Question

How find derivative of function: y=arcsin(2x+1)
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Answer 1

`( \arcsin{x})'= (1)/( √(1-x^2) ) \\ \\ (\arcsin{(2x+1)})'= (1)/( √(1-(2x+1)^2) ) (2x+1)'= (2)/( √(1-(4x^2+4x+1)) ) = \\ \\= (2)/( √(-4x^2-4x) ) =(2)/( √(4x(-x-1) )) =(2)/( 2√(x(-x-1) )) =(1)/( √(x(-x-1) ))`