Derivative of y=sinx/2cos^2x

1. Apply the quotient rule, which is:

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

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

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

   1. Apply the product rule:

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

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

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

      $g{\left(x \right)} = \sin{\left(x \right)}$; 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: $- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos^{3}{\left(x \right)}$

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

   1. The derivative of the constant $2$ is zero.

   Now plug in to the quotient rule:

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

2. Now simplify:

   $$\frac{\cos{\left(x \right)}}{8} + \frac{3 \cos{\left(3 x \right)}}{8}$$

The answer is:

$$\frac{\cos{\left(x \right)}}{8} + \frac{3 \cos{\left(3 x \right)}}{8}$$