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

Jun 2, 2015

One law of exponents states that ${a}^{-} n = \frac{1}{a} ^ n$

Thus, we can rewrite this whole expression as $f \left(x\right) = {\left({x}^{3} - {x}^{2}\right)}^{-} 1$

Now, using the chain rule, we cna rename $u = {x}^{3} - {x}^{2}$, thus making $f \left(x\right) = {u}^{-} 1$

The chain rule states that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

So,

$\frac{\mathrm{dy}}{\mathrm{du}} = - 1 \cdot {u}^{-} 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2} - 2 x$

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

Substituting $u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left({x}^{3} - {x}^{2}\right)}^{-} 2 \left(3 {x}^{2} - 2 x\right) = - \frac{3 {x}^{2} - 2 x}{{x}^{3} - {x}^{2}} ^ 2$=$\frac{3 {x}^{2} - 2 x}{{x}^{6} - 2 {x}^{5} + {x}^{4}} = \frac{\cancel{x} \left(3 x - 2\right)}{\cancel{x} \left({x}^{5} - 2 {x}^{4} + {x}^{3}\right)}$

$\frac{3 x - 2}{{x}^{5} - 2 {x}^{4} + {x}^{3}}$