Let $A\in\mathbb{R}^{n\times n}$ be an invertible matrix, $T\in\mathbb{R}^{n\times n}$ arbitrary.
Show $$(\det)'(A)(T)=\det(A)\operatorname{tr}(A^{-1}T)$$
Fist I need to determine $(\det)'(I)(T)=\lim_{\epsilon\to 0}\frac{\det(I+\epsilon T)-\det(I)}{\epsilon}$ and then use chain rule.
I have found many identical questions and a similar proof on Wikipedia https://en.wikipedia.org/wiki/Jacobi%27s_formula#Derivation (via chain rule) but I don't understand them. Why is $\lim_{\epsilon\to 0}\frac{\det(I+\epsilon T)-\det(I)}{\epsilon} = \operatorname{tr}(T)$? Thank you!
The key is to consider the polynomial $\det(I + \epsilon T)$. One way to get its linear term is to consider its relation to the characteristic polynomial. In particular, we have $$ \det(I + \epsilon T) = \det(\epsilon \cdot [\epsilon^{-1} I -(-T)]) = \epsilon^n \det(\epsilon^{-1} I - (-T)). $$ If the characteristic polynomial of $-T$ is $p(t) = \det(tI - (-T))$, then $\det(I + \epsilon T) = \epsilon^n p(\epsilon^{-1})$. Notably, this is simply the polynomial whose coefficients are the coefficients of $p$ in reverse order. Because the coefficient of $t^{n-1}$ in $p(t)$ is $\operatorname{tr}(T)$, it follows that the coefficient of $\epsilon^1$ in $\epsilon^{n}p(\epsilon^{-1})$ is also $\operatorname{tr}(T)$.
Alternatively, you can derive this fact directly using the Leibniz formula for the determinant $\det(I + \epsilon T)$.