# Difference between revisions of "Relationship between delta derivative and nabla derivative"

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− | + | ==Theorem== | |

− | + | Let $\mathbb{T}$ be a [[time scale]] and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is [[nabla derivative|$\nabla$-differentiable]] on $\mathbb{T}_{\kappa}$ and $g^{\nabla}$ is [[ld continuous]] on $\mathbb{T}_{\kappa}$, then $f$ is [[delta derivative|$\Delta$-differentiable]] on $\mathbb{T}^{\kappa}$ and | |

$$g^{\Delta}(t) = g^{\nabla}(\sigma(t)).$$ | $$g^{\Delta}(t) = g^{\nabla}(\sigma(t)).$$ | ||

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− | + | ==Proof== | |

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− | + | ==References== | |

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+ | [[Category:Theorem]] | ||

+ | [[Category:Unproven]] |

## Latest revision as of 00:31, 23 August 2016

## Theorem

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is $\nabla$-differentiable on $\mathbb{T}_{\kappa}$ and $g^{\nabla}$ is ld continuous on $\mathbb{T}_{\kappa}$, then $f$ is $\Delta$-differentiable on $\mathbb{T}^{\kappa}$ and $$g^{\Delta}(t) = g^{\nabla}(\sigma(t)).$$