The Probability Theory course I'm following starts to take derivatives of cumulative distribution functions to find derived distributions without much explanation. I want to understand this part in a bit more detail. I can see eight possible forms of such a derivative...
$\frac{d}{dx}F_X (g(x))$
$\frac{d}{dx}F_X (g(y))$
$\frac{d}{dx}F_Y (g(x))$
$\frac{d}{dx}F_Y (g(y))$
$\frac{d}{dy}F_X (g(x))$
$\frac{d}{dy}F_X (g(y))$
$\frac{d}{dy}F_Y (g(x))$
$\frac{d}{dy}F_Y (g(y))$
My guesses for the first four formulas would be...
$\frac{d}{dx}F_X (g(x)) = f_X(g(x)) * \frac{dg}{dx}$
$\frac{d}{dx}F_X (g(y)) = f_X(g(y)) * \frac{dg}{dy} * \frac{dy}{dx}$
$\frac{d}{dx}F_Y (g(x)) = f_Y(g(x)) * \frac{dg}{dx}$
$\frac{d}{dx}F_Y (g(y)) = f_Y(g(y)) * \frac{dg}{dy} * \frac{dy}{dx}$
From here, I make the assumption that which random variable is derived and which random variable is "primary" is invisible to the differentiation procedure. Thus, I proceed as though solving a CDF expression such as $\frac{d}{dx}F_X (g(y))$ is unchanged by whether $Y$ is derived from $X$, or $X$ is derived from $Y$. If true, then we can specify one (we'll say that $Y$ is derived from $X$), and the remaining four formulas would be...
$\frac{d}{dy}F_X (g(x)) = f_X(g(x)) * \frac{dg}{dx} * \frac{dx}{dy}$
$\frac{d}{dy}F_X (g(y)) = f_X(g(y)) * \frac{dg}{dy}$
$\frac{d}{dy}F_Y (g(x)) = f_X(g(x)) * \frac{dg}{dx} * \frac{dx}{dy}$
$\frac{d}{dy}F_Y (g(y)) = f_Y(g(y)) * \frac{dg}{dy}$
However, I'm not sure if the chain rule is allowed to be applied "uphill" like this.
Are these eight formulas correct? If not, what are the correct formulas for the derivatives of CDFs?