`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))).`

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