Question

differentiate the function.
y = ln(e^x + xe^x)
y =

Explanation:

Step1: Use chain - rule

The chain - rule states that if $y = \ln(u)$ and $u = e^{x}+xe^{x}$, then $y'=\frac{u'}{u}$. First, find the derivative of $u = e^{x}+xe^{x}$.

Step2: Differentiate $u = e^{x}+xe^{x}$

Using the sum - rule of differentiation $(f + g)'=f'+g'$ and the product - rule $(uv)' = u'v+uv'$ (for $uv = xe^{x}$ where $u = x$ and $v = e^{x}$). The derivative of $e^{x}$ is $e^{x}$, and for $xe^{x}$, $u' = 1$, $v = e^{x}$, $u = x$, $v' = e^{x}$, so $(xe^{x})'=e^{x}+xe^{x}$. Then $u'=e^{x}+(e^{x}+xe^{x})=2e^{x}+xe^{x}$.

Step3: Calculate $y'$

Since $y'=\frac{u'}{u}$ and $u = e^{x}+xe^{x}$, $u'=2e^{x}+xe^{x}$, we have $y'=\frac{2e^{x}+xe^{x}}{e^{x}+xe^{x}}$. Factor out $e^{x}$ from the numerator and denominator: $y'=\frac{e^{x}(2 + x)}{e^{x}(1 + x)}=\frac{2 + x}{1 + x}$.

Answer:

$\frac{2 + x}{1 + x}$