Derivative of the inverse of a parametric function with respect to the parameter Given a function
$y = f(x)$
its inverse with respect to the argument x is
$x = F(y) = f^{-1}(y)$
Now, suppose that function has a parameter p
$y = f(x;p)$
whose inverse is
$x = F(y;p) = f^{-1}(y;p)$
How do I compute the parametric derivative of the inverse function
$\partial{F}(y;p)/\partial{p}$
using the easily computable parametric derivative of the original function
$\partial{f}(x;p)/\partial{p}$
where p is actually a vector of parameters. This does not fit the mold of the inverse function theorem as usually presented. It seems too simple to just reciprocate each of the partial derivatives. Could it be something like
$\frac{\partial{F(y;p)}}{\partial{p}}=
-\frac{\partial{f(x;p)}}{\partial{p}} /
\frac{\partial{f(x;p)}}{\partial{x}}
$
?
@Narasimham suggested that I start with a simple problem, then generalize. A simplified version of the problem is
$y = x + k x^3$
with parametric derivative
$d y / d k = x^3$
This is a cubic equation, so luckily there is an analytic inverse -- actually three three of them, but we are only interested in the positive real one.
$x = \frac{2 \sqrt[3]{3} k-\sqrt[3]{2} \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y\right)^{2/3}}{6^{2/3} k
   \sqrt[3]{\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y}}$
Its parametric derivative is 
$dx / dk = \frac{-36\ 3^{5/6} k^3 y^2+9 \sqrt[3]{2} \sqrt{3} k^2 y^2 \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y\right)^{2/3}+2
   \sqrt[3]{2} \sqrt{3} k \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y\right)^{2/3}+12 \sqrt[3]{3} k y \sqrt{k^3 \left(27 k
   y^2+4\right)}-3 \sqrt[3]{2} y \sqrt{k^3 \left(27 k y^2+4\right)} \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2
   y\right)^{2/3}-4\ 3^{5/6} k^2}{6^{2/3} \sqrt{k^3 \left(27 k y^2+4\right)} \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2
   y\right)^{4/3}}$
But this gives me no insight. Substituting
$y \to x + k x^3$
doesn't simplify things nor provide any insight into extending this to a nonic polynomial. It seems like I could evaluate this expression from the $dy/dk$ derivative.
 A: We have the following relations
\begin{align}
y &= f(x; p), \tag{1}\\
x &= f^{-1}(y; p). \tag{2}
\end{align}
By taking derivative of (1) with respect to $p$ and by noting (2), we have
$$0 = \frac{\partial f}{\partial x} \frac{\partial x}{\partial p} + \frac{\partial f}{\partial p}$$
and hence
$$\frac{\partial x}{\partial p} = \frac{\partial f^{-1}(y; p)}{\partial p} = - \frac{\partial f}{\partial p}\cdot \frac{1}{\frac{\partial f}{\partial x}}. \tag{3}$$
Let us see three examples.
Example 1: $y = x + p^2$, its inverse $x = y - p^2$, $\frac{\partial x}{\partial p} = -2p$,
$\frac{\partial f}{\partial p} = 2p$, $\frac{\partial f}{\partial x} = 1$, (3) is valid.
Example 2: $y = x^p$ ($p > 0$, $x > 0$), its inverse $x = y^{1/p}$, 
$\frac{\partial x}{\partial p} = y^{1/p}\cdot \frac{-1}{p^2}\ln y$,
$\frac{\partial f}{\partial p} = x^p \ln x$,
$\frac{\partial f}{\partial x} = px^{p-1}$, 
$$- \frac{\partial f}{\partial p}\cdot \frac{1}{\frac{\partial f}{\partial x}}
= - x^p \ln x \cdot \frac{1}{px^{p-1}} = - y \cdot \frac{1}{p}\ln y \cdot \frac{1}{p y^{(p-1)/p}}
= - y^{1/p}\cdot \frac{-1}{p^2}\ln y,$$
(3) is valid.
Example 3: $y = x^2 + x(p^2 + p)$ ($p > 0$, $x > 0$), its inverse
$x = - \frac{1}{2}p^2 - \frac{1}{2}p + \frac{1}{2}\sqrt{p^4 + 2p^3 + p^2 + 4y}$, $\cdots$, (3) is valid.
