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Verified Answer :

\(( \arcsin{x})'= \frac{1}{ \sqrt{1-x^2} } \\ \\ (\arcsin{(2x+1)})'= \frac{1}{ \sqrt{1-(2x+1)^2} } (2x+1)'= \frac{2}{ \sqrt{1-(4x^2+4x+1)} } = \\ \\= \frac{2}{ \sqrt{-4x^2-4x} } =\frac{2}{ \sqrt{4x(-x-1} )} =\frac{2}{ 2\sqrt{x(-x-1} )} =\frac{1}{ \sqrt{x(-x-1} )}\)

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