Ok we have 2 problems here....one is just a slightly harder than the other
y=ln $\displaystyle x^2+1$
y=ln $\displaystyle (x2+1)/x^2$
Please post steps!!! Thanks
Remember: $\displaystyle \frac{d}{dx}ln(u)=\frac{1}{u}\cdot\frac{d}{dx}u$
So for the first problem, $\displaystyle u=x^{2}+1$,
$\displaystyle \frac{d}{dx}ln(x^{2}+1)=\frac{1}{x^{2}+1}\cdot\fra c{d}{dx}(x^{2}+1)=\frac{2x}{x^{2}+1}$
You should be able to do the second problem now.
What you wrote is somewhat unclear. If the problem is:
$\displaystyle \frac{ln(x^{2}+1)}{x^{2}}$ You'll need to invoke log differentiation AND the quotient rule:
$\displaystyle \frac{dy}{dx}=\frac{x^{2}\frac{2x}{x^{2}+1}-2x\,ln(x^{2}+1)}{x^{4}}$
And you can just simplify from there.
Maybe the problem was $\displaystyle \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right]$ instead of $\displaystyle \frac{d}{dx}\left[\frac{\ln(x^2+1)}{x^2}\right]$.
Because $\displaystyle \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] = \frac{d}{dx}[\ln(x^2+1)] - \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2+1}-\frac{2}{x}$ like you said.