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

Question

Luke Alvarez · Accepted Answer

#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))#

Luke Alvarez · Answer

undefined To differentiate $f(x) = \frac{1}{\ln(x^2 - x^3 + x^4)}$, use the chain rule. First, differentiate the function inside the natural logarithm, then apply the chain rule to the natural logarithm itself. The derivative is:

$$f'(x) = -\frac{1}{{\ln(x^2 - x^3 + x^4)}} \cdot \frac{{d}}{{dx}}(x^2 - x^3 + x^4) \cdot \frac{1}{{x^2 - x^3 + x^4}}$$

Compute $\frac{{d}}{{dx}}(x^2 - x^3 + x^4)$ to get $2x - 3x^2 + 4x^3$, then substitute into the expression to get the final derivative.