Differentiate the function. \[ y=\ln \left(e^{-x}+x e^{-x}\right) \] \( y^{\prime}=\square \)

To differentiate the function \( y = \ln \left(e^{-x} + x e^{-x}\right) \), we can apply the chain rule and the product rule. First, let's define the inside function as \( u = e^{-x} + x e^{-x} \). So, using the chain rule: \[ y' = \frac{1}{u} \cdot u' \] Now, we find \( u' \): \[ u' = \frac{d}{dx}(e^{-x} + x e^{-x}) = -e^{-x} + (e^{-x} - x e^{-x}) = e^{-x} - x e^{-x} - e^{-x} = -x e^{-x} \] Thus, substituting back, we have: \[ y' = \frac{-x e^{-x}}{e^{-x} + x e^{-x}} \] So the derivative simplifies to: \[ y' = \frac{-x}{1 + x} \] This yields the final answer for the derivative: \[ y' = \frac{-x}{1 + x} \]