Derivative of e^t*sin(2t)

1. Apply the product rule:

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

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

   1. The derivative of $e^{t}$ is itself.

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

   1. Let $u = 2 t$.

   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 $\frac{d}{d t} 2 t$:

      1. The derivative of a constant times a function is the constant times the derivative of the function.

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

         So, the result is: $2$

      The result of the chain rule is:

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

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

2. Now simplify:

   $$\left(\sin{\left(2 t \right)} + 2 \cos{\left(2 t \right)}\right) e^{t}$$

The answer is:

$$\left(\sin{\left(2 t \right)} + 2 \cos{\left(2 t \right)}\right) e^{t}$$