Proving a formula for the derivative of the determinant of an $n\times n$ matrix This time I'm having trouble proving the following question:
Assuming that the elements of an $n\times n$-matrix $Y(t) = (y_{ij}(t))_{i,j=1,...,n}$ are differentiable, I need to show the following equality:
$$\frac{d}{dt}(\det(Y(t))) = \sum\limits_{1=1}^n \det\left(\begin{array}{ccc}
        y_{11} & \cdots & y_{1n}\\
        \vdots &  & \vdots \\
        y_{i-1,1} & \cdots & y_{i-1,n}\\
        y'_{i,1} & \cdots & y'_{i,n}\\
        y_{i+1,1} & \cdots & y_{i+1,n}\\
        \vdots &  & \vdots \\
        y_{n1} & \cdots & y_{nn}\\
    \end{array}\right)$$
And as a hint I'm told that one way of proving this equality is to use the formula of determinant with n!, however I'm not sure I know this formula, and the only one I could think of was the one with  Levi-Civita symbol:
$$\det(Y(t)) = \frac{1}{n!}\sum\limits_{i_1,...,i_n = 1; j_1,...,j_n = 1}\epsilon_{i_1,...,i_n}\epsilon_{j_1,...,j_n}\alpha_{i_1j_1,...,i_nj_n}$$
which I don't know how to derivate. Could anyone help me derivate this determinant ? any lue to start would be welcome, I didn't find anything related to my problem on mathstack.
 A: You want to apply the product rule for $n$ factors:
$$
\frac{\mathrm d}{\mathrm dt} (f_1(t)\, f_2(t) \cdots f_n(t)) = \sum_{i=1}^n f_1(t) \cdots f_{i-1}(t) \, f_i'(t) \, f_{i+1}(t) \cdots f_n(t).
$$
Applying that to each summand of the Leibniz formula for $\det(Y(t))$ yields the desired result.
A: We have by definition
\begin{align}\det(Y(t))&=\sum_{\sigma\in\mathfrak S_n}\varepsilon(\sigma)\prod_{i=1}^n y_{i,\sigma(i)}\\
\text{explicitly: }\hskip 4em &=\sum_{\sigma\in\mathfrak S_n}\varepsilon(\sigma)\,y_{1,\sigma(1)}y_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y_{n,\sigma(n)},
\end{align}
and differentiating with the product rule, we obtain
\begin{align}
\bigl(\det(Y(t))\bigr)'=\sum_{\sigma\in\mathfrak S_n}\varepsilon(\sigma)\Bigl[&y'_{1,\sigma(1)}y_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y_{n,\sigma(n)}+y_{1,\sigma(1)}y'_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y_{n,\sigma(n)}\\
&+\dots+y_{1,\sigma(1)}y_{2,\sigma(2)}\dots y'_{i,\sigma(i)}\dots y_{n,\sigma(n)}+\cdots\\
&+\dots+y_{1,\sigma(1)}y_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y'_{n,\sigma(n)}\Bigr]
\end{align}
