Derivative of y=sin^3*2xcos*(8*x^5)

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

   1. Apply the product rule:

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

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

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

      $g{\left(x \right)} = \cos{\left(8 x^{5} \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

      1. Let $u = 8 x^{5}$.

      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 $\frac{d}{d x} 8 x^{5}$:

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

            1. Apply the power rule: $x^{5}$ goes to $5 x^{4}$

            So, the result is: $40 x^{4}$

         The result of the chain rule is:

         $$- 40 x^{4} \sin{\left(8 x^{5} \right)}$$

      The result is: $- 40 x^{5} \sin{\left(8 x^{5} \right)} + \cos{\left(8 x^{5} \right)}$

   So, the result is: $\left(- 40 x^{5} \sin{\left(8 x^{5} \right)} + \cos{\left(8 x^{5} \right)}\right) \sin^{3}{\left(2 \right)}$

The answer is:

$$\left(- 40 x^{5} \sin{\left(8 x^{5} \right)} + \cos{\left(8 x^{5} \right)}\right) \sin^{3}{\left(2 \right)}$$