Let $f\colon\mathbb{R} \rightarrow \mathbb{R}$ be an $\textbf{uniformly continuous}$ function. Then by definition $\forall\epsilon>0 \quad \exists\delta>0\quad$such that $\forall x,y\in\mathbb{R},\quad\mid{}x-y \mid{}<\delta\implies \mid{} f(x) -f(y) \mid{}<\epsilon$.
Define $\Delta_{\epsilon}=\{\delta>0\ \colon\;\mid{}x-y \mid{}<\delta\implies \mid{} f(x) -f(y) \mid{}<\epsilon\}$ for a fixed positive $\epsilon$.
Since $f$ is an uniformly continuous function $\Delta_{\epsilon}\ne\emptyset$. If $\Delta_{\epsilon}$ is bounded above, then by the $\textbf{least upper bound property}$ of the real numbers $\sup{\Delta_{\epsilon}}$ exists.
Now we define the following function:
\begin{align*} g \colon \mathbb{R^{+}} &\rightarrow \mathbb{R^{+}} \\ \epsilon &\mapsto \sup{\Delta_{\epsilon}}&\text{when} \ \Delta_{\epsilon}\ \text{is bounded above} \\ \epsilon &\mapsto 0 \ &\text{when}\ \Delta_{\epsilon}\ \text{isn't bounded above} \end{align*}
What can we say about function $g$? When is it continuous? When is it differentiable?
If the question is too broad, try giving an answer for the functions $sin(x)$ and $\sqrt{x}.$ I've tried but I haven't found an easy way to find the supremum.