How do you differentiate $f (x) = ln (x^{2} - x)$ $f(x)=ln(x^2-x)$ using the chain rule.?

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Answer 1 · 2016-03-04T12:23:36

$f'(x)=(2x-1)/(x^2-x)$

The chain rule for the specific case of a natural logarithm function can be derived as follows:

$d/dxln(x)=1/x" "=>" "d/dxln(g(x))=1/(g(x))*g'(x)$

So, for $f(x)=ln(x^2-x)$ , we see that

$f'(x)=1/(x^2-x)(d/dx(x^2-x))$

$f'(x)=1/(x^2-x)(2x-1)$

$f'(x)=(2x-1)/(x^2-x)$

How do you differentiate f(x)=ln(x^2-x)f(x)=ln(x2−x) f(x)=ln(x^2-x) using the chain rule.?