How do I differentiate the determinant like this column vector form

Question

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I cannot understand the first step of this prove. Maybe it is because of the differentiation of the determinate (also as column vector form). How do I differentiate like this?

Answer 1

View the determinant as a (multilinear) function $d(x_1,...,x_n) = \det \begin{bmatrix} x_1 & \cdots & x_n \end{bmatrix}$ and let $W(t) = d(x_1(t),\cdots, x_n(t))$.

Then the derivative is $W'(t) = \sum_{k=1}^n {\partial d(x_1(t),\cdots, x_n(t)) \over \partial x_k} x_k'(t)$.

We need to compute ${\partial d(x_1,\cdots, x_n) \over \partial x_k}$. Since $d$ is multilinear, note that $d(x_1,...,x_k+h,...,x_n) = d(x_1,...,x_k,...,x_n) + d(x_1,...,h,...,x_n)$, so it follows that ${\partial d(x_1,\cdots, x_n) \over \partial x_k} h = d(x_1,...,h,...,x_n)$.

Then $W'(t) = \sum_{k=1}^n d(x_1(t),\cdots,x_k'(t), \cdots x_n(t))$.