Derivative of (1-cos(8x))/(2x^2)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 1 - \cos{\left(8 x \right)}$ and $g{\left(x \right)} = 2 x^{2}$.

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

   1. Differentiate $1 - \cos{\left(8 x \right)}$ 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. Let $u = 8 x$.

         2. The derivative of cosine is negative sine:

            $$\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$$

         3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 8 x$:

            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: $8$

            The result of the chain rule is:

            $$- 8 \sin{\left(8 x \right)}$$

         So, the result is: $8 \sin{\left(8 x \right)}$

      The result is: $8 \sin{\left(8 x \right)}$

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

   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: $4 x$

   Now plug in to the quotient rule:

   $\frac{16 x^{2} \sin{\left(8 x \right)} - 4 x \left(1 - \cos{\left(8 x \right)}\right)}{4 x^{4}}$

2. Now simplify:

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

The answer is: