Delta derivative
Let $\mathbb{T}$ be a time scale. Define $\mathbb{T}^{\kappa} := \mathbb{T} \setminus \sup \mathbb{T}$. Let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. We define the delta-derivative of $f$ to be the function $f^{\Delta} \colon \mathbb{T}^{\kappa} \rightarrow \mathbb{R}$ by the formula $$f^{\Delta}(t) := \left\{ \begin{array}{ll} \dfrac{f(\sigma(t))-f(t)}{\mu(t)} &\colon \mu(t) > 0 \\ \displaystyle\lim_{s \rightarrow t} \dfrac{f(s) - f(t)}{s-t} &\colon \mu(t) = 0. \end{array} \right.$$
An equivalent definition to this defines $f^{\Delta}(t)$ to be the number (if it exists) with the property that for any $\epsilon > 0$ there exists $\delta >0$ such that for all $s \in (t-\delta,t+\delta)\cap \mathbb{T}$, $$|[f(\sigma(t))-f(s)]-f^{\Delta}(t)[\sigma(t)-s]| \leq \epsilon |\sigma(t)-s|$$
Properties of the $\Delta$-derivative
- $f(\sigma(t))=f(t)+\mu(t)f^{\Delta}(t)$
- Sum rule:
$$(f+g)^{\Delta}(t)=f^{\Delta}(t)+g^{\Delta}(t)$$
- Constant rule:if $\alpha$ is constant with respect to $t$, then
$$(\alpha f)^{\Delta}(t) = \alpha f^{\Delta}(t)$$
- Product Rule I
$$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t))$$
- Product Rule II
$$(fg)^{\Delta}(t) = f(t)g^{\Delta}(t)+ f^{\Delta}(t)g(\sigma(t))$$
- Quotient Rule:
$$\left( \dfrac{f}{g} \right)^{\Delta}(t) = \dfrac{f^{\Delta}(t)g(t)-f(t)g^{\Delta}(t)}{g(t)g(\sigma(t))}$$
Interesting Examples
The jump operator $\sigma$ is not $\Delta$-differentiable on all time scales. Consider $\mathbb{T}=
