Answer
$F'(x) = \frac{1}{2x}$
Work Step by Step
We begin by finding $g'(x)$ by applying the power and chain rule:
$g(x) = \sqrt{3x-1}$
$g'(x) = \frac{1}{2} (3x-1)^{-\frac{1}{2}} *( \frac{d}{dx} [3x-1]) = \frac{1}{2} (3x-1)^{-\frac{1}{2}} *3 = \frac{3}{2} (3x-1)^{-\frac{1}{2}} $
We apply the chain rule:
$F'(x) = \frac{d}{dx} f(g(x)) = f'(g(x)) *g'(x) =\frac{g(x)}{(g(x))^2+1}* g'(x) = \frac{ \sqrt{3x-1}}{( \sqrt{3x-1})^2+1}* ( \frac{3}{2} (3x-1)^{-\frac{1}{2}})= \frac{3}{2} * \frac{1}{3x} = \frac{1}{2x}$
