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Question Number 98816 by M±th+et+s last updated on 16/Jun/20 | ||
$${prove}\:{that} \\ $$$$ \\ $$$$\left({V}^{\:\mu} \right)_{;\mu} =\frac{\left(\sqrt{−{g}}{V}^{\:\mu} \right)_{;\mu} }{\sqrt{−{g}}} \\ $$ | ||