Answer
$y'=\frac{1}{\sqrt{-x^2-x}}$
Work Step by Step
Apply the Chain Rule:
$y=sin^{-1}(2x+1)\\
y'=\frac{1}{\sqrt{1-(2x+1)^2}}.2\\
y'=\frac{1}{\sqrt{1-4x^2-4x-1}}.2\\
y'=\frac{1}{\sqrt{-4x^2-4x}}.2\\
y'=\frac{1}{\sqrt{-x^2-x}}$
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