Derivative of (1+sin(2x))*cos(2x)

1. Apply the product rule:

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

   $f{\left(x \right)} = \sin{\left(2 x \right)} + 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

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

      2. Let $u = 2 x$.

      3. The derivative of sine is cosine:

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

      4. 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:

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

      The result is: $2 \cos{\left(2 x \right)}$

   $g{\left(x \right)} = \cos{\left(2 x \right)}$; 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: $- 2 \left(\sin{\left(2 x \right)} + 1\right) \sin{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}$

The answer is: