differentiate the function.\n$y = \\ln(e^{-x}+xe^{-x})$

Explanation:

Step1: Use chain - rule

Let $u = e^{-x}+xe^{-x}$. Then $y=\ln(u)$, and $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, $\frac{dy}{du}=\frac{1}{u}$.

Step2: Differentiate $u$

$u = e^{-x}+xe^{-x}=e^{-x}(1 + x)$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 1 + x$ and $v=e^{-x}$, we have $u^\prime=1$ and $v^\prime=-e^{-x}$. So $\frac{du}{dx}=e^{-x}(1 + x)^\prime+(1 + x)(e^{-x})^\prime=e^{-x}-(1 + x)e^{-x}=-xe^{-x}$.

Step3: Calculate $\frac{dy}{dx}$

Since $\frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}$ and $u = e^{-x}+xe^{-x}=e^{-x}(1 + x)$, then $\frac{dy}{dx}=\frac{-xe^{-x}}{e^{-x}(1 + x)}=\frac{-x}{1 + x}$.

Answer:

$\frac{-x}{1 + x}$