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How do you find the derivative of y=arcsin(2x+1)y=arcsin(2x+1)?

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Answer is in the attachment below.
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[tex]y=arcsin(2x+1)\\\\siny=2x+1\ (*)\\\\2x=siny-1\\\\x=\frac{1}{2}siny-\frac{1}{2}\\\\\frac{dx}{dy}=\frac{1}{2}cosy\Rightarrow\frac{dy}{dx}=\frac{2}{cosy}=\frac{2}{\sqrt{1-sin^2y}}\\\\substitute\ (*)\\\\=\frac{2}{\sqrt{1-(2x+1)^2}}=\frac{2}{\sqrt{1-4x^2-4x-1}}=\frac{2}{\sqrt{-4x^2-4x}}=\frac{2}{\sqrt{4(-x^2-x)}}\\\\=\frac{2}{\sqrt4\cdot\sqrt{-x^2-x}}=\frac{2}{2\sqrt{-x(x+1)}}\\\center\boxed{=\frac{1}{\sqrt{-x(x+1)}}}[/tex]
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