Differentiating the inverse of a function with respect to a parameter The derivative of a function's inverse is well understood, and it is explained in full detail here.
Say I have a function $\varphi_\varepsilon:\mathbb{R}\to\mathbb{R}$, where $\varepsilon$ is a parameter.  I also know that the function is smooth, invertible, and has smooth inverse.  This leads us to the question: do we have a representation for $\frac{\partial}{\partial\varepsilon}\varphi_\varepsilon^{-1}$ that is similar to the single variable case?
Here is a post that already addresses this question, but it is quite old and doesn't have any significant responses.  I'm also more interested in the theoretical case than a concrete example.
 A: After some further investigation, it appears we can't actually get away with what's stated in the comments.
Note that for a function $f$ of two variables $x$ and $y$, if $y$ is also a function of $x$, we have that
$$
\frac{df}{dx}(x,y(x))=\frac{\partial f}{\partial x}(x,y(x))+\frac{\partial f}{\partial y}(x,y(x))\frac{dy}{dx}(x).
$$
It is worth pointing out here that there is a significant difference between $\frac{df}{dx}$ and $\frac{\partial f}{\partial x}$.  The former is the total derivative of $f$ when we make $y$ dependent upon $x$, whereas the latter is the partial derivative of $f$ with respect to its "first" input variable.
Now say that $s$ and $t$ are the variables for the domain and codomain of $\varphi_\varepsilon$, respectively.  That is, $\varphi_\varepsilon(s)=t$.  As our function is invertible, we may write
$$
\varphi_\varepsilon(\varphi_\varepsilon^{-1}(t))=t.
$$
Differentiating both sides of this equation with respect to $\varepsilon$ then yields something similar to the above statement with $f$:
$$
\frac{\partial\varphi_\varepsilon}{\partial\varepsilon}(\varphi_\varepsilon^{-1}(t))+\frac{\partial\varphi_\varepsilon}{\partial s}(\varphi_\varepsilon^{-1}(t))\frac{\partial\varphi_\varepsilon^{-1}}{\partial\varepsilon}(t)=0.
$$
We can then solve this for $\frac{\partial\varphi_\varepsilon^{-1}}{\partial\varepsilon}(t)$, which yields
$$
\frac{\partial\varphi_\varepsilon^{-1}}{\partial\varepsilon}(t)=\frac{-\frac{\partial\varphi_\varepsilon}{\partial\varepsilon}(\varphi_\varepsilon^{-1}(t))}{\frac{\partial\varphi_\varepsilon}{\partial s}(\varphi_\varepsilon^{-1}(t))}.
$$
