Difference between revisions of "Delta derivative of product (2)"
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Revision as of 05:37, 10 June 2016
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 $$(fg)^{\Delta}(t)=f^{\Delta}(t)g(\sigma(t))+f(t)g^{\Delta}(t),$$ where $\sigma$ denotes the forward jump.
Proof
References
- Martin Bohner and Allan Peterson: Dynamic Equations on Time Scales (2001)... (previous)... (next): Theorem 1.20 (iii)
