Answer
$v' = \frac{2\sqrt x-1}{x^{2}}$
Work Step by Step
$v = \frac{1+x-4\sqrt x}{x}$
Quotient rule:
$v' = \frac{x(1-2x^{-\frac{1}{2}}) - (1+x-4\sqrt x)}{x^{2}}$
$v' = \frac{x-2x^{\frac{1}{2}} - 1-x+4\sqrt x}{x^{2}}$
$v' = \frac{2\sqrt x-1}{x^{2}}$
