Question

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

Answer 1

Use some logarithm properties and the fact that `d/dx(lnx)=1/x` to get `dy/dx=-x/(1+x)`.

Explanation:

Begin by factoring out an `e^-x` within the parenthesis:
`y=ln(e^-x(1+x))`

Now apply the property `ln(ab)=ln(a)+ln(b)` to get:
`y=ln(e^-x)+ln(1+x)`

Apply another property specific to the natural logarithm, `ln(e^a)=a`:
`y=-x+ln(1+x)`

We can now take the derivative with ease.

Using the sum rule, `d/dx(-x+ln(1+x))=d/dx(-x)+d/dx(ln(1+x))`. And using the fact that `d/dx(-x)=-1` and `d/dx(ln(1+x))=1/(1+x)`,
`(dy)/dx=-1+1/(1+x)`

Finally, add the fractions to get the final result:
`(dy)/dx=-(1+x)/(1+x)+1/(1+x)=-x/(1+x)`