Derivative of (x+1)e^-x

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = x + 1$ and $g{\left(x \right)} = e^{x}$.

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

   1. Differentiate $x + 1$ term by term:

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

      2. Apply the power rule: $x$ goes to $1$

      The result is: $1$

   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:

   $\left(- \left(x + 1\right) e^{x} + e^{x}\right) e^{- 2 x}$

2. Now simplify:

   $$- x e^{- x}$$

The answer is:

$$- x e^{- x}$$