Thus

$\dfrac{\partial}{\partial t}\left|_{t=0}\right. \dfrac{\mathrm{det}(A+t\,\Omega)}{\mathrm{det}(A)} = \mathrm{tr}(A^{-1}\Omega)$

and also implies

This comes from Jacobi's formula:

${\mathrm d} \det (F(t)) = \det (F(t)) \mathrm{tr} (F(t)^{-1} {\mathrm d}F(t))$

where $F(t)$ is a parameter dependent matrix

This is a special case of the product rule and generalizes

${\mathrm d}\left(a\cdot b\right) = a\,{\mathrm d}b+b\,{\mathrm d}a = a\cdot b\left(\dfrac{1}{a}{\mathrm d}a+\dfrac{1}{b}{\mathrm d}b\right)$.

which you get for

$F(t) := \mathrm{diag}(a(t),b(t))$,

which can be seen $a(t)\cdot b(t)=\det\,F(t)$ representing the changing area of a rectangle.

The expression $\dfrac{1}{a}{\mathrm d}a$ is the so called logarithmic derivative of $a$ and scale invariant.

The function of $t$ on the right acts like a generating function. Of course

$ \mathrm{tr}(\Omega) = \dfrac{\partial}{\partial t}\left|_{t=0}\right. \mathrm{tr}(K+t\,\Omega+{\mathcal O}(t^2)) $

but the det-formula involves a non-linear function.

Wikipedia: Jacobi's formula

Fréchet derivative chain rule

Determinant via multilinear functionals, Trace of square matrices