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

1 Answer
Mar 4, 2016

Answer:

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

Explanation:

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)#