Derivative of a rational function via the quotient rule

Identify $f(x) = 2x+1$ and $h(x) = x-3$, so $f'(x) = 2$ and $h'(x) = 1$. Applying the quotient rule gives $g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{h(x)^2} = \frac{2(x-3) - (2x+1)(1)}{(x-3)^2}.$

Simplify the numerator: $2(x-3) - (2x+1) = 2x - 6 - 2x - 1 = -7$, so $g'(x) = \dfrac{-7}{(x-3)^2}$. Substituting $x=2$ gives $g'(2) = \dfrac{-7}{(2-3)^2} = \dfrac{-7}{1} = -7$.

The numerator simplified to negative seven, which is negative. This came from expanding two times x minus three, minus the quantity two x plus one, giving negative seven. That negative numerator is why g prime of two came out negative.