Consider a function $f:\Theta \subseteq \mathbb{R}^l \rightarrow [0,\infty) $. Let
(1) $\sqrt{f(\theta)}$ is differentiable at $\theta_0 \in \Theta$
(2) $f(\theta)$ is differentiable at $\theta_0\in \Theta$.
Question: Are (1) and (2) equivalent, i.e. (1) holds if and only if (2) holds? Why?
When $l=1$ I think (1) ans (2) are equivalent; I'm confused about the case $l>1$.
Note that if $\theta _0 = 0$ and $\Theta$ is some open set around $0$, then the function $f(x) = \| x \|^2$ is differentiable at $0$ but $\sqrt f$ is not (the partial derivatives do not exist at $0$).
In general, $(1)$ implies $(2)$ because if $\sqrt{f}(\theta)$ is differentiable at $\theta_0$ then $f = (\sqrt{f})^2$ is differentiable at $\theta = 0$, being a composition of two differentiable functions.
If $f(\theta_0) \neq 0$, then $(2)$ implies $(1)$ because again, $\sqrt{f}$ is a composition of two differentiable functions (as $\sqrt{x}$ is differentiable at $f(\theta_0)$).
If $f(\theta_0) = 0$, then $(2)$ does not necessarily imply $(1)$ as the example of $f(\theta) = \theta^2$ shows.
