Directional Delta Derivative
Let $\mathbb{T}$ be a time scale with $0 \in \mathbb{T}$. Let $\omega=(\omega_1,\omega_2)$ be a unit vector. Define the time scales $\mathbb{T}_1=\{t=t^0+\xi \omega_1 \colon \xi \in \mathbb{T} \}$ and $\mathbb{T}_2=\left\{ s=s^0+\xi \omega_2\colon \xi \in \mathbb{T} \right\}$. Let $f \colon \mathbb{T}_1 \times \mathbb{T}_2 \rightarrow \mathbb{R}$. The directional delta derivative of $f$ at the point $(t^0,s^0)$ in the direction of the vector $\omega$ ("along $\omega$") is defined to be the number $$\dfrac{\partial f(t^0,s^0)}{\Delta \omega}=F^{\Delta}(0),$$ where $\dfrac{\partial f}{\Delta \omega}$ denotes a partial delta derivative and $F(\xi)=f(t^0+\xi \omega_1, s^0 + \xi \omega_2)$ for $\xi \in \mathbb{T}$.
Properties
Theorem: Suppose the function $f$ is $\sigma_1$ completely delta differentiable at $(t^0,s^0)$. Then the directional derivative of $f$ at $(t^0,s^0)$in the direction of the vector $\omega$ exists and is expressed by $$\dfrac{\partial f(t^0,s^0)}{\Delta \omega} = \dfrac{\partial f(t^0,s^0)}{\Delta_1 t} \omega_1 + \dfrac{\partial f(\omega_1(t^0),s^0)}{\Delta_2s}\omega_2.$$
Proof: █
