# Delta derivative of product (2)

Let $\mathbb{T}$ be a time scale and $f,g \colon \mathbb{T} \rightarrow \mathbb{R}$ delta differentiable. Then the product function $fg$ is delta differentiable with $$(fg)^{\Delta}(t)=f^{\Delta}(t)g(\sigma(t))+f(t)g^{\Delta}(t),$$ where $\sigma$ denotes the forward jump.