Comparing the order of growth of the two functions by taking a limit and using l'hospitals rule, it seems that $\sqrt{n}$ should be O($log^3n$). Here are the steps I took:
$$\lim_{n \to ∞} \frac{(\log_{10}n)^3}{\sqrt{n}}$$
L'hospitals: $$=\lim_{n \to ∞} \frac{3(\log_{10}n)^2 \frac{1}{n\ln(10)}}{\frac{1}{2n^2}}$$
$$=\lim_{n \to ∞} \frac{3(\log_{10}n)^2 2n^2}{n\ln(10)}$$
$$=\lim_{n \to ∞} \frac{3(\log_{10}n)^2 2n}{\ln(10)} = ∞$$
I.e., since $(\log_{10}n)^3$ was in the numerator, and the limit goes to infinity, it has the greater order of growth.
Where did I go wrong?
What went wrong is that you differentiated $\sqrt{n}$ incorrectly. The result should be $(1/2)n^{-1/2}$. Then manipulation gives you a $\log^2 n$ on top, and $\sqrt{n}$ at the bottom, and you can apply L'Hospital's Rule again, and then again.
By the way, you should take the cube root first. So you will be looking at $\frac{\log n}{n^{1/6}}$. Then one step of L'Hospital's Rule does it. The limit is $0$.