Derivative of (4x^2-1)(7x^3+x)

1. Apply the product rule:

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

   $f{\left(x \right)} = 4 x^{2} - 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $4 x^{2} - 1$ 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: $x^{2}$ goes to $2 x$

         So, the result is: $8 x$

      2. The derivative of the constant $\left(-1\right) 1$ is zero.

      The result is: $8 x$

   $g{\left(x \right)} = 7 x^{3} + x$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $7 x^{3} + x$ 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: $21 x^{2}$

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

      The result is: $21 x^{2} + 1$

   The result is: $8 x \left(7 x^{3} + x\right) + \left(4 x^{2} - 1\right) \left(21 x^{2} + 1\right)$

2. Now simplify:

   $$140 x^{4} - 9 x^{2} - 1$$

The answer is:

$$140 x^{4} - 9 x^{2} - 1$$