Differentiating matrix exponential

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

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.