I have the following polynomial function in $\mathbb{R}$:
$$f(x,a)=ax^3+x$$
However, $x$ also depends on $a$, so we should rather write:
$$f(x,a)=ax(a)^3+x(a)$$
Now I need derivative of $f$ and $f^{-1}$ with respect to $a$. The former is simple:
$${\partial\over \partial a}f'(x,a)=x^3+3x^2{\partial x\over\partial a}+{\partial x\over\partial a}$$
The inverse does not have simple analytical form, so I am trying to use inverse function theorem, i.e.:
$$(f^{-1})'(y)={1\over f'(f^{-1}(y))}$$
However, this works for single variable function, but the inverse is actually in two variables: $f^{-1}(y,a)$ ... I need however, just partial derivative of inverse w.r.t. $a$.
Please note that single root is assumed, therefore the inverse mapping is 1:1 and we don't need to care about multiple roots.
$f$ is actually a function of single variable ($a$), even though it is composed of another function inside (which also depends on $a$). i.e. $x(a)$. Therefore the inverse function theorem applies.
I will write the function and its derivative compactly as:
$$\begin{align}y &= ax^3+x \\ y' &= x^3+3ax^2x'+x'\end{align}$$
Since we can compute $x$ from the first equation numerically, obtaining $x'$ from second equation is straightforward:
$$x'={y'-x^3 \over 3ax^2+1}$$