## Calculus: Early Transcendentals 8th Edition

$y'=\frac{2}{\sqrt{1-(2x+1)^2}}$
Apply the Chain Rule: $y=sin^{-1}(2x+1)\\ y'=\frac{1}{\sqrt{1-(2x+1)^2}}.2\\ y'=\frac{2}{\sqrt{1-(2x+1)^2}}$