Difference between revisions of "Delta derivative of constant multiple"

From timescalewiki
Jump to: navigation, search
(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...")
 
Line 1: Line 1:
 
==Theorem==
 
==Theorem==
−
Let $\mathbb{T}$ be a [[time scale]] and $f,g \colon \mathbb{T} \rightarrow \mathbb{R}$ [[delta derivative|delta differentiable]]. Then the product function $fg$ is delta differentiable with
+
Let $\mathbb{T}$ be a [[time scale]], $\alpha \in \mathbb{R}$, and $f \colon \mathbb{T} \rightarrow \mathbb{R}$ [[delta derivative|delta differentiable]]. Then the function $\alpha f$ is delta differentiable with
−
$$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t),$$
+
$$(\alpha f)^{\Delta}(t)=\alpha f^{\Delta}(t).$$
−
where $\sigma$ denotes the [[forward jump]].
 
  
 
==Proof==
 
==Proof==

Revision as of 05:36, 10 June 2016

Theorem

Let $\mathbb{T}$ be a time scale, $\alpha \in \mathbb{R}$, and $f \colon \mathbb{T} \rightarrow \mathbb{R}$ delta differentiable. Then the function $\alpha f$ is delta differentiable with $$(\alpha f)^{\Delta}(t)=\alpha f^{\Delta}(t).$$

Proof

References

Retrieved from "http://timescalewiki.org/index.php?title=Delta_derivative_of_constant_multiple&oldid=1288"