# How do you calculator the derivative for (2x-5)/(x^(2)-4)?

The quotient rule says that for some $h \left(x\right) = f \frac{x}{g \left(x\right)}$, the derivative is:
$h ' \left(x\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$
$h ' \left(x\right) = \frac{\left({x}^{2} - 4\right) \left(2\right) - \left(2 x - 5\right) \left(2 x\right)}{{x}^{2} - 4} ^ 2 = \frac{2 {x}^{2} - 8 - 4 {x}^{2} + 10 x}{{x}^{2} - 4} ^ 2$
$= \frac{- 2 {x}^{2} + 10 x - 8}{{x}^{2} - 4} ^ 2$
$= \frac{\left(- 2 x + 2\right) \left(x - 4\right)}{{x}^{2} - 4} ^ 2 = \frac{2 \left(- x + 1\right) \left(x - 4\right)}{{x}^{2} - 4} ^ 2$