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

1 Answer
Sonnhard
Jul 1, 2018

#f'(x)=-(3*x^2-2x)/(ln(x^3-x^2))^2*1/(x^3-x^2)+(3*x^2-2x)/(x^3-x^2)#

Explanation:

At first we will simplify #f(x)#:
#ln(1/(x^3-x^2))=ln(1)-ln(x^3-x^2)=-ln(x^3-x^2)#
we use that

#(ln(x))'=1/x#
#f(x)=(ln(x^3-x^2))^(-1)+ln(x^3-x^2)#
so we get

#f'(x)=-(3x^2-2x)/(ln(x^3-x^2))^2*1/(x^3-x^2)+(3x^2-2x)/(x^3-x^2)#

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