Derivative of y=e^(-2x)sinx

1. Apply the quotient rule, which is:

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

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

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

   1. The derivative of sine is cosine:

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

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

   1. Let $u = 2 x$.

   2. The derivative of $e^{u}$ is itself.

   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:

      $$2 e^{2 x}$$

   Now plug in to the quotient rule:

   $\left(- 2 e^{2 x} \sin{\left(x \right)} + e^{2 x} \cos{\left(x \right)}\right) e^{- 4 x}$

2. Now simplify:

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

The answer is:

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