Let $f_{ij}(t)$ be a differentiable function, $$F(t)=\begin{vmatrix} f_{11}(t)&f_{12}(t)&\dots&f_{1n}(t)\\ f_{21}(t)&f_{22}(t)&\dots&f_{2n}(t)\\ \vdots&\vdots&&\vdots\\ f_{n1}(t)&f_{n2}(t)&\dots&f_{nn}(t)\\ \end{vmatrix},$$
Prove that $\frac d{dt}F(t)=\sum_{j=1}^n F_j(t)$, where
$$F_j(t)=\begin{vmatrix} f_{11}(t)&f_{12}(t)&\dots&\frac d{dt}f_{1j}(t)&\dots&f_{1n}(t)\\ f_{21}(t)&f_{22}(t)&\dots&\frac d{dt}f_{2j}(t)&\dots&f_{2n}(t)\\ \vdots&\vdots&&\vdots&&\vdots\\ f_{n1}(t)&f_{n2}(t)&\dots&\frac d{dt}f_{nj}(t)&\dots&f_{nn}(t)\\ \end{vmatrix},$$
I’ve just learnt determinants as my first chapter in linear algebra and stumbled upon this question.
I think it have something to do with the expansion of the determinant as $$\sum_{(k_1,k_2,\dots,k_n)} (-1)^{N(k_1,k_2,\dots,k_n)} a_{1k_1}a_{2k_2}\dots a_{nk_n}$$
but I’m really not sure how.
It would be very helpful if you start with a two by two determinant and use product rule of differentiation to see the result. Then you can easily generalize to higher dimension.