Easier way to show different forms of Jacobi's formula for the derivative of the determinant Jacobi's formula for the derivative of the determinant of a matrix A is
$$
{\displaystyle {\frac {d}{dt}}\det A=\det A\;\mathrm {tr} \left(A^{-1}{\frac {dA}{dt}}\right)=\mathrm {tr} \left(\mathrm {adj} \ A\;{\frac {dA}{dt}}\right)}
$$
A proof of this formula is given in the wikipedia and been asked before, link, but I was wondering is there a better way to prove it, probabily making use of the Leibniz formula for determinants ?
Probable Intuitive way
$$
\frac{d}{dt}\det A(t)=
\frac{d}{dt}\begin{vmatrix}
a_{11} & a_{12} & ... & a_{1n} \\
a_{21} & a_{22} & ... & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nn} \\
\end{vmatrix}\\
=\begin{vmatrix}
\dot{a}_{11} & \dot{a}_{12} & ... & \dot{a}_{1n} \\
a_{21} & a_{22} & ... & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nn} \\
\end{vmatrix}+\begin{vmatrix}
a_{11} & a_{12} & ... & a_{1n} \\
\dot{a}_{21} & \dot{a}_{22} & ... & \dot{a}_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nn} \\
\end{vmatrix}+...+\begin{vmatrix}
a_{11} & a_{12} & ... & a_{1n} \\
a_{21} & a_{22} & ... & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
\dot{a}_{n1} & \dot{a}_{n2} & ... & \dot{a}_{nn} \\
\end{vmatrix}\\
$$
If $A(t)=I$
$$
\frac{d}{dt}\det A(t)=\begin{vmatrix}
\dot{a}_{11} & \dot{a}_{12} & ... & \dot{a}_{1n} \\
0 & 1 & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & 1 \\
\end{vmatrix}+\begin{vmatrix}
1 & 0 & ... & 0 \\
\dot{a}_{21} & \dot{a}_{22} & ... & \dot{a}_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & 1 \\
\end{vmatrix}+...+\begin{vmatrix}
1 & 0 & ... & 0 \\
0 & 1 & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
\dot{a}_{n1} & \dot{a}_{n2} & ... & \dot{a}_{nn} \\
\end{vmatrix}\\
=\dot{a}_{11}+\dot{a}_{22}+...+\dot{a}_{nn}=Tr(\dot{A}(t))\\
\implies \boxed{\frac{d}{dt}\det A(t)=Tr(\dot{A}(t))\quad \text{when }A(t)=I}
$$
Let $B$ a constant and invertible matrix such that, $B.\psi(t)=I\implies\det\Big(B.\psi(t)\Big)=\det B.\det\psi(t)$
$$
\frac{d}{dt}\det B.\psi(t)=\det B.\frac{d}{dt}\det \psi(t)=Tr\;\big(B.\frac{d}{dt}{\psi(t)}\big)=Tr\;\Big(B.\dot{\psi}(t)\Big)\\
\boxed{ \det B.\frac{d}{dt}\det \psi(t)=Tr\;\Big(B.\dot{\psi}(t)\Big)\quad\text{when }B.\psi(t)=I\implies B=\psi^{-1}(t) }\\
\det B.\frac{d}{dt}\det\psi(t)=\frac{1}{\det{\psi(t)}}\frac{d}{dt}\det\psi(t)=Tr\;\Big(\psi^{-1}(t).\dot{\psi}(t)\Big)\\
\boxed{\frac{d}{dt}\det\psi(t)=\det\psi(t).Tr\;\Big(\psi^{-1}(t).\dot{\psi}(t)\Big)}
$$
Is there atleast an intuitive way to prove that, $\det\psi(t).Tr\;\Big(\psi^{-1}(t).\dot{\psi}(t)\Big)=Tr\;\Big(adj \;\psi(t).\dot{\psi}(t)\Big)$ ?
 A: $\DeclareMathOperator{\sign}{sgn}$
$\DeclareMathOperator{\trace}{tr}$
$\DeclareMathOperator{\adjugate}{adj}$
Hi Sooraj. Your proof starts fine. Basically you are using the definition
$$
\det(A(t)) = \sum_{\sigma \in S_n} 
\sign(\sigma) \prod_{i=1}^{n} a_{i,\sigma(i)}(t),
$$
and then you are taking a derivative to obtain
$$
\begin{align}
\frac{d}{dt}\det(A(t)) &= \sum_{\sigma \in S_n} 
\sign(\sigma) \sum_{k=1}^{n}\dot{a}_{k,\sigma(k)}\prod_{i\neq k}^{n} a_{i,\sigma(i)}(t) \\
&= \sum_{k=1}^{n} \sum_{\sigma \in S_n} 
\sign(\sigma) \dot{a}_{k,\sigma(k)}\prod_{i\neq k}^{n} a_{i,\sigma(i)}(t) \\
&= \sum_{k=1}^{n} \sum_{\sigma \in S_n} 
\sign(\sigma) \prod_{i=1}^{n} b_{i,\sigma(i)}^{(k)}(t),
\end{align}
$$
where
$$
b^{(k)}_{i,j}=
\left\{
\begin{array}{rl}
\dot{a}_{i,j}, & i=k \\
{a}_{i,j}, & \text{else}.
\end{array}
\right.
$$
This proves your formula
$$
\frac{d}{dt}\det A(t)=
\frac{d}{dt}\begin{vmatrix}
a_{11} & a_{12} & ... & a_{1n} \\
a_{21} & a_{22} & ... & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nn} \\
\end{vmatrix}\\
=\begin{vmatrix}
\dot{a}_{11} & \dot{a}_{12} & ... & \dot{a}_{1n} \\
a_{21} & a_{22} & ... & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nn} \\
\end{vmatrix}+\begin{vmatrix}
a_{11} & a_{12} & ... & a_{1n} \\
\dot{a}_{21} & \dot{a}_{22} & ... & \dot{a}_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nn} \\
\end{vmatrix}+\cdots+\begin{vmatrix}
a_{11} & a_{12} & ... & a_{1n} \\
a_{21} & a_{22} & ... & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
\dot{a}_{n1} & \dot{a}_{n2} & ... & \dot{a}_{nn} \\
\end{vmatrix}.
$$
However, as soon as you say $A(t)=I$, all your matrices become
zero and at the end you are basically proving that $0=0$. Let me suggest an alternative approach which combines your idea with was already mentioned in the link that you provided by Jean Van Schaftingen.
Since the determinant is a function from the space of $n\times n$ matrices to $\mathbb{R}$, i.e., $\det:M_n(\mathbb{R})\to\mathbb{R}$, we can ask the question: how does $\det$ change as we move in the direction of another matrix $U$, when we are located at the identity matrix, i.e., at the position $X=I$? This is a directional derivative
$$
\left.\frac{\partial \det}{\partial U}\right|_{X=I}
= \lim_{t\to 0} \frac{\det(I+tU)-\det(I)}{t}
$$
Define
$$g(t)=\det(I+tU).$$
Then
$$
\left.\frac{dg}{dt}\right|_{t=0} = 
\left.\frac{\partial \det}{\partial U}\right|_{X=I}
$$
So we can use your formula to obtain
$$
\left.\frac{dg}{dt}\right|_{t=0}=\begin{vmatrix}
{u}_{11} & {u}_{12} & ... & {u}_{1n} \\
0 & 1 & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & 1 \\
\end{vmatrix}+\begin{vmatrix}
1 & 0 & ... & 0 \\
{u}_{21} & {u}_{22} & ... & {u}_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & ... & 1 \\
\end{vmatrix}+\cdots+\begin{vmatrix}
1 & 0 & ... & 0 \\
0 & 1 & ... & 0 \\
\vdots & \vdots & \ddots & \vdots \\
{u}_{n1} & {u}_{n2} & ... & {u}_{nn} \\
\end{vmatrix}\\
={u}_{11}+{u}_{22}+...+{u}_{nn}=\trace(U),
$$
and conclude that
$$
\left.\frac{dg}{dt}\right|_{t=0}  
%=\left.\frac{\partial \det}{\partial U}\right|_{X=I}
=\left.\frac{d\det(I+tU)}{dt}\right|_{t=0} = \trace(U).
$$
Now let $A(t) \in M_n(\mathbb{R})$, fix a time $s$, and assume that $A(t)$ is invertible and differentiable in a neighborhood of $t=s$. Furthermore, consider the function
$$
H(X(t)) = \det(A(s))\cdot\det(A^{-1}(s)X(t)).
$$
Note that $H(A(s)) = \det(A(s))$. Basically, what we want to do is observe how $H$ changes as we move from $X=A(s)$ to $X=A(s+t)$. Formally,
we are interested in the limit
$$
\begin{align}
\left.\frac{dH(A(t))}{dt}\right|_{t=s}
= \lim_{t\to 0} \frac{H(A(s+t))-H(A(s))}{t}
\end{align}
$$
Since $A(t)$ is differentiable,
write its Taylor expansion about $t=s$ to obtain
$$
A(t+s)=A(s)+t\dot{A}(s)+\mathcal{O}(t^2)
$$
and define the trajectory
$$
X(t) = A(s)+t\dot{A}(s) = A(t+s)+\mathcal{O}(t^2),
$$
which describes the behavior $A$ near $A(s)$.
Furthermore, note that
$$
\begin{align}
H(X(t)) &= \det(A(s))\cdot\det(A^{-1}(s)(A(s)+t\dot{A}(s)))\\
 &= \det(A(s))\cdot\det(I+tA^{-1}(s)\dot{A}(s))),\\
H(X(0)) &= \det(A(s)).
\end{align}
$$
Hence,
$$
\begin{align}
\left.\frac{d H}{dt}\right|_{t=0}
&= \lim_{t\to 0} \frac{H(X(t))-H(X(0))}{t} \\
&= \lim_{t\to 0} 
\frac{
\det(A(s))\cdot\det(I+tA^{-1}(s)\dot{A}(s)))
-
\det(A(s))
}{t} \\
&= \det(A(s)) \lim_{t\to 0} 
\frac{\det(I+tA^{-1}(s)\dot{A}(s)))
-
\det(I)
}{t} \\
&= \det(A(s))
\trace(A^{-1}(s)\dot{A}(s)).
\end{align}
$$
Lastly, using the fact that the trace is a linear operator, we can bring $\det(A(s))$ into the argument and use the identity
$$
\det(A) A^{-1} = \adjugate(A),
$$
to obtain
$$
\left.\frac{d\det(A(t))}{dt}\right|_{t=s}
=\trace(\adjugate(A)(s)\dot{A}(s)).
$$
A: Let $E_{ij}$ be the matrix with a $1$ at the $(i,j)$-th position and zeroes elsewhere. Denote the $(i,j)$-th minor of $A$ by $M_{ij}$. Jacobi's formula holds in the following special case:
$$
\frac{d\det(A(t)+hE_{ij})}{dh}
=\frac{d\left(\det(A)+h(-1)^{i+j}M_{ij}\right)}{dh}
=(-1)^{i+j}M_{ij}
=(\operatorname{adj}(A))_{ji}
=\operatorname{tr}\left(\operatorname{adj}(A)E_{ij}\right).
$$
Denote the determinant function by $F$. The general case now follows from the total derivative formula:
\begin{aligned}
&\frac{d\det(A(t))}{dt}
=\frac{dF(A(t))}{dt}=\sum_{i,j}\frac{\partial F}{\partial a_{ij}}\frac{da_{ij}}{dt}
=\sum_{i,j}\frac{d\det(A(t)+hE_{ij})}{dh}\frac{da_{ij}}{dt}\\
&=\sum_{i,j}\operatorname{tr}\left(\operatorname{adj}(A)E_{ij}\right)\frac{da_{ij}}{dt}
=\operatorname{tr}\left(\operatorname{adj}(A)\sum_{i,j}\frac{da_{ij}}{dt}E_{ij}\right)=\operatorname{tr}\left(\operatorname{adj}(A)\frac{dA}{dt}\right).
\end{aligned}
