Derivative of y=sin^2(x)+tg(2x)*cos(x)

1. Differentiate $\sin^{2}{\left(x \right)} + \cos{\left(x \right)} \tan{\left(2 x \right)}$ term by term:

   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:

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

   4. Apply the product rule:

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

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

      1. Rewrite the function to be differentiated:

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

      2. Apply the quotient rule, which is:

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

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

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

         1. Let $u = 2 x$.

         2. The derivative of sine is cosine:

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

         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:

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

         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:

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

         Now plug in to the quotient rule:

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

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

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

2. Now simplify:

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

The answer is:

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