Derivative of (3x-1)/(2x+1)^2

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = 3 x - 1$ and $g{\left(x \right)} = \left(2 x + 1\right)^{2}$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $3 x - 1$ term by term:

      1. The derivative of the constant $-1$ is zero.

      2. The derivative of a constant times a function is the constant times the derivative of the function.

         1. Apply the power rule: $x$ goes to $1$

         So, the result is: $3$

      The result is: $3$

   To find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x + 1$.

   2. Apply the power rule: $u^{2}$ goes to $2 u$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(2 x + 1\right)$:

      1. Differentiate $2 x + 1$ term by term:

         1. The derivative of the constant $1$ is zero.

         2. The derivative of a constant times a function is the constant times the derivative of the function.

            1. Apply the power rule: $x$ goes to $1$

            So, the result is: $2$

         The result is: $2$

      The result of the chain rule is:

      $$8 x + 4$$

   Now plug in to the quotient rule:

   $\frac{3 \left(2 x + 1\right)^{2} - \left(3 x - 1\right) \left(8 x + 4\right)}{\left(2 x + 1\right)^{4}}$

2. Now simplify:

   $$\frac{7 - 6 x}{\left(2 x + 1\right)^{3}}$$

The answer is:

$$\frac{7 - 6 x}{\left(2 x + 1\right)^{3}}$$