Answer
The Derivative is:
$y'=\frac{x^2+\space x\space +\space 4}{(x\space +\space 0.5)^2}$
Work Step by Step
$y=\frac{x^2\space -\space 4}{x\space +\space 0.5}$
Using the Quocient Rule:
$y'=\frac{f'(x)\cdot g(x)\space - \space f(x)\cdot g'(x)}{g^2(x)}$
$y'=\frac{((2)x^{2-1} -\space 0)(x\space +\space 0.5)\space -\space (x^2-\space 4)((1)x^{1-1}+\space 0)}{(x\space +\space 0.5)^2}$
$y'=\frac{2x(x\space +\space 0.5)\space -\space (1)(x^2-\space 4)}{(x\space +\space 0.5)^2}$
$y'=\frac{2x^2- x^2+\space x\space+4}{(x\space +\space 0.5)^2}$
$y'=\frac{x^2+\space x\space +\space 4}{(x\space +\space 0.5)^2}$