TITLE: Differentiating matrix exponential
QUESTION [7 upvotes]: I know that
$$\frac{d}{dt}e^{At} = Ae^{At}$$ 
However, in one lecture, I found the following
$$\frac{d}{dt}e^{A^Tt} = e^{A^Tt}A^T$$ 
The lecture is as follows



How to show the second case? Why not $A^Te^{A^Tt}$?

REPLY [14 votes]: Note, that for any matrix $C$, $C$ and $e^C$ commute. One can see this as follows: 
$$ Ce^C = C \cdot \sum_{k=0}^\infty \frac 1{k!} C^k = \sum_{k=0}^\infty \frac 1{k!} C^{k+1} = \sum_{k=0}^\infty \frac 1{k!} C^k \cdot C = e^C C $$
Hence, 
$$ A^T e^{A^T} = e^{A^T} A^T. $$

REPLY [5 votes]: Both $A^Te^{A^Tt}$ and $e^{A^Tt}A^T$ are correct as $A^T$ commutes with itself, therefore it will also commute with the exponential of itself. To prove this, write the exponential as series.