Derivative of 2^x*sin^2x

1. Apply the product rule:

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

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

   1. $\frac{d}{d x} 2^{x} = 2^{x} \log{\left(2 \right)}$

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

   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:

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

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

2. Now simplify:

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

The answer is: