Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 28

f(x)=sqrt(2+ln(x)) . To find the domain of f(x) we first need to find out what values does ln(x)take. Given below is a plot of ln(x)

We see that its domain is x>0 . Since sqrt(x) has a domain for x>=0 , 2+ln(x)>=0 . That is ln(x)>=-2 , which is x>=e^(-2)~~0.135 . Here is a plot of f(x),

Recall the Chain rule: d/dx f(g(x)) = f'(g(x)) * g'(x) . With this we can find the derivative of f(x). By now you should know that d/dx sqrt(f(x))=(1)/(2 sqrt(f(x))) f'(x) . And that d/dx ln(x)=1/x . Then it is straight forward to apply the Chain rule,
d/dx f(x)=(1)/(2 sqrt(2+ln(x)))(2+ln(x))'=
d/dx f(x) = (1)/(2x sqrt(2+ln(x))).