How to calculate the derivative of the angular momentum vector? $$ d\vec L = d(\hat I \vec \omega)$$ or $$ d\vec L = d(\vec r \times m \vec v)$$
I would like to note that the change of the moment of inertia $d \hat I$ must result in change vector $\vec L$ and there must be a second element $d \vec \omega$ that compensates for this change that the angular momentum vector remains unchanged.
When I asked this question, I didn't know the full answer but now I know the full answer in several ways.
These calculations are performed only for reference systems where the center of the reference system is aligned with the center of mass of the system
The angular momentum vector $\vec L$ is defined as the vector product of the position vector $\vec r$ over the derivative of the position vector $\frac {d \vec r} {dt}$ times the mass $ m $ , i.e. the momentum vector $m \vec v$
$$\vec r \times m \vec v = \vec L \tag 1$$
The position vector can be changed in two directions:
The vector product of two parallel vectors is zero, which gives us
$$\vec r \times m \vec v_{\bot} + \vec r \times m \vec v_{{}={}} = \vec L = \vec r \times m \vec v_{\bot} \tag {1.a}$$
To be more precise, the angular momentum vector is the vector product of the position vector on the perpendicular component of the velocity vector times the mass.
The parallel component of the velocity vector $\vec v_{{}={}}$ is not part of the angular momentum vector.
The perpendicular component of the velocity vector can be calculated as the vector product of the angular velocity vector times the position vector
$$ \vec \omega \times \vec r = \vec v_{\bot} \tag 2$$
We can develop a pattern (1.a)
$$ \vec L = \vec r \times m \vec v_{\bot} = m(\vec r \times (\vec \omega \times \vec r)) \tag 3$$
Derivative of the formula (3)
$$ \frac {d \vec L} {dt} = m(\frac {d \vec r} {dt} \times (\vec \omega \times \vec r) + \vec r \times (\frac {d \vec \omega} {dt} \times \vec r) + \vec r \times (\vec \omega \times \frac {d \vec r} {dt})) \tag {3.a}$$
simpler version
$$ \frac {d \vec L} {dt} = m( \vec v \times (\vec \omega \times \vec r) + \vec r \times (\vec \omega \times \vec v)+ \vec r \times (\vec \varepsilon \times \vec r) ) \tag {3.b}$$
Let us also derive the derivative of the formula (1)
$$ \frac { d (\vec r \times m \vec v)} {dt} = \vec r \times m \vec a \tag 4$$
The angular momentum vector can also be written as the product of the moment of inertia tensor on the angular velocity vector
$$ \vec L = \hat I \vec \omega \tag 5$$
and its derivative
$$ \frac {d \vec L} {dt} = \frac {d \hat I} {dt} \vec \omega + \hat I \vec \varepsilon \tag {5.a}$$
I have an additional formula which is used to derive Euler's formulas for the rotation of a rigid body
$$ \hat I \frac {d \vec \omega} {dt} + \vec \omega \times \hat I \vec \omega = \vec M \tag 6$$
This formula is based on the formula for transforming the vector derivative to a non-inertial frame of reference
$$ \frac {d \vec A} {dt}= \frac {d \vec A `} {dt} + \vec \omega_o \times \vec A \tag {6.a}$$
When transforming the derivative of a vector to a non-inertial frame of reference (6.a), we use the angular velocity of the non-inertial frame of reference $ \vec \omega_o $. In deriving Euler's formulas (6), however, we use the inertial angular velocity $ \vec \omega$ of a rigid body.
Details of the calculation of the proof of equality (3.a)=(4) for the point can be found on my blog in Polish
https://www.salon24.pl/u/przestrz/1192364,mechanika-obrotu-punktu-pochodna-pedu
Since this proof is very extensive, I will only present the results here.
The acceleration vector resulting from the equation (2)
$$ \frac {d \vec v_{\bot}} {dt} = \vec \omega \times \vec v + \vec \varepsilon \times \vec r = \vec a_{\omega \varepsilon } \tag 7$$
Because there are such situations when the angular acceleration vector is zero, therefore
$$ \exists \vec \varepsilon (0,0,0) ; [ \vec \varepsilon \times \vec r = (0,0,0) \to \vec \omega \times \vec v = \vec a_{\omega } ] \tag {7.a}$$
this is why
$$\vec \omega \times \vec v = \vec a_{\omega } \tag {7.b}$$
so the angular acceleration vector will give the following acceleration vector
$$ \vec \varepsilon \times \vec r = \vec a_{\varepsilon \omega } - \vec a_{\omega } = \vec a_{\varepsilon } \tag {7.c}$$
The angular velocity vector of a point
$$ \frac {\vec r_n \times \vec v_n} { \vert \vec r_n \vert ^2 } = \vec \omega_n \tag {8}$$
the derivative of the vector of the angular velocity of a point
$$ \frac {(\vec r_n \times \vec a_n) \vert \vec r_n \vert ^2 - (\vec r_n \times \vec v_n)(2 r_x v_x + 2 r_y v_y + 2 r_z v_z) } { \vert \vec r_n \vert ^4 } = \frac {d\vec \omega_n} {dt} \tag {8.a} $$
I split the angular acceleration vector into two components
$$ \vec \varepsilon _1 = \frac {(\vec r_n \times \vec a_n) } { \vert \vec r_n \vert ^2 } \tag {8.b.1}$$
$$\vec \varepsilon _2 = - 2 \frac { (\vec r_n \times \vec v_n)(\vec r_n \circ \vec v_n) } { \vert \vec r_n \vert ^4} \tag {8.b.2} $$
After substituting the angular velocity vector of the point (8) into the formula (3.b) and calculating the value, I obtained the following equations
$$ \vec v \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times \vec r) = \vec r \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times \vec v) = 2 \frac { (\vec r_n \times \vec v_n)(\vec r_n \circ \vec v_n) } { \vert \vec r_n \vert ^4} \tag {9.a}$$
and I was able to reset part of the angular velocity vector to zero
$$ \vec v \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times \vec r) + \vec r \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times \vec v) + \vec r \times (\vec \varepsilon _2 \times \vec r) =0 \tag {9.b}$$
and
$$ \vec r \times (\vec \varepsilon _1 \times \vec r) = \vec r \times \vec a \tag {9.b}$$
by which I obtained equation between the equations (3.b)=(4)
$$ m( \vec v \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times \vec r) + \vec r \times (( \frac {\vec r \times \vec v} { \vert \vec r \vert ^2 } ) \times \vec v) + \vec r \times (\vec \varepsilon _2 \times \vec r) + \vec r \times (\vec \varepsilon _1 \times \vec r) )=\vec r \times m \vec a \tag {9.e}$$
Now let's count (5.a).
The first moment of inertia tensor
$$ \left [ \begin{matrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \\ \end{matrix} \right ] \tag {10} $$
elements of the moment of inertia tensor
$$ I_{xx}= \sum _n {m_n (r_{ny} ^2 + r_{nz} ^2)} ;\qquad I_{yy} = \sum _n {m_n (r_{nz} ^2 + r_{nx} ^2)}; \qquad I_{zz} = \sum _n {m_n (r_{nx} ^2 + r_{ny} ^2)}; \tag {10.a} $$
$$ I_{xy}= I_{yx}= - \sum _n {m_n r_{nx} r_{ny}} ;\qquad I_{yz}= I_{zy}= - \sum _n {m_n r_{ny} r_{nz}} ; \qquad I_{zx}= I_{xz}= - \sum _n {m_n r_{nz} r_{nx}} ; \tag {10.a} $$
derivative of the moment of inertia tensor
$$ \left [ \begin{matrix} \frac {d I_{xx}} {dt} & \frac {d I_{xy}} {dt} & \frac {d I_{xz}} {dt} \\ \frac {d I_{yx}} {dt} & \frac {d I_{yy}} {dt} & \frac {d I_{yz}} {dt} \\ \frac {d I_{zx}} {dt} & \frac {d I_{zy}} {dt} & \frac {d I_{zz}} {dt} \\ \end{matrix} \right ] \tag {10.b} $$
derivative of the elements of the moment of inertia tensor
$$ \frac {d I_{xx}} {dt}= \sum _n {2 m_n (v_{ny} r_{ny} + v_{nz} r_{nz} )} ;\quad \frac {d I_{yy}} {dt}= \sum _n {2 m_n (v_{nz} r_{nz} + v_{nx} r_{nx})}; \quad \frac {d I_{zz}} {dt}= \sum _n {2 m_n (v_{nx} r_{nx} + v_{ny} r_{ny} )}; \tag {10.c} $$
$$ \frac {d I_{xy}} {dt}= \frac {d I_{yx}} {dt}= - \sum _n {m_n (v_{nx} r_{ny} + r_{nx} v_{ny})} ;\qquad \frac {d I_{yz}} {dt}= \frac {d I_{zy}} {dt}= - \sum _n {m_n (v_{ny} r_{nz} + r_{ny} v_{nz})} ; \qquad \frac {d I_{zx}} {dt}= \frac {d I_{xz}} {dt}= - \sum _n {m_n (v_{nz} r_{nx} + r_{nz} v_{nx})} ; \tag {10.c} $$
We count the first element (5.a) $ \frac {d \hat I} {dt} \vec \omega $
$$ \frac {d \hat I} {dt} \vec \omega =(\omega_x \frac {d I_{xx}} {dt}+ \omega_y \frac {d I_{xy}} {dt}+ \omega_z \frac {d I_{xz}} {dt}, \omega_x \frac {d I_{yx}} {dt}+ \omega_y \frac {d I_{yy}} {dt}+ \omega_z \frac {d I_{yz}} {dt}, \omega_x \frac {d I_{zx}} {dt}+ \omega_y \frac {d I_{zy}} {dt}+ \omega_z \frac {d I_{zz}} {dt}) \tag {11}$$
where
$$ \omega_x \frac {d I_{xx}} {dt}+ \omega_y \frac {d I_{xy}} {dt}+ \omega_z \frac {d I_{xz}} {dt} = \sum _n {2 m_n \omega_x (v_{ny} r_{ny} + v_{nz} r_{nz} )}- \sum _n {m_n \omega_y (v_{nx} r_{ny} + r_{nx} v_{ny})} - \sum _n {m_n \omega_z (v_{nz} r_{nx} + r_{nz} v_{nx})} = \sum _n { m_n( 2 \omega_x v_{ny} r_{ny} + 2 \omega_x v_{nz} r_{nz} - \omega_y v_{nx} r_{ny} - \omega_y r_{nx} v_{ny} - \omega_z v_{nz} r_{nx} - \omega_z r_{nz} v_{nx})} = \sum _n { m_n( r_{ny} (\omega_x v_{ny} - \omega_y v_{nx}) + r_{nz} (\omega_x v_{nz} - \omega_z v_{nx}) + v_{ny} (\omega_x r_{ny} - \omega_y r_{nx}) + v_{nz} (\omega_x r_{nz} - \omega_z r_{nx} ))} \tag {11.a} $$
$$ \omega_x \frac {d I_{yx}} {dt}+ \omega_y \frac {d I_{yy}} {dt}+ \omega_z \frac {d I_{yz}} {dt} = \sum _n { m_n (- \omega_x v_{nx} r_{ny} - \omega_x v_{ny} r_{nx} + 2 \omega_y v_{nz} r_{nz} + 2 \omega_y r_{nx} v_{nx} - \omega_z v_{ny} r_{nz} + r_{nz} v_{ny})} = \sum _n { m_n( r_{nz} (\omega_y v_{nz} - \omega_z v_{ny}) + r_{nx} (\omega_y v_{nx} - \omega_x v_{ny}) + v_{nz} (\omega_y r_{nz} - \omega_z r_{ny}) + v_{nx} (\omega_y r_{nx} - \omega_x r_{ny} ))} \tag {11.b} $$
$$ \omega_x \frac {d I_{zx}} {dt}+ \omega_y \frac {d I_{zy}} {dt}+ \omega_z \frac {d I_{zz}} {dt} = \sum _n { m_n (- \omega_x v_{nx} r_{nz} - \omega_x v_{nz} r_{nx} - \omega_y v_{ny} r_{nz} - \omega_y r_{nz} v_{ny} + 2 \omega_z v_{nx} r_{nx} + 2 \omega_z r_{ny} v_{ny})} = \sum _n { m_n( r_{nx} (\omega_z v_{nx} - \omega_x v_{nz}) + r_{ny} (\omega_z v_{ny} - \omega_y v_{nz}) + v_{nx} (\omega_z r_{nx} - \omega_x r_{nz}) + v_{ny} (\omega_z r_{ny} - \omega_y r_{n} ))} \tag {11.c} $$
let's write out the details of the formula (7.b)
$$\vec \omega \times \vec v = (\omega_y v_z - \omega_z v_y , \omega_z v_x - \omega_x v_z , \omega_x v_y - \omega_y v_x ) = \vec a_{\omega } \tag {12.a}$$
and the details of the formula (2)
$$\vec \omega \times \vec r = (\omega_y r_z - \omega_z r_y , \omega_z r_x - \omega_x r_z , \omega_x r_y - \omega_y r_x ) = \vec v_{\bot} \tag {12.b}$$
Now we can simplify the formulas (11)
$$ \frac {d \hat I} {dt} \vec \omega = \sum _n { m_n (\vec v_n \times (\vec \omega \times \vec r_n) + \vec r_n \times (\vec \omega \times \vec v_n))} = \sum _n { m_n (\vec v_n \times \vec v_{n \bot} + \vec r_n \times \vec a_{n \omega })}\tag {13}$$
So, we have a fragment of the pattern (3.b) $ m (\vec v \times (\vec \omega \times \vec r) + \vec r \times (\vec \omega \times \vec v))$. And naw let's check out the rest of this pattern $m (\vec r \times (\vec \varepsilon \times \vec r)) $
$$\sum _n { m_n (\vec r_n \times (\vec \varepsilon \times \vec r_n))} = \sum _n {m_n (\varepsilon_x (r_{ny}^2 + r_{nz}^2) - \varepsilon_y r_{nx} r_{ny} - \varepsilon_z r_{nx} r_{nz},- \varepsilon_x r_{ny} r_{nx} + \varepsilon_y (r_{nz}^2 + r_{nx}^2) - \varepsilon_z r_{ny} r_{nz}, - \varepsilon_x r_{nz} r_{nx} - \varepsilon_y r_{nz} r_{ny} + \varepsilon_x (r_{nx}^2 + r_{ny}^2))} \tag {14} $$
that is
$$\sum _n { m_n (\vec r_n \times (\vec \varepsilon \times \vec r_n))} = \hat I \vec \varepsilon \tag {14.a} $$
On the basis of equality (13) and (14.a) we got another equality (3.b)=(5.a)
$$\frac {d \hat I} {dt} \vec \omega + \hat I \vec \varepsilon= \sum _n { m_n( \vec v_n \times (\vec \omega \times \vec r_n) + \vec r_n \times (\vec \omega \times \vec v_n)+ \vec r_n \times (\vec \varepsilon \times \vec r_n) )} \tag {15}$$
The properties of a rigid body show that the distance of a point from the center of mass is constant over time that is, for a rotating rigid body, a point dont has velocity vector along its position vector
$$ \vec v_{{}={}}=0 \land \vec v = \vec v_{\bot} \tag {16}$$
that is, always for a rigid body
$$ \vec v_n \times (\vec \omega \times \vec r_n)=0 \tag{16.a}$$
and the formula for the velocity vector is ours (2). So the formula (3.b) for a rigid body has the following form
$$ \frac {d \vec L} {dt} = \sum _n { m_n( \vec r_n \times (\vec \omega \times (\vec \omega \times \vec r_n))+ \vec r_n \times (\vec \varepsilon \times \vec r_n) )} \tag {16.b}$$
so let's count $m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times \vec r_n)))$
$$ \sum _n {m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times \vec r_n))) =(M_{nx},M_{ny},M_{nz})} \tag {16.c}$$
$$M_{nx} = \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_z r_{ny} - \omega_y r_{nz}))}$$
$$M_{ny} = \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_x r_{nz} - \omega_z r_{nx}))}$$
$$M_{nz} = \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_y r_{nx} - \omega_x r_{ny}))}$$
after simplification
$$ \sum _n {m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times \vec r_n))) }=- \sum _n {m_n (\vec \omega \circ \vec r) (\vec \omega \times \vec r)} \tag {16.d}$$
we check the missing element of the formula (6) $ \vec \omega \times \hat I \vec \omega$
$$ \vec \omega \times \hat I \vec \omega = \vec \omega \times (\omega_x I_{xx} + \omega_y I_{xy} + \omega_z I_{xz}, \omega_x I_{yx} + \omega_y I_{yy}+ \omega_z I_{yz}, \omega_x I_{zx} + \omega_y I_{zy} + \omega_z I_{zz}) = (\sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_z r_{ny} - \omega_y r_{nz}))},\sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_x r_{nz} - \omega_z r_{nx}))}, \sum _n { m_n ((r_{nx} \omega_x + r_{ny} \omega_y + r_{nz} \omega_z)(\omega_y r_{nx} - \omega_x r_{ny}))}) \tag {16.e}$$
that is
$$\vec \omega \times \hat I \vec \omega =- \sum _n {m_n (\vec \omega \circ \vec r) (\vec \omega \times \vec r)}= \sum _n {m_n(\vec r_n \times (\vec \omega \times (\vec \omega \times \vec r_n))) }\tag {16.f}$$
So for a rigid body with assumptions (16) and (16.a) we obtain equality (5.a)=(6)
$$ \frac {d \hat I} {dt} \vec \omega + \hat I \vec \varepsilon = \hat I \frac {d \vec \omega} {dt} + \vec \omega \times \hat I \vec \omega \tag {16.g}$$
In this answer I presented four different forms of the angular momentum vector derivative (3.b), (4), (5.a) and (6) which, as I showed, are contable with each other.