# How do you find the derivative of 1/(x^3-x^2)?

Apr 7, 2016

Use either the quotient rule or the power rule (and chain) to get $- \frac{3 {x}^{2} - 2 x}{{x}^{3} - {x}^{2}} ^ 2$

#### Explanation:

Quotient

$\frac{d}{\mathrm{dx}} \left(\frac{1}{{x}^{3} - {x}^{2}}\right) = \frac{\left(0\right) \left({x}^{3} - {x}^{2}\right) - \left(1\right) \left(3 {x}^{2} - 2 x\right)}{{x}^{3} - {x}^{2}}$

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

Power (and chain)

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

$= \left(- 1\right) {\left({x}^{3} - {x}^{2}\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} - {x}^{2}\right)$

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

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

A different way to write the answer

-(3x^2-2x)/(x^3-x^2)^2 = -(x(3x-2))/(x^4(x-1)^2

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