Question

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

Answer 1

`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)`