Prove $n^\frac{1+2}{\sqrt{\log n}} = O(n \log n)$

Question

Prove that $$n^\frac{1+2}{\sqrt{\log n}} = O(n\ \log n).$$

I want to compute the two growth rates by using L'Hôpital's rule:

$$\lim_{n\to \infty} \frac{f(n)}{g(n)}$$

so I get something like this:

$$\lim_{n\to \infty}\frac{n^\frac{1+2}{\sqrt{\log n}}}{n\ \log n}$$

However the main trouble I'm having is differentiating $$ n^\frac{1+2}{\sqrt{\log n}}$$

What would be the best way to approach $f(n)$?

Thanks in advance, guys.

Answer 1

It is same as to show $n^{\frac{2}{\sqrt{\log n}}}=O(\log n)$ I think it's wrong. Let $n=e^{x^2}$, then it becomes $e^{2x}=O(x^2)$, But it's false..