Preliminaries
Suppose we have a random variable $X$ with density $f$ and a suitably smooth function $g: \mathbb{R} \mapsto \mathbb{R}$. The random variable $Y = g(X)$ also has a density function $h$.
Examples
I've been imagining the graphs of functions, and the resulting densities $h$, when $f$ is uniform for various choices of $g$. I only use the uniform distribution because I find it easy to interpret and illustrate the basic pattern, but in general I have noticed that both the input density $f$ and the choice of function $g$ have an influence on what $h$ will look like. I also sample over a symmetric interval about zero, as this choice also has an effect. Unlike many actual regression models, I have not included any error terms in $g$ because this doesn't seem to affect the result beyond introducing some instability into the resulting distribution shape. At the core, what I want to show below is that $\frac{dg}{dx}$ seems to be related to $h$.
Linear
For example, if $g$ is a line then $h$ will be uniform. Below is a plot of a such a sample where $g(x) = 3x+1$. Note that $\frac{dg}{dx} = 3$, a constant.
Quadratic
Now let us consider $Y = X^2$. Notice that $\frac{dg}{dx} = 2x$, and that the highest density appears to be around $x=0$ where this derivative is at its minimum.
Sigmoidal
Now consider this non-convex function, the standard sigmoid function, $Y = \frac{1}{1 + e^{-X}}$. Here the derivative $\frac{dg}{dx} = \frac{e^X}{(1 + e^{X})^2}$. The density $h$ seems to be at its largest when the derivative is at its lowest in absolute value, and vice versa.
Hypothesis
Some kind of antimonotonic relationship exists between $\frac{dg}{dx}$ and $h$. Maybe even something similar to $h = \left| \frac{dg}{dx} \right|^{-1}$, but that is unfettered speculation.
Question
I would be shocked if this has not been noticed by others. Rather I suspect it is simply a gap in my knowledge of mathematical statistics. So I humbly ask for some instruction on the relationship between $\frac{dg}{dx}$ with $h$.