Suppose that you have an $n$-dimensional vector space $V$, equipped with a standard volume n-form $(x_1,\dots,x_n)$. Let $\{b_1,\dots,b_n\}$ be a standard basis such that $(b_1,\dots,b_n)=1$.

For a linear operator $A:V\to V$ we have:
$$
(x_1,\dots,x_n)\mathrm{tr}A=\sum_{i=1}^n(x_1,\dots,Ax_i,\dots,x_n).
$$
You can check that it agrees with the standard $\mathrm{tr}A=\sum_iA_{ii}$ by setting $x=b$ and expanding $Ab_i$ in the basis.

Also, we have
$$
(x_1,\dots,x_n)\mathrm{det}A=(Ax_1,\dots,Ax_n),
$$
can be easily compared with your favorite definition of $\det$.

Now, consider an $(n-1)$-form $\omega$. There exists a unique vector $x_\omega$ s.t.
$$
\omega(x_1,\dots,x_{n-1})=(x_\omega,x_1,\dots,x_{n-1}).
$$
Fix some $x_0$ and let $\omega$ be
$$
\omega(x_1,\dots,x_{n-1})=(x_0,Ax_1,\dots,Ax_{n-1}),
$$
which defines some $x_\omega$
$$
(x_\omega,x_1,\dots,x_{n-1})=(x_0,Ax_1,\dots,Ax_{n-1}).
$$
This procedure defines a linear map $A^{*}:V\to V$ by $x_0\mapsto x_\omega$. Let me write this as
$$
(A^*x_1,x_1,\dots,x_n)=(x_1,Ax_2,\dots,Ax_n),
$$
now if you let $x_1=Ax_0$ you immediately obtain $AA^* =I\det A$, so by continiuty $A^*$ coincides with your favourite definition of the adjugate $\mathrm{adj}A$.

All that said, here is a nice proof of the Jacobi's formula:

For any set of $x_1,\dots,x_n$,
\begin{align}
\frac{\partial}{\partial\alpha}(Ax_1,\dots,Ax_n)=&\sum_{i=1}^n(Ax_1,\dots,\frac{\partial A}{\partial\alpha}x_i,\dots,Ax_n)=\\\\
\sum_{i=1}^n(x_1,\dots,A^*\frac{\partial A}{\partial\alpha}x_i,\dots,x_n)=&\mathrm{tr}\left(A^*\frac{\partial A}{\partial\alpha}\right)(x_1,\dots,x_n).
\end{align}
Now we only have to recall the above definition of $\det$, and immediately obtain
$$
\frac{\partial}{\partial\alpha}\det A=\mathrm{tr}\left(A^* \frac{\partial A}{\partial\alpha}\right).
$$

Of course, for an invertible $A$ we have $A^* = A^{-1}\det A$.