Question

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

Answer 1

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.