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

1. Apply the product rule:

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

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

   1. Apply the product rule:

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

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

      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: $1$

      $g{\left(x \right)} = x + 1$; to find $\frac{d}{d x} g{\left(x \right)}$:

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

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

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

         The result is: $1$

      The result is: $2 x - 1$

   $g{\left(x \right)} = 3 x + 1$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $3 x + 1$ 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: $3$

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

      The result is: $3$

   The result is: $3 \left(x - 2\right) \left(x + 1\right) + \left(2 x - 1\right) \left(3 x + 1\right)$

2. Now simplify:

   $$9 x^{2} - 4 x - 7$$

The answer is:

$$9 x^{2} - 4 x - 7$$