Delta derivative of product (1)
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Theorem
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 $$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t),$$ where $\sigma$ denotes the forward jump.
Proof
References
- {{BookReference|Dynamic Equations on Time Scales|2001|Martin Bohner|author2=Allan Peterson|prev=Delta derivative of constant multiple|next=Delta derivative of product (2)}: Theorem 1.20 (iii)
