Answer
$F'(x) = 2x\sqrt{3x^2+1} $
Work Step by Step
We begin by finding $g'(x)$ by applying the power rule:
$g(x) = x^2-1$
$g'(x) = 2x$
We apply the chain rule:
$F'(x) = \frac{d}{dx} f(g(x)) = f'(g(x)) *g'(x) = \sqrt{3g(x)+4} * g'(x) = \sqrt{3(x^2-1)+4} * (2x) = \sqrt{3x^2+1} * (2x) = 2x\sqrt{3x^2+1} $
