Answer
$f'(x) = \frac{-4\sqrt x-4+ \frac{1}{\sqrt x}}{(4x+1)^2}$
Work Step by Step
$f(x) = (2\sqrt x-1)(4x+1)^{-1} = \frac{2\sqrt x - 1}{4x+1}$
$f'(x) = \frac{(\frac{1}{\sqrt x})(4x+1)-(2\sqrt x-1)(4)}{(4x+1)^2} = \frac{4\sqrt x + \frac{1}{\sqrt x} -8\sqrt x-4}{(4x+1)^2} = \frac{-4\sqrt x-4+ \frac{1}{\sqrt x}}{(4x+1)^2}$
