I've searched around a little more and found a paper deriving the guiding wave equation directly from the strodinger equation. Link: https://www.imperial.ac.uk/media/imperial-college/research-centres-and-groups/theoretical-physics/msc/dissertations/2009/Robert-Dabin-Dissertation.pdf
The argument starts from orthodox quantum mechanics and states:
$$P=\int_\Gamma d^3r|\Psi|^2$$
Where $\Psi$ is the universal wave function, depends on $(q,t)$ where $\ q \ $ is the collection of starting points for each particle $(\vec{\mu_1}, \vec{\mu_2},...,\vec{\mu_N})$ and follows the universal strodinger equation:
$$i\hbar\partial_t \Psi=\sum_{r=1}^N-\frac{\hbar^2}{2m_{r}}\ \partial_{\mu_r} \partial^{\mu_r}\Psi+V\Psi$$
If we differentiate $P$ with respect to time and assume the volume of the integral is constant:
$$\frac{\partial P}{\partial t}=\frac{d}{dt}\int_\Gamma d^3r|\Psi|^2=\int_\Gamma d^3r\frac{\partial|\Psi|^2}{\partial t}$$
$$|\Psi|^2=\Psi\Psi^\dagger$$
$$\frac{\partial |\Psi|^2}{\partial t}=\frac{\partial}{\partial t}(\Psi\Psi^\dagger)=\Psi^\dagger \frac{\partial \Psi}{\partial t} + \Psi \frac{\partial \Psi^\dagger}{\partial t}$$
Substituting:
$$\frac{\partial P}{\partial t}=\frac{d}{dt}\int_\Gamma d^3r|\Psi|^2=\int_\Gamma d^3r \left( \Psi^\dagger \frac{\partial \Psi}{\partial t} + \Psi \frac{\partial \Psi^\dagger}{\partial t} \right) $$
If we solve the strodinger equation for $\partial_t \Psi$ and $\partial_t \Psi^\dagger$ we can substitute this back into the integral:
$$i\hbar\partial_t \Psi=\sum_{r=1}^N-\frac{\hbar^2}{2m_{r}}\ \partial_{\mu_r} \partial^{\mu_r}\Psi+V\Psi$$
$$\partial_t \Psi=\sum_{r=1}^N-\frac{\hbar^2}{i2m_{r}\hbar}\ \partial_{\mu_r} \partial^{\mu_r}\Psi+\frac{V}{i\hbar}\Psi$$
$$\partial_t \Psi=\sum_{r=1}^N\frac{\hbar}{2m_{r}}i\ \partial_{\mu_r} \partial^{\mu_r}\Psi-i\frac{V}{\hbar}\Psi$$
We know:
$$\frac{\partial A^\dagger}{\partial F} = \left( \frac{\partial A}{\partial F} \right)^\dagger$$
So:
$$\partial_t \Psi^\dagger=\sum_{r=1}^N-\frac{\hbar}{2m_{r}}i\ \partial_{\mu_r} \partial^{\mu_r}\Psi^\dagger+i\frac{V}{\hbar}\Psi^\dagger$$
Substituting:
$$\frac{\partial P}{\partial t}=\int_\Gamma d^3r \left( \Psi^\dagger \left(\sum_{r=1}^N\frac{\hbar}{2m_{r}}i\ \partial_{\mu_r} \partial^{\mu_r}\Psi-i\frac{V}{\hbar}\Psi \right) + \Psi \left( \sum_{r=1}^N-\frac{\hbar}{2m_{r}}i\ \partial_{\mu_r} \partial^{\mu_r}\Psi^\dagger+i\frac{V}{\hbar}\Psi^\dagger \right) \right) $$
Factorizing:
$$\frac{\partial P}{\partial t}=\int_\Gamma d^3r \sum_{r=1}^N\ \left( \Psi^\dagger \left(\frac{\hbar}{2m_{r}}i \partial_{\mu_r} \partial^{\mu_r}\Psi-i\frac{V}{\hbar}\Psi \right) + \Psi \left(-\frac{\hbar}{2m_{r}}i\partial_{\mu_r} \partial^{\mu_r}\Psi^\dagger+i\frac{V}{\hbar}\Psi^\dagger \right) \right) $$
Distribute $\Psi$ and $\Psi^\dagger$ into the parenthesis:
$$\frac{\partial P}{\partial t}=\int_\Gamma d^3r \sum_{r=1}^N \left( \frac{\hbar}{2m_{r}}i \Psi^\dagger\ \partial_{\mu_r} \partial^{\mu_r}\Psi-i\frac{V}{\hbar}\Psi^{\dagger}\Psi - \frac{\hbar}{2m_{r}}i \Psi\partial_{\mu_r} \partial^{\mu_r}\Psi^\dagger+i\frac{V}{\hbar}\Psi^\dagger\Psi \right) $$
$-i\frac{V}{\hbar}\Psi^\dagger\Psi$ and $i\frac{V}{\hbar}\Psi^\dagger\Psi$ cancel and $\frac{\hbar}{2m_{r}}i$ can be factorized to give:
$$\frac{\partial P}{\partial t}=\int_\Gamma d^3r \sum_{r=1}^N \frac{\hbar}{2m_{r}}i \left( \Psi^\dagger\ \partial_{\mu_r} \partial^{\mu_r}\Psi-\Psi\partial_{\mu_r} \partial^{\mu_r}\Psi^\dagger\right) $$
We can factorize out a $\partial^{\mu_r}$:
$$\frac{\partial P}{\partial t}=\int_\Gamma d^3r \sum_{r=1}^N \frac{\hbar}{2m_{r}}i \partial^{\mu_r}\left( \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger\right) $$
$$\frac{\partial P}{\partial t}=\int_\Gamma d^3r \sum_{r=1}^N \partial^{\mu_r} \frac{\hbar}{2m_{r}}i \left( \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger\right) $$
We are going to name $-\frac{\hbar}{2m_{r}}i\left( \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger\right)$ the current or $j_{\mu_r}$ so when we substitute we get:
$$\frac{\partial P}{\partial t}=-\int_\Gamma d^3r \sum_{r=1}^N \partial^{\mu_r} j_{\mu_r} $$
Lets substitute $\frac{\partial P}{\partial t}=\int_\Gamma d^3r\frac{\partial|\Psi|^2}{\partial t}$ into the left side of the equation:
$$\int_\Gamma d^3r\frac{\partial|\Psi|^2}{\partial t}$=-\int_\Gamma d^3r \sum_{r=1}^N \partial^{\mu_r} j_{\mu_r} $$
Rearrange:
$$\int_\Gamma d^3r\frac{\partial|\Psi|^2}{\partial t} + \int_\Gamma d^3r \sum_{r=1}^N \partial^{\mu_r} j_{\mu_r}=0 $$
$$\int_\Gamma d^3r \left( \frac{\partial|\Psi|^2}{\partial t} + \sum_{r=1}^N \partial^{\mu_r} j_{\mu_r}\right)=0 $$
Since the region chosen to be integrated is completely arbitrary, the only way to make sure the integral is equal to $0$ is:
$$\frac{\partial|\Psi|^2}{\partial t} + \sum_{r=1}^N \partial^{\mu_r} j_{\mu_r}=0$$
Lets set $\rho=|\Psi|^2$ (rho is the density):
$$\frac{\partial\rho}{\partial t} + \sum_{r=1}^N \partial^{\mu_r} j_{\mu_r}=0$$
This is just the continuity equation for a incomprehensible fluid.
We know that $j_{\mu_r} = \rho v_{\mu_r}$ so the velocity is:
$$v_{\mu_r}=\frac{j_{\mu_r}}{\rho}$$
We can now substitute in the current and density:
$$v_{\mu_r}=-\frac{\hbar}{2m_{r}}i\left( \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger\right)\frac{1}{|\Psi|^2} $$
Rearranging and considering that $-i=\frac{1}{i}$
$$v_{\mu_r}=\frac{1}{2i}\left( \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger\right)\frac{\hbar}{m_{r}|\Psi|^2} $$
Suppose we have a complex number $z$ and $z = a + ib$. The imaginary part of z is $b$ so $\operatorname{Im}z = b$. If we know $z = a + ib$ then we know its hermissian conjugate $z^\dagger = a - ib$. Lets subtract these two from each-other $z - z^\dagger = 2ib$. Now divide by $2i$, $b = \frac{z - z^\dagger}{2i}$ so $\operatorname{Im} z = \frac{z - z^\dagger}{2i}$. Lets suppose $z = \Psi^\dagger\ \partial_{\mu_r} \Psi$, its hermissian conjugate is $z^\dagger = \Psi \partial_{\mu_r} \Psi^\dagger $. If we take $z$ and subtract $z^\dagger$ we get $z - z^\dagger = \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger $ divide by $2i$ and we get $\operatorname{Im}z = \frac{1}{2i} \left( \Psi^\dagger\ \partial_{\mu_r} \Psi-\Psi\partial_{\mu_r} \Psi^\dagger \right) $.
We can substitute this into v_{\mu_r} and we get:
$$v_{\mu_r}= \operatorname{Im}(z) \frac{\hbar}{m_{r}|\Psi|^2} $$
$$v_{\mu_r}= \frac{\hbar}{m_{r}|\Psi|^2} \operatorname{Im}z $$
But we know $z = \Psi^\dagger\ \partial_{\mu_r} \Psi$ so if we substitute that we get:
$$v_{\mu_r}= \frac{\hbar}{m_{r}|\Psi|^2} \operatorname{Im}\Psi^\dagger\ \partial_{\mu_r} \Psi $$
If we rearrange:
$$v_{\mu_r}= \frac{\hbar}{m_{r}} \operatorname{Im}\frac{\Psi^\dagger\ \partial_{\mu_r} \Psi}{|\Psi|^2} $$
We notice $|\Psi|^2 = \Psi\Psi^\dagger$:
$$v_{\mu_r}= \frac{\hbar}{m_{r}} \operatorname{Im}\frac{\Psi^\dagger\ \partial_{\mu_r} \Psi}{\Psi\Psi^\dagger} $$
We are now left with:
$$v_{\mu_r}= \frac{\hbar}{m_{r}} \operatorname{Im}\frac{\ \partial_{\mu_r} \Psi}{\Psi} $$
But $v_{\mu_r}=\frac{\partial Q_{\mu_r}}{\partial t}$
So finally:
$$\frac{\partial Q_{\mu_r}}{\partial t}= \frac{\hbar}{m_{r}} \operatorname{Im}\frac{\ \partial_{\mu_r} \Psi}{\Psi} $$
This rank 1 tensor is in L dimensions since $\mu = x = y = z = ... $ and has r amount of particles. If we assume the particles are in 1 dimension the equation boils down to:
$$\frac{\partial Q_{x_r}}{\partial t}= \frac{\hbar}{m_{r}} \operatorname{Im}\frac{\ \partial_{x_r} \Psi}{\Psi} $$ or down to $$\frac{\partial Q_{r}}{\partial t}= \frac{\hbar}{m_{r}} \operatorname{Im}\frac{\ \partial_{x_r} \Psi}{\Psi}$$
