Difference between revisions of "Delta derivative of constant multiple"

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==Theorem==
 
==Theorem==
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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
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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
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$$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t),$$
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$$(\alpha f)^{\Delta}(t)=\alpha f^{\Delta}(t).$$
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where $\sigma$ denotes the [[forward jump]].
 
  
 
==Proof==
 
==Proof==
  
 
==References==
 
==References==
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* {{BookReference|Dynamic Equations on Time Scales|2001|Martin Bohner|author2=Allan Peterson|prev=Delta derivative of sum|next=Delta derivative of product (1)}}: Theorem 1.20 (iii)
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* {{BookReference|Dynamic Equations on Time Scales|2001|Martin Bohner|author2=Allan Peterson|prev=Delta derivative of sum|next=Delta derivative of product (1)}}: Theorem 1.20 (ii)
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 05:45, 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

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