Derivative of cosx*sinx*cos2x

1. Apply the product rule:

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

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

   1. Apply the product rule:

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

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

      1. The derivative of cosine is negative sine:

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

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

      1. The derivative of sine is cosine:

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

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

   $g{\left(x \right)} = \cos{\left(2 x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x$.

   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 $\frac{d}{d x} 2 x$:

      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: $2$

      The result of the chain rule is:

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

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

2. Now simplify:

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

The answer is:

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