# How do you differentiate  y=1/ln(x^3-x^2)-ln(1/(x^3-x^2)) using the chain rule?

Jul 1, 2018

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

#### Explanation:

At first we will simplify $f \left(x\right)$:
$\ln \left(\frac{1}{{x}^{3} - {x}^{2}}\right) = \ln \left(1\right) - \ln \left({x}^{3} - {x}^{2}\right) = - \ln \left({x}^{3} - {x}^{2}\right)$
we use that

$\left(\ln \left(x\right)\right) ' = \frac{1}{x}$
$f \left(x\right) = {\left(\ln \left({x}^{3} - {x}^{2}\right)\right)}^{- 1} + \ln \left({x}^{3} - {x}^{2}\right)$
so we get

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