Derivative of y=((e)^(-x))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 $g{\left(x \right)} = e^{x}$.

   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. The derivative of $e^{x}$ is itself.

   Now plug in to the quotient rule:

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

2. Now simplify:

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

The answer is:

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