Differential in Maxwell velocity distrubution derivation I'm trying to understand how Maxwell velocity distribution is derived.
I'm using this PDF as a base. It's in Russian, the closest English analogue is Chapter 2 of this PDF.
I'm stuck on this derivation and equality:

(Russian text is: Variables Vx, Vy, Vz are independent. Let's derive both equations with respect to Vx... Thus...)
Function $f$ accepts a velocity vector, $v$ is magnitude of the velocity.
If I need to show the equality from the last line, I need to somehow get differential of magnitude from first equation's derivative. But how?
P.S: I do not understand why we can derive $f(\vec{v})$ (vector function) by means of $v$ (magnitude) in the first line. And where we get $\partial v$ from in the first equation.
 A: I suspect there is a typo at the russian reference since the notation $f'(\vec{v})$ is typically used to represent the Jacobian matrix of a multivariable function $f$, which in this case have dimension 3x1 while $\frac{\phi'(v_x)}{\phi(v_x)}$ is a scalar, so the dimensions are mismatched. Very likely the russian author wanted to write $f'(v)$ that would be equivalent to $\Omega(\nu)$ of the english reference, which I will use from now on.
The confusion seems to stem of fact that the symbol $\Omega$ (as well as the russian $f$) is overloaded with two meanings. One as a function of spatial coordinates
$$\Omega(v_x, v_y, v_z) = \Omega(\vec{v}) \tag{1}$$
and the second one as a function of the speed $\nu = \sqrt{v_x^2 + v_y^2 + v_z^2}$
$$\Omega(\nu) = \Omega\left(\sqrt{v_x^2 + v_y^2 + v_z^2}\right) \tag{2}$$
alternated when conveniently.
This is arguably not good math notation since it leads to unnecessary ambiguity. It becomes clearer what is going if we write $\Omega$ as a function of speed (2) as lowercase omega instead, namely $\omega$. For that consider
$$\Omega :\mathbb{R}^3 \to \mathbb{R} : (v_x, v_y, v_z) \mapsto f(v_x)f(v_y)f(v_z)$$
$$\omega :\mathbb{R}_{\geq 0} \to \mathbb{R} : (v_x, v_y, v_z) \mapsto \nu(v_x, v_y, v_z)$$
$$\omega \circ \nu : \mathbb{R}^3 \to \mathbb{R} : (v_x, v_y, v_z) \mapsto \omega(\nu(v_x, v_y, v_z))$$
Then simply name $\omega \circ \nu$ as $\Omega$. So
$$\Omega(v_x, v_y, v_z) = \omega(\nu(v_x, v_y, v_z))$$
Note that $\nu$ is acting like a one-way change of coordinates of $\mathbb{R}^3$ (spatial) to $\mathbb{R}$ (radial).
Now looking to the notation
$$\frac{d \ln \Omega(\nu)}{d \nu}$$
with the proposed notation we can infer that the author actually means something like
$$\frac{d \left( x \mapsto \ln (\omega(x)) \right)}{dx}$$
in Leibniz' notation, or
$$\left(\ln \circ\ \omega\right)'$$
in Lagrange's notation. Again overloading symbols by also using $\nu$ to mean the independent variable of the function $\omega$. That is, it means the derivative with respect to $x$ of the funtion
$$\ln \circ\ \omega : \mathbb{R}_{> 0} \to \mathbb{R} : x \mapsto \ln (\omega(x))$$
While by writing
$$\frac{\partial \ln \Omega(\nu)}{\partial v_x}$$
it means the partial derivative with respect to $v_x$ of the function
$$\ln \circ\ \omega \circ \nu : \mathbb{R}^3 \to \mathbb{R} : (v_x, v_y, v_z) \mapsto \ln(\omega(\nu(v_x, v_y, v_z)))$$
Thus, for every point $(a, b, c) \in \mathbb{R}^3$, by the chain rule it follows that
$$\frac{\partial \ln \Omega(\nu)}{\partial v_x} \biggr\rvert_{(a, b, c)} = \frac{d \left( \ln \circ\ \omega \right)}{dx} \biggr\rvert_{\nu(a, b, c)} \frac{\partial \nu}{ \partial v_x} \biggr\rvert_{(a, b, c)}$$
in Leibniz' notation, and
$$\frac{\partial \ln \Omega(\nu)}{\partial v_x} (a, b, c) = \left( \ln \circ\ \omega \right)' (\nu(a, b, c)) \frac{\partial \nu}{ \partial v_x}(a, b, c)$$
in Lagrange's notation.
