# How do you differentiate y=(5-2x+3x^2)/(x^2+3)?

Nov 6, 2016

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{2 \left({x}^{2} + 4 x - 3\right)}{{\left({x}^{2} + 3\right)}^{2}}$

#### Explanation:

The quotient rule says:
if $y = \setminus \frac{a}{b}$ then $\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{b \setminus \cdot \setminus \frac{d}{\mathrm{dx}} \left[a\right] - a \setminus \cdot \setminus \frac{d}{\mathrm{dx}} \left[b\right]}{{b}^{2}}$
so
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\left({x}^{2} + 3\right) \setminus \cdot \setminus \frac{d}{\mathrm{dx}} \left(5 - 2 x + 3 {x}^{2}\right) - \left(5 - 2 x + 3 {x}^{2}\right) \setminus \cdot \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} + 3\right)}{{\left({x}^{2} + 3\right)}^{2}}$
calculate the derivatives
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\left({x}^{2} + 3\right) \left(6 x - 2\right) - \left(5 - 2 x + 3 {x}^{2}\right) \left(2 x\right)}{{\left({x}^{2} + 3\right)}^{2}}$
expand
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\left(6 {x}^{3} + 18 x - 2 {x}^{2} - 6\right) - \left(10 x - 4 {x}^{2} + 6 {x}^{3}\right)}{{\left({x}^{2} + 3\right)}^{2}}$

more expanding
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{6 {x}^{3} + 18 x - 2 {x}^{2} - 6 - 10 x + 4 {x}^{2} - 6 {x}^{3}}{{\left({x}^{2} + 3\right)}^{2}}$

group like terms
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{6 {x}^{3} - 6 {x}^{3} + 18 x - 10 x - 2 {x}^{2} + 4 {x}^{2} - 6}{{\left({x}^{2} + 3\right)}^{2}}$

simplify
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{8 x + 2 {x}^{2} - 6}{{\left({x}^{2} + 3\right)}^{2}}$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{2 \left({x}^{2} + 4 x - 3\right)}{{\left({x}^{2} + 3\right)}^{2}}$