y = sin4(2x+1)

The derivative of $\sin^{-1}(u)$ with respect to $u$ is given by:

$$\frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1-u^2}}$$

Let $u = 2x+1$. Then $\frac{du}{dx} = \frac{d}{dx}(2x+1) = 2$.
Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

$$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$

$$\frac{dy}{dx} = \frac{1}{\sqrt{1-(2x+1)^2}} \cdot 2$$ $$\frac{dy}{dx} = \frac{2}{\sqrt{1-(4x^2+4x+1)}}$$ $$\frac{dy}{dx} = \frac{2}{\sqrt{1-4x^2-4x-1}}$$ $$\frac{dy}{dx} = \frac{2}{\sqrt{-4x^2-4x}}$$

$$\frac{dy}{dx} = \frac{2}{\sqrt{-4(x^2+x)}}$$ $$\frac{dy}{dx} = \frac{2}{2\sqrt{-(x^2+x)}}$$ $$\frac{dy}{dx} = \frac{1}{\sqrt{-(x^2+x)}}$$

The final answer is $\frac{1}{\sqrt{-x^2-x}}$.