How do you differentiate f(x) =(1-x)/(x^3-x) using the quotient rule?

Feb 7, 2017

see below

Explanation:

Using the quotient rule.

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

 = (-(x^2+x)(x-1) + (x-1)(3x^2-1))/((x(x-1)(x+1))^2

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

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

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

$= \frac{2 x + 1}{{x}^{2} + x} ^ 2$

Simplifying first

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

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

$f ' \left(x\right) = 1 {\left({x}^{2} + x\right)}^{-} 2 \left(2 x + 1\right)$ $\text{ }$ (Chain rule)

$= \frac{2 x + 1}{{x}^{2} + x} ^ 2$