Derivative of 8sin^2xcosx

1. Apply the product rule:

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

   $f{\left(x \right)} = 8 \sin^{2}{\left(x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

      1. Let $u = \sin{\left(x \right)}$.

      2. Apply the power rule: $u^{2}$ goes to $2 u$

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \sin{\left(x \right)}$:

         1. The derivative of sine is cosine:

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

         The result of the chain rule is:

         $$2 \sin{\left(x \right)} \cos{\left(x \right)}$$

      So, the result is: $16 \sin{\left(x \right)} \cos{\left(x \right)}$

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

   1. The derivative of cosine is negative sine:

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

   The result is: $- 8 \sin^{3}{\left(x \right)} + 16 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$

2. Now simplify:

   $$- 24 \sin^{3}{\left(x \right)} + 16 \sin{\left(x \right)}$$

The answer is:

$$- 24 \sin^{3}{\left(x \right)} + 16 \sin{\left(x \right)}$$