Derivative of (6x³-x)(10-20x)

1. Apply the product rule:

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

   $f{\left(x \right)} = 6 x^{3} - x$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $6 x^{3} - x$ 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: $x^{3}$ goes to $3 x^{2}$

         So, the result is: $18 x^{2}$

      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: $-1$

      The result is: $18 x^{2} - 1$

   $g{\left(x \right)} = 10 - 20 x$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $10 - 20 x$ term by term:

      1. The derivative of the constant $10$ is zero.

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

         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: $20$

         So, the result is: $-20$

      The result is: $-20$

   The result is: $- 120 x^{3} + 20 x + \left(10 - 20 x\right) \left(18 x^{2} - 1\right)$

2. Now simplify:

   $$- 480 x^{3} + 180 x^{2} + 40 x - 10$$

The answer is:

$$- 480 x^{3} + 180 x^{2} + 40 x - 10$$