# How do you find the derivative for Y=(2x^7+3x)/(X^3-8)?

$\frac{d}{\mathrm{dx}} \left[\frac{2 {x}^{7} + 3 x}{{x}^{3} - 8}\right] = \frac{\left({x}^{3} - 8\right) \left(14 x + 3\right) - \left(2 {x}^{7} + 3 x\right) \left(3 {x}^{2}\right)}{{x}^{3} - 8} ^ 2$
$\therefore \frac{d}{\mathrm{dx}} \left[\frac{2 {x}^{7} + 3 x}{{x}^{3} - 8}\right] = \frac{\left({x}^{3} - 8\right) \left(14 x + 3\right) - \left(2 {x}^{7} + 3 x\right) \left(3 {x}^{2}\right)}{{x}^{3} - 8} ^ 2$