Find $\frac{d \rho}{d x}$ for $\rho = \rho(t,x(t),p(t))$ I got a question relating to this thread difference between implicit, explicit, and total time dependence
Considering the reply in the top by Kostya, I konw what is the difference between $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho}{d t}$. What I want to know is what is $\frac{d \rho}{d x}$ for a function $\rho = \rho(t,x(t),p(t))$?
In my opinion, we have $$\frac{{d\rho }}{{dx}} = \frac{{\partial \rho }}{{\partial t}}\frac{{dt}}{{dx}} + \frac{{\partial \rho }}{{\partial x}}$$, if $x$ and $p$ are independent variables.
But if that is the case, I get another confusion about integration by parts in calculating an integration in calculate time evolution of ensemble average
so i guess I got some misunderstanding
 A: It seems that $x$ and $p$ are functions of the unique "primary" independent variable $t$. Furthermore $\rho$ is a function with three "formal" entries, numbered $1$ to $3$. For the function $$\hat\rho(t):=\rho\bigl(t,x(t),p(t)\bigr)$$ (again denoted by $\rho$ in your source) we have by the chain rule:
$$\hat\rho'(t)=\rho_{.1}\bigl(t,x(t),p(t)\bigr)\cdot 1+\rho_{.2}\cdot x'(t)+\rho_{.3}\bigl(t,x(t),p(t)\bigr)p'(t)\ .\tag{1}$$
Now it could be that the function $t\mapsto x(t)$ can be inverted, so that $t$ appears as a function of $x$ in some interval $\>]a,b[\>$:
$$t=\psi(x)\qquad(a< x< b)\ .$$
In such a case we can replace $\hat\rho$ by the new function
$$x\mapsto\tilde\rho(x):=\hat\rho\bigl(\psi(x)\bigr)\qquad(a< x< b)\ .$$
This function $\tilde\rho$ still represents the original "physical quantity" $\rho$, but as a function of $x$. The derivative of $\tilde\rho$ computes as follows:
$$\tilde\rho'(x)\biggr|_{x=\psi(t)}=\hat\rho'\bigl(\psi(x)\bigr)\cdot\psi'(x)={\rho_{.1}\bigl(t,x(t),p(t)\bigr)\cdot 1+\rho_{.2}\cdot x'(t)+\rho_{.3}\bigl(t,x(t),p(t)\bigr)p'(t)\over x'(t)}\ .$$
Here we have used $(1)$ and the formula for the derivative of the inverse function, in order to express everything in terms of the data present at the beginning.
