How do you differentiate (-x^2 -5x-4 )/ (2x^2-1) using the quotient rule?

Jan 7, 2016

$\frac{- 8 {x}^{3} - 30 {x}^{2} - 14 x + 5}{2 {x}^{2} - 1} ^ 2$

Explanation:

Applying the quotient rule gives :

$f ' \left(x\right) = \frac{\left(\left(2 {x}^{2} - 1\right) . \frac{d}{\mathrm{dx}} \left(- {x}^{2} - 5 x - 4\right) - \left(- {x}^{2} - 5 x - 4\right) . \frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 1\right)\right)}{2 {x}^{2} - 1} ^ 2$

$\Rightarrow f ' \left(x\right) = \frac{\left(2 {x}^{2} - 1\right) \left(2 x - 5\right) - \left(- {x}^{2} - 5 x - 4\right) \left(4 x\right)}{2 {x}^{2} - 1} ^ 2$

now 'tidying' the numerator ie. multiply out brackets and collect like terms :

$f ' \left(x\right) = \frac{- 4 {x}^{3} + 2 x - 10 {x}^{2} + 5 - 4 {x}^{3} - 20 {x}^{2} - 16 x}{2 {x}^{2} - 1} ^ 2$

$\Rightarrow f ' \left(x\right) = \frac{- 8 {x}^{3} - 30 {x}^{2} - 14 x + 5}{2 {x}^{2} - 1} ^ 2$