How to derive the relation between Euler angles and angular velocity and get the same form as mentioned in the bellow figure

How to derive the relation between euler angles and angular velocity begin{align*} &text{The equations to calculate the angular velocity $vec{omega}$ for a given transformation matrix $S$ are: } \ &left[_B^Idot{S}right]=left[tilde{vec{omega}}_Iright]left[_B^I Sright],quad Rightarrow left[tilde{vec{omega}}_Iright]=left[_B^Idot{S}right]left[_I^B Sright] &text{or} &left[_B^Idot{S}right]=left[_B^I Sright]left[tilde{vec{omega}}_Bright],quad Rightarrow left[tilde{vec{omega}}_Bright]=left[_I^B Sright]left[_B^Idot{S}right] &text{with} &left[_B^I Sright],left[_I^B S right]= begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{bmatrix}\ &text{$left[_B^I{S}right]$ Transformation matrix between B- System and I-System } &text{$left[vec{omega}right]_B$ Vector components B-System} &text{$left[vec{omega}right]_I$ Vector components I-System and} &tilde{vec{omega}}= begin{bmatrix} 0 & -omega_z &omega_y omega_z & 0 & -omega_x -omega_y & omega_x & 0 end{bmatrix}\ &textbf{Example: Transformation matrix Euler angle} &left[_B^I{S}right]=S_z(psi),S_y(vartheta),S_z(varphi) &S_z(psi)=left[ begin {array}{ccc} cos left( psi right) &-sin left( psi right) &0 sin left( psi right) &cos left( psi right) &0 0&0&1end {array} right] &S_y(vartheta)=left[ begin {array}{ccc} cos left( vartheta right) &0&sin left( vartheta right) 0&1&0 -sin left( vartheta right) &0&cos left( vartheta right) end {array} right] &S_z(varphi)=left[ begin {array}{ccc} cos left( varphi right) &-sin left( varphi right) &0 sin left( varphi right) &cos left( varphi right) &0 0&0&1end {array} right] &Rightarrow &vec{omega}_B= left[ begin {array}{ccc} 0&sin left( varphi right) &-cos left( varphi right) sin left( vartheta right) 0&cos left( varphi right) &sin left( varphi right) sin left( vartheta right) 1 &0&cos left( vartheta right) end {array} right] begin{bmatrix} dot{varphi} dot{vartheta} dot{psi} end{bmatrix} &begin{bmatrix} dot{varphi} dot{vartheta} dot{psi} end{bmatrix}= left[ begin {array}{ccc} {frac {cos left( varphi right) cos left( vartheta right) }{sin left( vartheta right) }}&-{ frac {sin left( varphi right) cos left( vartheta right) }{ sin left( vartheta right) }}&1 sin left( varphi right) &cos left( varphi right) &0 -{frac {cos left( varphi right) }{sin left( vartheta right) }}&{frac {sin left( varphi right) }{sin left( vartheta right) }}&0end {array} right] begin{bmatrix} omega_x omega_y omega_z end{bmatrix}_B &vec{omega}_I= left[ begin {array}{ccc} cos left( psi right) sin left( vartheta right) &-sin left( psi right) &0 sin left( psi right) sin left( vartheta right) &cos left( psi right) &0 cos left( vartheta right) &0& 1end {array} right] begin{bmatrix} dot{varphi} dot{vartheta} dot{psi} end{bmatrix} &begin{bmatrix} dot{varphi} dot{vartheta} dot{psi} end{bmatrix}= left[ begin {array}{ccc} {frac {cos left( psi right) }{sin left( vartheta right) }}&{frac {sin left( psi right) }{sin left( vartheta right) }}&0 -sin left( psi right) &cos left( psi right) &0 -{frac { cos left( vartheta right) cos left( psi right) }{sin left( vartheta right) }}&-{frac {cos left( vartheta right) sin left( psi right) }{sin left( vartheta right) }}&1end {array} right] begin{bmatrix} omega_x omega_y omega_z end{bmatrix}_I end{align*}

Here is how to evaluate the rotational kinematics of a rigid body from the Euler angles

$$boldsymbol{omega} = hat{imath} dot{Phi} + mathrm{R}_X ( hat{jmath} dot{Theta} + mathrm{R}_Y hat{k} dot{Psi}) tag{1} $$

Here is how to derive the above:

Consider the orientation to be defined as sequence of thre elementary rotations $$ mathrm{R} = mathrm{R}_X mathrm{R}_Y mathrm{R}_Z tag{2}$$

where $mathrm{R}_X = mathrm{rot}(hat{imath},,Phi)$, $mathrm{R}_Y = mathrm{rot}(hat{jmath},,Theta)$ and $mathrm{R}_Z = mathrm{rot}(hat{k},,Psi)$.

Now the derivative on a rotating frame dictates that $$ begin{aligned} dot{mathrm{R}}_X & = (hat{imath} dot{Phi}) times mathrm{R}_X dot{mathrm{R}}_Y & = (hat{jmath} dot{Theta}) times mathrm{R}_Y dot{mathrm{R}}_Z & = (hat{k} dot{Psi}) times mathrm{R}_Z end{aligned} $$

and

$$dot{mathrm{R}} = boldsymbol{omega} times mathrm{R} tag{3} $$ which is used to derive $boldsymbol{omega}$, the rotational velocity of the rigid body.

Starting from the product rule on the left-hand side of (3)

$$ dot{mathrm{R}} = dot{mathrm{R}}_Xmathrm{R}_Y mathrm{R}_Z + mathrm{R}_Xdot{mathrm{R}}_Y mathrm{R}_Z + mathrm{R}_X mathrm{R}_Ydot{mathrm{R}}_Z$$

and substitute the derivatives from rotating frames to equate to the right-hand side of (3)

$$ boldsymbol{omega} times mathrm{R} = left((hat{imath} dot{Phi}) times mathrm{R}_X right) (mathrm{R}_Y mathrm{R}_Z) + mathrm{R}_X left( (hat{jmath} dot{Theta}) times mathrm{R}_Y right) mathrm{R}_Z + (mathrm{R}_X mathrm{R}_Y) left( (hat{k} dot{Psi}) times mathrm{R}_Z right) $$

now start grouping and distribute the rotations. Note that $mathrm{R} (a times b) = (mathrm{R} a) times (mathrm{R} b)$ is used below.

$$begin{aligned} boldsymbol{omega} times mathrm{R} & = (hat{imath} dot{Phi}) times mathrm{R} + (mathrm{R}_X hat{jmath} dot{Theta}) times mathrm{R} + (mathrm{R}_X mathrm{R}_Y hat{k} dot{Psi}) times mathrm{R} & = left( hat{imath} dot{Phi}+mathrm{R}_X hat{jmath} dot{Theta}+mathrm{R}_X mathrm{R}_Y hat{k} dot{Psi} right) times mathrm{R} end{aligned} $$

or

$$ boxed{ boldsymbol{omega} = hat{imath} dot{Phi}+mathrm{R}_X hat{jmath} dot{Theta}+mathrm{R}_X mathrm{R}_Y hat{k} dot{Psi} } $$

Using linear algebra the above is

$$boldsymbol{omega} = pmatrix{1 0 0} dot{Phi}+pmatrix{0 cosPhi sinPhi } dot{Theta}+ pmatrix{ sinTheta -sinPhicosTheta cosPhi cosTheta } dot{Psi} $$