What is the derivative of 1/(x^3-x^2)?

May 29, 2015

One exponential rule states that $\frac{1}{a} ^ n = {a}^{-} n$, thus, we can rewrite the expression as

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

Now, we can rename $u = {x}^{3} - {x}^{2}$ and use the chain rule to derivate it, as follows:

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

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

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

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

Now, substituting $u$:

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

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