There is a formula in the bilateral filter thesis
$$
h(x)=k^{-1}(x)\int_{-\infty}^\infty\int_{-\infty}^\infty f(ξ)c(ξ,x)s(f(ξ),f(x))dξ\tag{1}
$$
$$
k(x)=\int_{-\infty}^\infty\int_{-\infty}^\infty c(ξ,x)s(f(ξ),f(x))dξ\tag{2}
$$
Is there any way to express $k^{-1}(x)$?
BTW, I find a article, but something I still don't understand.
The article point out that $$y'=\frac{1}{g(y)}\tag{3}$$.
While $y$ is $f^{-1}(x)$, the (3) becomes $$(f^{-1})'(x)=\frac{1}{g(f^{-1}(x))}\tag{4}$$.
But I want to know is $f^{-1}(x)$ instead of $(f^{-1})'(x)$, isn't it?