Find the derivative of the function $$ y=\cos ^{-1}(2 x+1) $$ a) $$ y^{\prime}=\frac{2}{\sqrt{1-2 x}} $$ b) $$ y^{\prime}=\frac{-2}{\sqrt{1-(2 x+1)^{2}}} $$ c) $$ y^{\prime}=\frac{4}{\sqrt{1-2 x}} $$ d) $$ y^{\prime}=\frac{1}{\sqrt{1-(2 x+1)^{2}}} $$

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Let $$ u = 2x + 1 $$ so that $$ y = \cos^{-1}(u) $$.

First, differentiate using the chain rule:
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
$$

The derivative of $$ \cos^{-1}(u) $$ with respect to $$ u $$ is:
$$
\frac{dy}{du} = -\frac{1}{\sqrt{1 - u^2}}
$$

Since $$ u = 2x + 1 $$, we have:
$$
\frac{du}{dx} = 2
$$

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

So, the correct answer is:
$$
\boxed{\frac{dy}{dx} = -\frac{2}{\sqrt{1 - (2x+1)^2}}}
$$

This corresponds to option (b).
The derivative of $$ y = \cos^{-1}(2x + 1) $$ is $$ y' = -\frac{2}{\sqrt{1 - (2x + 1)^2}} $$.

Find the derivative of the function \( y=\cos ^{-1}(2 x+1) \) a) \( y^{\prime}=\frac{2}{\sqrt{1-2 x}} \) b) \( y^{\prime}=\frac{-2}{\sqrt{1-(2 x+1)^{2}}} \) c) \( y^{\prime}=\frac{4}{\sqrt{1-2 x}} \) d) \( y^{\prime}=\frac{1}{\sqrt{1-(2 x+1)^{2}}} \)

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