Showing that $f_0 (x_1, \ldots, x_m) \mathrm tr A =\displaystyle{ \sum_{i=1}^n} f_0(x_1, \ldots, Ax_i,\ldots, x_m)$ Question:
Consider $f: (-\epsilon, \epsilon) \to \mathbb R^{m^2}$ a differentiable path of matrices $m \times m$ such that $f(0) = I_m$ and the function $g: I \to \mathbb R$ is defined by $$g(t) = \det f(t)$$
Show that $g'(0) = \mathrm{tr} \,a $ (trace of the matrix $a$), where $a = f'(0)$. 
Attempt:
First we use that $\det f(t) = \det (f_1(t), \ldots, f_m(t)) $, where $f_i$ is the $i$-th row of the matrix $f(t)$ $m \times m$. Considering the operator $': \mathbb R^m \to \mathbb R^m$, $f(t) \mapsto f'(t)$
With this in mind $$\begin{align}g'(t) = \det'f(t) &= \sum_{i=1}^{m} \det (f_1(t),\ldots,f'_i(t),\ldots, f_m(t)) \\&\underbrace{=}_{(*)}\det(f_1(t), \ldots, f_m(t))\, \mathrm tr f'(t) \end{align}$$
For $t = 0$ we have 
$$\begin{align}g'(0) &=\det(e_1, \ldots, e_m) \,\mathrm tr f'(0) \\&= 1 \dot\, \mathrm tr \,a \\& = \mathrm tr \, a\end{align}$$
I'm trying to come up with a proof using the fact that $\det$ is a $m$-linear form. 
In order to prove $(*)$ I need the following equality, which according to this answer holds. 
If $A: V \to V$ is a linear operator and $f_0$ is a alternated $m$-linear form, such that $f_0 (e_1, \ldots, e_m) = 1$ then 
$$f_0 (x_1, \ldots, x_m) \mathrm tr A = \sum_{i=1}^m f_0(x_1, \ldots, Ax_i,\ldots, x_m) $$
Could someone give a proof to this?
 A: Recall that $\det$ is the product of eigenvalues and that $\mathrm{tr}$ is the sum of the eigenvalues. WLOG you can assume that $f(t) = I + tA$.
Given the eigenvalues $\lambda_i$ of $A$, the eigenvalues of $f(t)$ are $1+t\lambda_i$. Thus
$$g(t) = \prod_{i=1}^n (1+t\lambda_i)$$
Using the trick of logarithmic derivative, we get
$$\frac{g'(t)}{g(t)} = (\ln g)'(t) = \partial_t \sum_{i=1}^n \ln(1+t\lambda_i) = \sum_{i=1}^n \frac{\lambda_i}{1+t\lambda_i}$$
and thus
$$g'(0) = (\ln g)'(0) \cdot g(0) = \sum_{i=1}^n \frac{\lambda_i}{1+0\lambda_i} \cdot \prod_{i=1}^n (1+0\lambda_i) = \sum_{i=1}^n \lambda_i = \mathop{\rm tr}(A)$$
The WLOG assumption allows us to infer that for any $f$ with $f'(0) = A$, $g'(0) = \mathop{\rm tr}(A)$, since the derivative of a composition at a fixed point only depends on the derivative at that point (here it is $0$).
A: Use Taylor.
$g(t)=det(I+f'(0)t+o(t^2))$.
Then $$g'(0)=\lim_{t\to0}\frac{\det(I+f'(0)t)-1}{t}=\lim_{t\to0}\frac{t^{n}\cdot\det(t^{-1}I+f'(0))-1}{t}$$
which is the second-to-leading coefficient of the characteristic polynomial of $f'(0)$. This is because, in general, if $p(x)$ is a monic polynomial of degree $n$, then $$\lim_{t\to0}\frac{t^np(t^{-1})-1}{t}$$ is equal to that coefficient.
A: I think I have the answer. 
If $Ax_i = \displaystyle\sum_{j=1}^{k} a_{ij} x_j$ then  as $f_0$ is alternated we have 
 $$\begin{align}f_0(x_1, \ldots, Ax_i, \ldots, x_m) &= \sum_{j=1}^k a_{ij}f_0 (x_1, \ldots,x_i, \ldots,x_j, \ldots, x_m ) \\&=a_{ii} f_0 (x_1,\ldots, x_i,\ldots, x_m)\end{align} $$
Because $f_0(x_1, \ldots, x_i, \ldots, x_j, \ldots, x_m) =0$ whenever there is an element repeating. Thus it follows that 
$$\begin{align}\sum_{i=1}^mf_0(x_1, \ldots, Ax_i, \ldots, x_m) &= \sum_{i=1}^m a_{ii} f_0(x_1,  \ldots, x_m)\\&=\mathrm tr A \,f_0(x_1,  \ldots, x_m)\end{align}$$
Now seeing $\det' A$ as the functional $\det' A: \mathbb R^{m^2} \to \mathbb R$ then $(*)$ follows trivially. 
A: We will use the chain rule: $D_a(f\circ g)=D_{g(a)}f \circ D_a g$.
In our case, $a$ is $t$, $g:t\mapsto A_t$ (using subscript to reduce brackets clutter), $f:A\mapsto \det A$, $F=f \circ g = \det A_t$.
We will need the formula for the derivative (linear) map of determinant: $D_A(\det A): H\mapsto \det(A) \operatorname{tr}(A^{-1}H)$;
and the map $D_t A_t: \tau \mapsto A_t'\tau$
Now we apply the chain rule as follows,
$$
[D_tF](\tau)=[D_{A_t}\det A_t \circ D_t A_t](\tau)=\det(A_t) \operatorname{tr}(A^{-1}_tA'_t\tau)
$$
Taking the argument $\tau\in\mathbb R$ out of the trace we can write the map itself as
$$
D_t\left(\det A_t\right)=\det(A_t) \cdot \operatorname{tr}(A^{-1}_tA'_t).
$$
A: Here is yet another way to solve this. We can write for the $i$-th line $$\det (f(t)) = \sum_{j=1}^{n} (-1)^{i+j} f_{ij}(t)F_{ij}(t)$$
where $F_{ij}(t)$ is the determinant of the matrix obtained by crossing out the $i$-th line and $j$-th column. Then for the $1$-st line development we get 
$$\begin{align}\frac{d}{dt}\bigg|_{t=0} \det (f(t)) &= \frac{d}{dt}\bigg|_{t=0} \sum_{j=1}^{n} (-1)^{1+j} f_{1j}(t)F_{1j}(t)\\&=\sum_{j=1}^{n} (-1)^{1+j} f'_{1j}(0)F_{1j}(0) + f_{1j}(0)\frac{d}{dt}\bigg|_{t=0}F_{1j}(t)\\&=f'_{11}(0) + \frac{d}{dt}\bigg|_{t=0} F_{11}(t)\end{align}$$
Since $f(0) = I$. Using the same argument by inductively to $\frac{d}{dt}\bigg|_{t=0}F_{11}(t)$ should give us
$$\frac{d}{dt}\bigg|_{t=0} \det (f(t)) = f'_{11}(0) + \ldots + f'_{nn}(0)$$
which is the desired result.  
