(Spivak 2-15, c)
If $a_{i,j}$ are differentiable and $f(t)=\det\left(a_{i,j}(t)\right)$ show that $f'(t)= \sum_{j=0}^n\det\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ \vdots& \vdots & \ddots & \vdots \\ a_{j,1}' & a_{j,2}' & \cdots & a_{j,n}' \\ \vdots& \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \end{pmatrix}$.
What I've already done, is to prove that indeed $\det\colon\Bbb R^n\times\dots\times \Bbb R^n \to \Bbb R$ is differentiable, and:
(+) $\Bbb D \det(a_1, ..., a_n)(x_1,...,x_n)= \sum_{i=0}^n\det\begin{pmatrix} a_{1}\\ \vdots\\ x_{i}\\ \vdots\\ a_{n}\end{pmatrix}$
Which came from that for bilinear (generalized for multilinear) $f$ $\Bbb D f(a,b)(x,y)=f(a,y)+f(x,b)$
What I've also proven. I know that what I am looking for comes from (+) and a chain rule, but I am a bit stuck with computing it in a right way. So I would be truly grateful, if someone could help me with that. Thanks in advance!
This can be done in the following way: we write the determinant as $$f(t)=\sum_{\sigma\in\mathcal S_n}\varepsilon\left(\sigma\right) \prod_{i=1}^n a_{i,\sigma\left(i\right)} , $$ and use the fact that the derivative of $t\mapsto \prod_{l=1}^ng_l(t)$ is $$\sum_{l=1}^ng'_l(t)\sum_{ i=1,i\neq l}^ng_i(t).$$