# How do you differentiate y=(3x)/(x^2-3)?

Dec 4, 2016

The derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 \left(3 + {x}^{2}\right)}{{x}^{2} - 3} ^ 2$

#### Explanation:

This thequotient of two functions

The derivative is $\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

Here. $u = 3 x$, $u ' = 3$

and $v = {x}^{2} - 3$, $v ' = 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left({x}^{2} - 3\right) - \left(3 x\right) \left(2 x\right)}{{x}^{2} - 3} ^ 2$

$= \frac{3 {x}^{2} - 9 - 6 {x}^{2}}{{x}^{2} - 3} ^ 2$

$= \frac{- 9 - 3 {x}^{2}}{{x}^{2} - 3} ^ 2$

$= \frac{- 3 \left(3 + {x}^{2}\right)}{{x}^{2} - 3} ^ 2$