Answer
The Derivative is:
$v'=-\frac{1}{x^2}+2x^{-\frac{3}{2}}$
Work Step by Step
$v=\frac{1+x-4\sqrt{x}}{x}$
Using Quotient Rule:
$y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$
$v'=\frac{(0+(1)x^{1-1}-(4)\frac{1}{2\sqrt{x}})(x)-(1+x-4\sqrt{x})((1)x^{1-1})}{(x)^2}$
$v'=\frac{(1-\frac{2}{\sqrt{x}})(x)-(1+x-4\sqrt{x})}{x^2}$
$v'=\frac{\frac{(\sqrt{x}-2)(x)}{\sqrt{x}}-1-x+4\sqrt{x}}{x^2}$
$v'=\frac{\frac{x\sqrt{x}-2x-\sqrt{x}-x\sqrt{x}+4x}{\sqrt{x}}}{x^2}$
$v'=\frac{\frac{2x-\sqrt{x}}{\sqrt{x}}}{x^2}$
$v'=\frac{2x-\sqrt{x}}{\sqrt{x}}\cdot \frac{1}{x^2}$
$v'=\frac{\sqrt{x}(2\sqrt{x}-1)}{x^2\sqrt{x}}$
$v'=\frac{2\sqrt{x}-1}{x^2}$
$v'=-\frac{1}{x^2}+2x^{-\frac{3}{2}}$
