Derivative of determinant of a matrix

Think I can provide a proof for Matias' formula.

So, let

$$ A(t) = \mathrm{det}\left( A_1(t), \dots , A_n(t) \right) \ . $$

By definition,

$$ \frac{dA(t)}{dt} = \mathrm{lim}_{h\rightarrow 0} \frac{A(t+h) - A(t)}{h} = \mathrm{lim}_{h\rightarrow 0} \frac{\det (A_1(t+h), \dots, A_n(t+h)) - \det(A_1(t), \dots , A_n(t))}{h} $$

Now, we subtract and add

$$ \det(A_1(t), A_2(t+h), \dots , A_n(t+h)) $$

obtaining:

$$ \frac{dA(t)}{dt} = \mathrm{lim}_{h\rightarrow 0} \frac{\det (A_1(t+h), A_2(t+h),\dots, A_n(t+h)) - \det(A_1(t), A_2(t+h), \dots , A_n(t+h))}{h} + \mathrm{lim}_{h\rightarrow 0}\frac{ \det(A_1(t), A_2(t+h), \dots , A_n(t+h))-\det(A_1(t), \dots , A_n(t))}{h} $$

Now we focus on the first addend, which is

$$ \det \left( \mathrm{lim}_{h\rightarrow 0} \frac{A_1(t+h) - A_1(t)}{h}, \mathrm{lim}_{h\rightarrow 0} A_2(t+h), \dots,\mathrm{lim}_{h\rightarrow 0} A_n(t+h) \right) $$

That is,

$$ \det (A_1'(t), A_2(t), \dots , A_n(t)) \ . $$

Now, let's go for the second addend to which we substract and add

$$ \det(A_1(t), A_2(t), A_3(t+h), \dots , A_n(t+h)) \ . $$

From which we will obtain the term

$$ \det (A_1(t), A'_2(t), A_3(t), \dots , A_n(t)) \ . $$

Keep on doing analogous operations till you get

$$ \det (A_1(t), A_2(t), \dots , A_{n-1}(t), A_n'(t)) \ . $$

The formula is $$d(\det(m))=\det(m)Tr(m^{-1}dm)$$ where $dm$ is the matrix with $dm_{ij}$ in the entires. The derivation is based on Cramer's rule, that $m^{-1}=\frac{Adj(m)}{\det(m)}$. It is useful in old-fashioned differential geometry involving principal bundles.

I noticed Terence Tao posted a nice blog entry on it. So I probably do not need to explain more at here.

Like product rule:

$$\dfrac{d}{dt}\det(A_1(t),A_2(t),...,A_n(t))=\det(A_1^{'}(t),A_2(t),A_n(t))+\det(A_1(t),A_2^{'}(t),...,A_n(t))+...+\det(A_1(t),A_2(t),...,A_n^{'}(t)) $$