Answer
$f'(x)=\dfrac{3}{4\sqrt{3x}\sqrt{4+\sqrt{3x}}}$
Work Step by Step
First, lets make an «u» substitution in order to make it easier.
$f(u) = \sqrt{u}$
$u = 4 + \sqrt{3x}$
Then lets derivate using the chain rule
$f'(u) = \dfrac{1}{2\sqrt{u}} (u') $
Now let's find u'
$u'=\dfrac{3}{2\sqrt{3x}}$
Now let's undo the substitution and simplify
$f'(x) = (\dfrac{1}{2\sqrt{4+\sqrt{3x}}})(\dfrac{3}{2\sqrt{3x}})$
And you got the answer:
$f'(x)=\dfrac{3}{4\sqrt{3x}\sqrt{4+\sqrt{3x}}}$