# How do you differentiate f(x)=1/ln(x^2-x^3+x^4)?

Dec 5, 2015

$f ' \left(x\right) = - \frac{4 {x}^{2} - 3 x + 2}{\left({x}^{3} - {x}^{2} + x\right) {\ln}^{2} \left({x}^{4} - {x}^{3} + {x}^{2}\right)}$

#### Explanation:

Rewrite $f \left(x\right) = {\left(\ln \left({x}^{2} - {x}^{3} + {x}^{4}\right)\right)}^{- 1}$.

Use the chain rule to differentiate from here:

$f ' \left(x\right) = - {\left(\ln \left({x}^{2} - {x}^{3} + {x}^{4}\right)\right)}^{- 2} \frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2} - {x}^{3} + {x}^{4}\right)\right]$

Find just $\frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2} - {x}^{3} + {x}^{4}\right)\right]$.

Using the chain rule, remembering that $\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right] = \frac{1}{x}$, recall that $\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{u '}{u}$.

$\frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2} - {x}^{3} + {x}^{4}\right)\right] = \frac{\frac{d}{\mathrm{dx}} \left[{x}^{2} - {x}^{3} + {x}^{4}\right]}{{x}^{2} - {x}^{3} + {x}^{4}} = \frac{2 x - 3 {x}^{2} + 4 {x}^{3}}{{x}^{2} - {x}^{3} + {x}^{4}}$

Plug back in.

$f ' \left(x\right) = - {\left(\ln \left({x}^{2} - {x}^{3} + {x}^{4}\right)\right)}^{- 2} \left(\frac{2 x - 3 {x}^{2} + 4 {x}^{3}}{{x}^{2} - {x}^{3} + {x}^{4}}\right)$

$f ' \left(x\right) = - \frac{1}{{\ln}^{2} \left({x}^{2} - {x}^{3} + {x}^{4}\right)} \left(\frac{2 x - 3 {x}^{2} + 4 {x}^{3}}{{x}^{2} - {x}^{3} + {x}^{4}}\right)$

$f ' \left(x\right) = - \frac{x \left(4 {x}^{2} - 3 x + 2\right)}{x \left({x}^{3} - {x}^{2} + x\right) {\ln}^{2} \left({x}^{4} - {x}^{3} + {x}^{2}\right)}$

$f ' \left(x\right) = - \frac{4 {x}^{2} - 3 x + 2}{\left({x}^{3} - {x}^{2} + x\right) {\ln}^{2} \left({x}^{4} - {x}^{3} + {x}^{2}\right)}$