Is root of a function differentiable? Let's assume a function $f(\alpha,\theta)$ always has a single zero wrt $\alpha$:
$\forall \theta, \exists \hat\alpha_\theta$ such that $f(\hat\alpha_\theta,\theta)=0$.
Let's now consider this root as a function of $\theta$:
$$g(\theta)=\hat\alpha_\theta$$
Assuming $f$ is differentiable wrt both its variables, is $g(\theta)$ differentiable ?
Thanks a lot for your hints !
 A: Assuming you mean functions from  $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$, a counterexample is
$$
f(\alpha,\theta) = \alpha |\alpha| - \theta^2.
$$
(As the OP requested for hints, proof is left as an exercise.)

Edit: And $g$ does not even have to be continuous. My idea was to consider two different functions, one of which has a root only for one $\theta$, and the other has roots always except at one point. Then multiplying them would give an example with noncontinuous $g$.
After some thinking I figured out that one way to get a function that has a root always when $\theta\neq 0$ but does not have one when $\theta=0$, is to find a function whose root goes to infinity as $\theta$ approaches $0$.
After some drawing, this lead me to the following example:
$$
f(\alpha,\theta) = (\theta^2 e^\alpha - 1) (\alpha^2 + \theta^2)
$$
When $\theta\neq 0$, the second part is always positive but the first part has the unique root $\alpha=\ln(1/\theta^2)$. When $\theta=0$, the first part does not have a root but the second part has a unique root.
