Answer
$g' = \frac{x^{2}+x+4}{(x+0.5)^{2}}$
Work Step by Step
$g = \frac{x^{2}-4}{x+0.5}$
$g' = \frac{(x+0.5)(2x)- (x^{2}-4)(1)}{(x+0.5)^{2}}$
$g' = \frac{x^{2}+x+4}{(x+0.5)^{2}}$
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