How do you differentiate $f (x) = \frac{1}{ln (x^{2} - x^{3} + x^{4})}$ $f(x)=1/ln(x^2-x^3+x^4)$ ?

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Answer 1 · 2015-12-05T02:11:06

$f'(x)=-(4x^2-3x+2)/((x^3-x^2+x)ln^2(x^4-x^3+x^2))$

Rewrite $f(x)=(ln(x^2-x^3+x^4))^(-1)$ .

Use the chain rule to differentiate from here:

$f'(x)=-(ln(x^2-x^3+x^4))^(-2)d/dx[ln(x^2-x^3+x^4)]$

Find just $d/dx[ln(x^2-x^3+x^4)]$ .

Using the chain rule, remembering that $d/dx[ln(x)]=1/x$ , recall that $d/dx[ln(u)]=(u')/u$ .

$d/dx[ln(x^2-x^3+x^4)]=(d/dx[x^2-x^3+x^4])/(x^2-x^3+x^4)=(2x-3x^2+4x^3)/(x^2-x^3+x^4)$

Plug back in.

$f'(x)=-(ln(x^2-x^3+x^4))^(-2)((2x-3x^2+4x^3)/(x^2-x^3+x^4))$

$f'(x)=-1/(ln^2(x^2-x^3+x^4))((2x-3x^2+4x^3)/(x^2-x^3+x^4))$

$f'(x)=-(x(4x^2-3x+2))/(x(x^3-x^2+x)ln^2(x^4-x^3+x^2))$

$f'(x)=-(4x^2-3x+2)/((x^3-x^2+x)ln^2(x^4-x^3+x^2))$

How do you differentiate f(x)=1/ln(x^2-x^3+x^4)f(x)=1ln(x2−x3+x4)f(x)=1/ln(x^2-x^3+x^4)?