Differentiating matrix exponential

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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 $A^Te^{A^Tt}$?

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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, $$ A^T e^{A^T} = e^{A^T} A^T. $$

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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.