Jt})}{dt} = \begin{bmatrix} -\sin t & \cos t \\ -\cos t & -\sin t \end{bmatrix}; \tag{10}$
I leave the matrix multiplication to my readership, but (7) is again verified in this manner.
So the above shows how things can work, in theory and practice.  Of course, as the commentators have observed, taking $t = 0$ after differentiation saves some work, since then $e^{At} = I$.  And in general things won't be as nice as my example, but sometimes there are formulas for $e^{At}$ which allow one to skirt the power series, which tends to generate a lot of arithmetic if directly computed.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
