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

1 Answer
Nov 16, 2015

#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!