Find the derivative and simplify. $$ \begin{array}{l}\frac{d}{d x} \sec ^{-1}(2 x+1) \ =\frac{[?]}{|2 x+1| \sqrt{x^{2}+x}}\end{array} $$

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We start with the function
$$
y=\sec^{-1}(2x+1).
$$
Let
$$
u=2x+1.
$$
The derivative of $$ y=\sec^{-1}(u) $$ with respect to $$ u $$ is given by
$$
\frac{dy}{du}=\frac{1}{|u|\sqrt{u^2-1}}.
$$
Since
$$
u=2x+1,
$$
its derivative with respect to $$ x $$ is
$$
\frac{du}{dx}=2.
$$

By the chain rule,
$$
\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{2}{|2x+1|\sqrt{(2x+1)^2-1}}.
$$

Next, we simplify the expression inside the square root:
$$
(2x+1)^2-1=4x^2+4x+1-1=4x^2+4x.
$$
This can be factored as
$$
4x(x+1)=4(x^2+x).
$$

Thus, the square root becomes
$$
\sqrt{4(x^2+x)}=2\sqrt{x^2+x}.
$$

Substitute this back into our derivative:
$$
\frac{dy}{dx}=\frac{2}{|2x+1|\cdot 2\sqrt{x^2+x}}=\frac{1}{|2x+1|\sqrt{x^2+x}}.
$$

So, the simplified derivative is:
$$
\frac{d}{dx} \sec^{-1}(2x+1)=\frac{1}{|2x+1|\sqrt{x^2+x}}.
$$
The derivative of $$ \sec^{-1}(2x+1) $$ with respect to $$ x $$ is $$ \frac{1}{|2x+1|\sqrt{x^2+x}} $$.

Find the derivative and simplify.

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