Derivative of (cos^2)x

1. Apply the product rule:

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

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

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

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

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

      1. The derivative of cosine is negative sine:

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

      The result of the chain rule is:

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

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

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

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

2. Now simplify:

   $$- x \sin{\left(2 x \right)} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$$

The answer is:

$$- x \sin{\left(2 x \right)} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$$