If we take $n$ vectors $w_1,w_2,...,w_n \in \mathbb R^n$ then $\det(w_1,w_2,...,w_n)=M_{w_1,w_2,...,w_n}(t)$.
Let $g_1,g_2,...,g_n: \mathbb R \to \mathbb R^n$. Knowing what is $M_{w_1,w_2,...,w_n}(t)$ prove that: $$\frac{d}{dt} M_{g_1,g_2,...,g_n}=M_{g'_1,g_2,...,g_n}+M_{g_1,g'_2,...,g_n}+...+M_{g_1,g_2,...,g'_n}$$
Assume that:
$$g_1(x)=w_1=(w_{11},w_{12},...,w_{1n})$$
$$g_2(x)=w_2=(w_{21},w_{22},...,w_{2n})$$
$$...$$
$$g_n(x)=w_n=(w_{n1},w_{n2},...,w_{nn})$$
Then we must prove that:
$$\frac{d}{dt} \det \begin{bmatrix} w_{11} & w_{21} & ... & w_{n1} \\ w_{12} & w_{22} & ... & w_{n2} \\ ... & ... & ... & ... \\ w_{1n} & w_{2n} & ... & w_{nn} \end{bmatrix}= \det \begin{bmatrix} 1 & w_{21} & ... & w_{n1} \\ 1 & w_{22} & ... & w_{n2} \\ ... & ... & ... & ... \\ 1 & w_{2n} & ... & w_{nn} \end{bmatrix}+ ...+\det \begin{bmatrix} w_{11} & w_{21} & ... & 1 \\ w_{12} & w_{22} & ... & 1 \\ ... & ... & ... & ... \\ w_{1n} & w_{2n} & ... & 1 \end{bmatrix}$$
However I don't have any idea how to show this. I thought about integral because on the left of equality I have a derivative but I don't know if it can be helpful.
Can you give me some tips?
I'll do the thing for $n=2$ and leave it to you to generalise. Note that $\det$ is multilinear. Hence, we get $$ \frac{1}{h} \left( \det(v(t+h), w(t+h)) - \det(v(t), w(t)) \right) = \frac{1}{h} \left( \det (v(t+h), w(t+h)) - \det(v(t), w(t+h)) \right) + \frac{1}{h} \left( \det( v(t), w(t+h)) - \det(v(t), w(t)) \right) = \det\left( \frac{1}{h}(v(t+h)-v(t)), w(t+h)\right) + \det\left( v(t), \frac{1}{h} ( w(t+h)-w(t))\right) .$$ Now use the continuity of the determinant and conclude. The general case works with the same idea, is just more annoying to write down.