Proof of the derivative of a matrix exponential $\frac d {dt} e^{tA} = Ae^{tA}$ I have the proposition in my book that $$\frac d {dt} e^{tA} = Ae^{tA}$$ 
The proof provided is somewhat terse. I think I've proved it using games with indices, but the book's preferred proof uses the defn of the derivative.
In particular, I'm annoyed by not understanding this line.
$$\frac d {dt} e^{tA}=e^{tA} \lim_{h\rightarrow 0} \frac {e^{hA} -I}{h}$$ (we're fine and good) 
$$= e^{tA}A$$
(huh?) 
This would imply that the numerator of the limit is equal to $hA$, which I am uncertain of how to prove. I have tried expanding $e^{hA}$ according to the defn of the matrix exponential, but see no further steps.
 A: Here is a very explicit solution in terms of calculating the matrix exponential.We'll take a real square matrix $A$ and work entirely over the real numbers. There exists a real invertible matrix $P$ such that $A=P\check DP^{-1}$ where $\check D$ is almost diagonal,more precisely, $\check D$ has elements $\lambda_i$ on the leading diagonal, 'rotation' blocks of form $$\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$  on the leading diagonal, Jordan blocks with a fixed $\lambda_*$ on the leading diagonal and 1's on the super-diagonal and block-Jordan blocks with 'rotation' blocks on the leading diagonal and 2x2 identity matrices $I$ on the super diagonal. Then $$\exp(A)=P\exp(\check D)P^{-1}\text{ and }\exp(At)=P\exp(\check Dt)P^{-1}$$ and $$A\exp(At)=P\check D\exp(\check Dt)P^{-1}$$ When we calculate $\exp(\check D)\text{ and }\exp(\check Dt)$ there is no problem with elements $\lambda_i$ on the leading diagonal-they give $e^{\lambda_i}$ and $e^{\lambda_it}.$ The desired identity is easily checked for those elements. For 'rotation' blocks we use $$\exp(\begin{bmatrix}a&-b\\b&a\end{bmatrix})=\begin{bmatrix}e^a\cos b&-e^a\sin b\\e^a\sin b&e^a\cos b\end{bmatrix}$$ and $$\exp(t\begin{bmatrix}a&-b\\b&a\end{bmatrix})=\begin{bmatrix}e^{at}\cos bt&-e^{at}\sin bt\\e^{at}\sin bt&e^{at}\cos bt\end{bmatrix}$$. Again, the desired identity is easily checked for these blocks.Now we look at Jordan blocks $J$ with with a fixed $\lambda_*$ on the leading diagonal and 1's on the super-diagonal. The $m-$th power of $J$ has $\lambda_*^m$ on the leading diagonal and ${m \choose i}\lambda_*^{m-i}$ on the $i-$th super-diagonal. Using the power series for $\exp(J)$ , it follows that $\exp(J)$ is upper-triangular with every element in the upper triangle equal to $e^{\lambda_*}.$ Using the power series for $\exp(tJ)$ , it follows that $\exp(tJ)$ has $e^{\lambda_*t}$ on the leading diagonal and $t^ie^{\lambda_it}$ on the $i-$th super-diagonal. Taking the derivative of this expression it is easy to check that exactly the same result is obtained by multiplying $\exp(tJ)$ by$J.$ Combining these Jordan block results with our earlier result tor 'rotation' blocks, we can establish the desired formula for Jordan block Jordan blocks. 
