How do you differentiate # f(x)=sqrt(ln(1/sqrt(xe^x))# using the chain rule.?

1 Answer
Mar 15, 2016

Answer:

Just chain rule over and over again.

#f'(x)=e^x(1+x)/4sqrt((xe^x)/(ln(1/sqrt(xe^x))(xe^x)^3))#

Explanation:

#f(x)=sqrt(ln(1/sqrt(xe^x)))#

Okay, this is gonna be hard:

#f'(x)=(sqrt(ln(1/sqrt(xe^x))))'=#

#=1/(2sqrt(ln(1/sqrt(xe^x))))*(ln(1/sqrt(xe^x)))'=#

#=1/(2sqrt(ln(1/sqrt(xe^x))))*1/(1/sqrt(xe^x))(1/sqrt(xe^x))'=#

#=1/(2sqrt(ln(1/sqrt(xe^x))))*sqrt(xe^x)(1/sqrt(xe^x))'=#

#=sqrt(xe^x)/(2sqrt(ln(1/sqrt(xe^x))))(1/sqrt(xe^x))'=#

#=sqrt(xe^x)/(2sqrt(ln(1/sqrt(xe^x))))((xe^x)^-(1/2))'=#

#=sqrt(xe^x)/(2sqrt(ln(1/sqrt(xe^x))))(-1/2)((xe^x)^-(3/2))(xe^x)'=#

#=sqrt(xe^x)/(4sqrt(ln(1/sqrt(xe^x))))((xe^x)^-(3/2))(xe^x)'=#

#=sqrt(xe^x)/(4sqrt(ln(1/sqrt(xe^x))))1/sqrt((xe^x)^3)(xe^x)'=#

#=sqrt(xe^x)/(4sqrt(ln(1/sqrt(xe^x))(xe^x)^3))(xe^x)'=#

#=1/4sqrt((xe^x)/(ln(1/sqrt(xe^x))(xe^x)^3))(xe^x)'=#

#=1/4sqrt((xe^x)/(ln(1/sqrt(xe^x))(xe^x)^3))[(x)'e^x+x(e^x)']=#

#=1/4sqrt((xe^x)/(ln(1/sqrt(xe^x))(xe^x)^3))(e^x+xe^x)=#

#=e^x(1+x)/4sqrt((xe^x)/(ln(1/sqrt(xe^x))(xe^x)^3))#

P.S. These exercises should be illegal.