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Question Number 98816 by  M±th+et+s last updated on 16/Jun/20

prove that    (V^( μ) )_(;μ) =((((√(−g))V^( μ) )_(;μ) )/(√(−g)))

$${prove}\:{that} \\ $$$$ \\ $$$$\left({V}^{\:\mu} \right)_{;\mu} =\frac{\left(\sqrt{−{g}}{V}^{\:\mu} \right)_{;\mu} }{\sqrt{−{g}}} \\ $$

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