How do you find the derivative using quotient rule of [x(3x+5)] / (1-x^2)?

Jan 24, 2016

$f ' \left(x\right) = \frac{5 {x}^{2} + 6 x + 5}{1 - {x}^{2}} ^ 2$

Explanation:

First, simplify the numerator.

$f \left(x\right) = \frac{3 {x}^{2} + 5 x}{1 - {x}^{2}}$

Now, according to the quotient rule,

$f ' \left(x\right) = \frac{\left(1 - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 5 x\right) - \left(3 {x}^{2} + 5 x\right) \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)}{1 - {x}^{2}} ^ 2$

Find each derivative through the power rule.

$f ' \left(x\right) = \frac{\left(1 - {x}^{2}\right) \left(6 x + 5\right) - \left(3 {x}^{2} + 5 x\right) \left(- 2 x\right)}{1 - {x}^{2}} ^ 2$

Distribute and simplify.

$f ' \left(x\right) = \frac{- 6 {x}^{3} - 5 {x}^{2} + 6 x + 5 + 6 {x}^{3} + 10 {x}^{2}}{1 - {x}^{2}} ^ 2$

$f ' \left(x\right) = \frac{5 {x}^{2} + 6 x + 5}{1 - {x}^{2}} ^ 2$