I have a function
$h(x) = f(x-\beta b)-g(x)$
and I want to compute $\frac{\partial}{\partial\beta}(h^{-1}(x)[0])$
meaning that if $h(x_m)=0$, I want to compute $\frac{\partial x_m}{\partial \beta}$.
Can I use the inverse function theorem to do it like this? $$\frac{\partial}{\partial\beta}(h^{-1}(x)[0])=\frac{1}{\frac{\partial h(x)}{\partial\beta}|_{x=x_m}}=\frac{1}{-b \frac{df(x)}{dx}|_{x=x_m-\beta b}} $$
I tried to do it by Wolfram Mathematica and got $$ \frac{\partial}{\partial\beta}(h^{-1}(x)[0]) = (-bf'(x-\beta b)(h^{-1})'(f(x-\beta b)-g(x))[0]$$
but I would really like to simplify it further to express in in derivatives of $f$. I assume differentiability of $f$ and $g$.
You can use the implicit function theorem (which is closely related to the inverse function theorem) to calculate this. First write $h$ explicitly as function of two variables: $$ h(x,\beta) = f(x - \beta b) - g(x). $$ Now, you're looking for a function $x = x_m(\beta)$ that solves $$ h(x_m(\beta),\beta) = 0. $$ The implicit function theorem tells you that $x_m$ exists in a small neighborhood of a point $x_0,\beta_0$ where $h(x_0,\beta_0) = 0$, that $x_m$ is $C^1$ (assuming $f,g$ are $C^1$, not just differentiable), and finally that the derivative of $x_m$ is given by $$ x_m'(\beta) = -\big(D_1 h(x_m(\beta),\beta)\big)^{-1} \cdot D_2 h(x_m(\beta),\beta), $$ where $D_i h$ denotes the derivative of $h$ with respect to the $i$-th argument (this notation creates less ambiguity than partial derivatives). With this formula you can explicitly calculate $x_m'$ (although you may not find an explicit expression for $x_m$ itself).