I have two function T(n), how do i compare which are asymptotically better

    1 - T(n) = n^(1/2) T(n^(1/2)) + 3 n, T(1) = 1, T(2) = 1;


    2 - T(n) = 3 T(n/3) + 2n log n, T(1) = 1, T(2) = 1.

For the first function i guess there is `O(sqrt(n)*sqrt(n))` for loop, and `O(n)` for the `c`; which becomes `O(n^2)` totally
For the second one, the Master's theorem is usable, but as I assume the complexity becomes O(n*nlogn) <==> `O(n^2 * logn)` => `O(n)` for loop, a `O(nlogn)` for `c`

So if comparing both `O()'s` we can totally see that

    O(n^2) < O(n^2 * logn)

Was there some mistakes, or this is not how recurrance unrolling is being done?