# How do you find the derivative of (x^3-3x^2+4)/x^2?

Be $y = f \frac{x}{g} \left(x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(3 {x}^{2} - 6 x\right) {x}^{\cancel{2}} - \left({x}^{3} - 3 {x}^{2} + 4\right) \left(2 \cancel{x}\right)}{x} ^ \left(\cancel{4} 3\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{3} \cancel{- 6 {x}^{2}} - 2 {x}^{3} \cancel{+ 6 {x}^{2}} - 8}{x} ^ 3 = \frac{{x}^{3} + 8}{x} ^ 3$