The problem asks to differentiate the function $y = (2x+1)^{-1}$.

1. Problem Description

The problem asks to differentiate the function $y = (2x+1)^{-1}$.

2. Solution Steps

We need to find $\frac{dy}{dx}$. We will use the chain rule.

The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In our case, $y = (2x+1)^{-1}$. We can consider $f(u) = u^{-1}$ and $g(x) = 2x+1$.

Then $f'(u) = -u^{-2}$ and $g'(x) = 2$.

Therefore, $\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = -(2x+1)^{-2} \cdot 2 = -2(2x+1)^{-2}$.

Alternatively, we can write $y = \frac{1}{2x+1}$. Using the quotient rule, if $y = \frac{u}{v}$, then

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Here, $u = 1$ and $v = 2x+1$. Then $\frac{du}{dx} = 0$ and $\frac{dv}{dx} = 2$.

So, $\frac{dy}{dx} = \frac{(2x+1)(0) - (1)(2)}{(2x+1)^2} = \frac{-2}{(2x+1)^2} = -2(2x+1)^{-2}$.

3. Final Answer

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