# Difference between revisions of "Relationship between nabla derivative and delta 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 [[Delta derivative|$\Delta$-differentiable]] and $f^{\Delta}$ is [[rd continous]] on $\mathbb{T}^{\kappa}$, then $f$ is [[nabla derivative|$\nabla$-differentiable]] on $\mathbb{T}_{\kappa}$ and | |

$$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$ | $$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$ | ||

− | + | ||

− | + | ==Proof== | |

− | + | ||

− | + | ==References== | |

+ | |||

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

+ | [[Category:Unproven]] |

## Latest revision as of 06:02, 10 June 2016

## Theorem

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