How do you differentiate # f(x)=ln(sqrt(e^x-x))# using the chain rule.?

1 Answer
Sep 18, 2016

Answer:

#f'(x)= (e^x-1)/(2(e^x-x))#

Explanation:

#f(x)=ln(sqrt(e^x-x))#

#f'(x) = 1/(sqrt(e^x-x)) * d/dx(sqrt(e^x-x))# (Standard differential and chain rule)

#= 1/(sqrt(e^x-x)) * 1/2(e^x-x)^(-1/2) * d/dx(e^x-x)# (Power rule and chain rule)

#=1/(sqrt(e^x-x)) * 1/(2 (sqrt(e^x-x))) *(e^x-1)# (Standard differential and power rule)

#= (e^x-1)/(2(e^x-x))#