f(x)=ln[(squareroot(x^2+1)) / (x(2x^3-1)^2)]
Quotient Rule is where im starting off at but i can't get past that. Need help on this problem.Thanks.
Remember that $\displaystyle \frac{d}{dx}ln(f(x))=\frac{f'(x)}{f(x)}$.$\displaystyle
\ln{\sqrt{x^2+1}}-\ln{(x(2x^3-1)^2)} = \frac{1}{2}\ln{(x^2+1)}-\ln{(x)}-2\ln{(2x^3-1)}
$
For example (here's the first one!) $\displaystyle \frac{d}{dx} \frac{1}{2}ln(x^2+1)=\frac{1}{2}. \frac{2x}{x^2+1}=\frac{x}{x^2+1}$.
All you need to do now is repeat this process for each of the other ones.