Delta derivative of product (1)

From timescalewiki
Revision as of 05:35, 10 June 2016 by Tom (talk | contribs) (Created page with "==Theorem== Let $\mathbb{T}$ be a time scale and $f,g \colon \mathbb{T} \rightarrow \mathbb{R}$ delta differentiable. Then the product function $fg$ i...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

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)