Question

What are the first and second derivatives of `f(x)=ln(e^-x + xe^-x)`?

Answer 1

`f'(x) = -1 + 1/(1+x)`
`f''(x) = -1/(1+x)^2`

Explanation:

In this case, you could build the derivative immediately using the chain rule.
However, it is much easier if you simplify the function beforehand.

Let me break it down to you. :)

First, you can factor `e^(-x)` in the argument of the `log` expression:

`ln(e^(-x) + x e^(-x)) = ln( e^(-x) ( 1 + x))`

Now, you can use the logarithm rule `log_a(x * y ) = log_a(x) + log_a(y)` leading to:

`ln(e^(-x) + x e^(-x)) = ln( e^(-x) ( 1 + x)) = ln(e^(-x)) + ln(1+x)`

As next, please remember that `ln(x)` and `e^x` are inverse functions, so `ln(e^x) = x` and `e^ln(x) = x` holds. With this knowledge, we can simplify further:

`ln(e^(-x) + x e^(-x)) = ln( e^(-x) ( 1 + x)) = ln(e^(-x)) + ln(1+x) = -x + ln(1+x)`

This means that our function can be simplified in:

`f(x) = - x + ln(1+x)`

Now we can compute the derivatives. :)

`f'(x) = -1 + 1/(1+x) = -1 + (1+x)^(-1)`

`f''(x) = -1/(1+x)^2 = - (1+x)^(-2)`

Hope that this helped!