$$f(x) = \ln(x^2 − 18x)$$
So first I used the natrual log rules of $\ln(A-B)=\frac{\ln(A)}{\ln(B)}$ and $\ln(x^n)=n\ln(x)$ to rewrite the equation as:
$$f(x)=\frac{2\ln(x)}{\ln(18x)}$$
Then I used the quotient rule, but I think that's where I messed up. Is there an easier way to solve this? And if not, how do you use the quotient rule correctly here? My last question is about solving for the domain. I know the denominator can't be zero, but is there anything else it can't be? Help would be much appreciated.
Best Answer
So for the interval of definition, you need $\log$ to be defined, which requires $$ x^2-18x>0\Rightarrow x\in (-\infty,0,)\cup (18,+\infty) $$ You can just differentiate directly, but do review your log rules. I think those are the source of your difficulties. $$ f(x)=\ln(x^2-18x)\Rightarrow f'(x)=\frac{2x-18}{x^2-18x} $$