Delta derivative

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Revision as of 20:13, 19 May 2014 by Tom (talk | contribs) (Properties of the $\Delta$-derivative)
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Let $\mathbb{T}$ be a time scale. Define $\mathbb{T}^{\kappa} := \mathbb{T} \setminus \sup \mathbb{T}$. Let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. We define the delta-derivative of $f$ to be the function $f^{\Delta} \colon \mathbb{T}^{\kappa} \rightarrow \mathbb{R}$ by the formula $$f^{\Delta}(t) := \left\{ \begin{array}{ll} \dfrac{f(\sigma(t))-f(t)}{\mu(t)} &\colon \mu(t) > 0 \\ \displaystyle\lim_{s \rightarrow t} \dfrac{f(s) - f(t)}{s-t} &\colon \mu(t) = 0. \end{array} \right.$$

Properties of the $\Delta$-derivative

  • $f(\sigma(t))=f(t)+\mu(t)f^{\Delta}(t)$
  • Sum rule:$(f+g)^{\Delta}(t)=f^{\Delta}(t)+g^{\Delta}(t)$ *Constant rule:if $\alpha$ is constant with respect to $t$, then

$$(\alpha f)^{\Delta}(t) = \alpha f^{\Delta}(t)$$

  • Product Rule I

$$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t))$$

  • Product Rule II

$$(fg)^{\Delta}(t) = f(t)g^{\Delta}(t)+ f^{\Delta}(t)g(\sigma(t))$$

  • Quotient Rule:

$$\left( \dfrac{f}{g} \right)^{\Delta}(t) = \dfrac{f^{\Delta}(t)g(t)-f(t)g^{\Delta}(t)}{g(t)g(\sigma(t))}$$