How to show $\frac{d}{dt}(\det(I+At)=\text{Tr}(A)$? How to show $\frac{d}{dt}\det(I+At)\mid_{t=0}=\DeclareMathOperator{\Tr}{Tr}\Tr(A)$? I can't find a easy way to prove that.
You just need to look at the coefficient before $t$ in the polynomial obtained by expanding the determinant. Now try to identify where the terms $at$ come from. You should note that all the off-diagonal elements contain $t$, so there should not be too many such terms.

You can start with $2\times 2$ or $3\times 3$ to get the idea if you still stuck.
Hint: Note that the characteristic polynomial of $A$ is

$$\det(A-tI)=t^n-\mbox{Tr}(A)t^{n-1}+O(t^{n-2}) \; .$$