Lemma for Jacobi's formula proof In the wikipedia article on Jacobi's formula
$$\frac{d}{dt}\det A(t) = \textrm{tr}\left(\textrm{adj}(A(t))\frac{dA(t)}{dt}\right)$$ they offer a proof via the chain rule that i am trying to understand.
The lemma states that $\det'(I) = \textrm{tr}$, where $I$ is the identity matrix and $\det'$ is the differential of $\det$. This is where I'm confused. What is the differential of $\det$? I figured it meant the derivative of the determinant of a matrix with respect to the matrix's indices, but then how does it make sense to apply it to the specific matrix $I$? 
 A: The differential of a function $\mathrm{d}f$ is its best linear approximation at some point $x_0$. In this case $\det$ is a function $\det:M_{n\times n} \rightarrow\mathbb{R}$ where $M_{n\times n}$ is the space of $n\times n$ square matrices. Therefore, a matrix is the equivalent of a point for real functions. The best linear approximation to $\det$ near the identity is given by:
$$ \det(\mathbf{I}+\mathbf{M})=\det(\mathbf{I})+\mathrm{d}(\det(\mathbf{I}))\mathbf{M}+R(\mathbf{I},\mathbf{M}),\qquad \lim_{\|\mathbf{M}\|\rightarrow 0} \frac{R(\mathbf{I},\mathbf{M})}{\|\mathbf{M}\|}= 0 \tag{1}\label{eq1}.$$
The analogous expression for $f: \mathbb{R} \rightarrow\mathbb{R}$ is 
$$ f(x_0+\epsilon)=f(x_0)+\mathrm{d}(f(x_0))\epsilon+R(x_0,\epsilon),\qquad \lim_{|\epsilon|\rightarrow 0} \frac{R(x_0,\epsilon)}{|\epsilon|}= 0 \tag{2}\label{eq2}.$$
Regarding the original question, $\det'(\mathrm{I})=\mathrm{tr}$ is equivalent to the following:
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \det(\mathbf{I}+t\mathbf{B})\right] \right|_{t=0} = \mathrm{tr}(\mathbf{B}) \tag{3} \label{eq3}$$
You can use induction to prove $\eqref{eq3}$. The first step is to prove it for $n=2$:
$$\frac{\mathrm{d}}{\mathrm{d}t}  \left . \begin{bmatrix}
1+tB_{11} & tB_{12}  \\
tB_{21} & 1+tB_{22}  
\end{bmatrix} \right|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \left. \left[ 
(1+tB_{11})(1+tB_{22}) - t^2B_{12}B_{21}    \right] \right|_{t=0}=B_{11}+B_{22}.  \tag{4} \label{eq4}$$
Assuming that $\eqref{eq3}$ is valid for $n=k-1$, we will prove it for $n=k$. 
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \det(\mathbf{I}+t\mathbf{B})\right] \right|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ 
\sum_{i=1}^k (\delta_{i,k}+tB_{i,k})M_{i,k}(t)
\right] \right|_{t=0}  \tag{5} \label{eq5}$$
here $M_{i,k}(t)$ represent the $(i,k)$ minor of $\mathbf{I}+t\mathbf{B}$.
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ 
\sum_{i=1}^k (\delta_{i,k}+tB_{i,k})M_{i,k}(t)
\right] \right|_{t=0}=\sum_{i=1}^k\left. \left [ 
 B_{i,k}M_{i,k}(t)+(\delta_{i,k}+tB_{i,k})\frac{\mathrm{d}}{\mathrm{d}t} M_{i,k}(t)
\right] \right|_{t=0}  \tag{6} \label{eq6}$$
evaluating the first term in the RHS of $\eqref{eq6}$ at $t=0$ gives zero except when $i=k$ since each minor has lower order proportional to $t$. On the other hand, $M_{k,k}(0)=\det(\mathbf{I})=1$. Therefore,
$$ \sum_{i=1}^k\left. \left [ 
 B_{i,k}M_{i,k}(t)+(\delta_{i,k}+tB_{i,k})\frac{\mathrm{d}}{\mathrm{d}t} M_{i,k}(t)
\right] \right|_{t=0}=B_{k,k}+\frac{\mathrm{d}}{\mathrm{d}t} M_{k,k}(0)= \sum_{j=1}^k B_{j,j}\tag{7} \label{eq7}$$
In the last step we use the hypothesis $n=k-1$ since $M_{k,k}$ is  $\det(\mathbf{I^{(k-1)\times(k-1)}}+t\mathbf{B^{(k-1)\times(k-1)}})$. 
A: I probably can't help with the original question, but I can offer a proof that could be perhaps more understandable.
First, let's expand the determinant. We know that $\det A = \sum_\sigma (-1)^{\sigma} A^{\sigma_1}_1 A^{\sigma_2}_2 \ldots A^{\sigma_n}_n$, with the sum going over all permutations $\sigma$ of natural numbers from 1 to $n$. The derivative is then obtained using the Leibniz rule:
$$ \frac{\mathrm d}{\mathrm d t} \det A = \sum_\sigma (-1)^{\sigma} \sum_k A^{\sigma_1}_1 A^{\sigma_2}_2 \ldots \frac{\mathrm d A^{\sigma_k}_k}{\mathrm d t} \ldots A^{\sigma_n}_n. $$
When you use the determinant definition again, you get a rule that is best expressed with this pseudo-formula
$$ \frac{\mathrm d}{\mathrm d t} \det A = \\ \sum_k \det(\text{$A$ with all elements in the $k$-th column replaced by their derivatives}). $$
Now when you have a look at the RHS: $\operatorname{adj} A \frac{\mathrm d A}{\mathrm d t}$. Let's now consider any matrix $R = \operatorname{adj} A \cdot B$. Using the Laplace's expansion rule, we will see that the element in the $i$-th row and $j$-th column of this matrix is exactly determinant of $A$ with $i$-th column replaced by the $j$-th column of $B$.
And the trace just takes the diagonal elements and adds them up, so you get a sum of determinant of $A$ with first column replaced by the first column of $\mathrm d A / \mathrm d t$, i.e. its derivatives, plus $\det A$ with 2nd column replaced with derivatives and so on. Exactly what we found on the left.
This line of reasoning is really clumsy to explain, but hopefully should be easy to follow.
