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

1. Apply the product rule:

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

   $f{\left(x \right)} = \left(2 x - 1\right)^{2}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   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 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: $2$

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

         The result is: $2$

      The result of the chain rule is:

      $$8 x - 4$$

   $g{\left(x \right)} = x + 2$; to find $\frac{d}{d x} g{\left(x \right)}$:

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

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

      2. The derivative of the constant $2$ is zero.

      The result is: $1$

   The result is: $\left(x + 2\right) \left(8 x - 4\right) + \left(2 x - 1\right)^{2}$

2. Now simplify:

   $$\left(2 x - 1\right) \left(6 x + 7\right)$$

The answer is:

$$\left(2 x - 1\right) \left(6 x + 7\right)$$