Answer
$$
y^{\prime}=\frac{d}{dx} \left[ y \right] = \frac{d}{dx} \left[ \sin ^{-1}(2 x+1) \right] =\frac{1}{\sqrt{-x^{2}-x}}
$$
Work Step by Step
$$
y=\sin ^{-1}(2 x+1)
$$
Differentiating both sides of this equation we have
$$
\begin{aligned}
\frac{d}{dx} \left[ y \right] &= \frac{d}{dx} \left[ \sin ^{-1}(2 x+1) \right] \\
y^{\prime}&=\frac{1}{\sqrt{1-(2 x+1)^{2}}} \cdot \frac{d}{d x}(2 x+1) \\
&=\frac{1}{\sqrt{1-\left(4 x^{2}+4 x+1\right)}} \cdot 2 \\
&=\frac{2}{\sqrt{-4 x^{2}-4 x}} \\
&=\frac{1}{\sqrt{-x^{2}-x}}
\end{aligned}
$$
