# How do I find the derivative of y=ln(e^-x + xe^-x) ?

Mar 31, 2016

Use some logarithm properties and the fact that $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$ to get $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{1 + x}$.

#### Explanation:

Begin by factoring out an ${e}^{-} x$ within the parenthesis:
$y = \ln \left({e}^{-} x \left(1 + x\right)\right)$

Now apply the property $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$ to get:
$y = \ln \left({e}^{-} x\right) + \ln \left(1 + x\right)$

Apply another property specific to the natural logarithm, $\ln \left({e}^{a}\right) = a$:
$y = - x + \ln \left(1 + x\right)$

We can now take the derivative with ease.

Using the sum rule, $\frac{d}{\mathrm{dx}} \left(- x + \ln \left(1 + x\right)\right) = \frac{d}{\mathrm{dx}} \left(- x\right) + \frac{d}{\mathrm{dx}} \left(\ln \left(1 + x\right)\right)$. And using the fact that $\frac{d}{\mathrm{dx}} \left(- x\right) = - 1$ and $\frac{d}{\mathrm{dx}} \left(\ln \left(1 + x\right)\right) = \frac{1}{1 + x}$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1 + \frac{1}{1 + x}$

Finally, add the fractions to get the final result:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1 + x}{1 + x} + \frac{1}{1 + x} = - \frac{x}{1 + x}$