Solving a recurrence relation with √n as parameter
In your comment you mentioned that you tried substitution but got stuck. Here's a derivation that works. The motivation is that we'd like to get rid of the $\sqrt{n}$ multiplier on the right hand side, leaving us with something that looks like $U(n) = U(\sqrt{n}) + something$. In this case, things work out very nicely:

$$\begin{align}
T(n) &amp;= \sqrt{n}\ T(\sqrt{n}) + n &amp; \text{so, dividing by $n$ we get}\\ 
\frac{T(n)}{n} &amp;= \frac{T(\sqrt{n})}{\sqrt{n}} + 1 &amp;\text{and letting $n = 2^m$ we have}\\
\frac{T(2^m)}{2^m} &amp;= \frac{T(2^{m/2})}{2^{m/2}} + 1 
\end{align}$$
Now let's simplify things even further, by changing to logs (since $\lg \sqrt{n} = (1/2)\lg{n}$). Let 
$$\begin{align}
S(m) &amp;= \frac{T(2^m)}{2^m} &amp; \text{so our original recurrence becomes}\\
S(m) &amp;= S(m/2)+1
\end{align}$$
Aha! This is a well-known recurrence with solution 
$$
S(m)=\Theta(\lg m)
$$ 
Returning to $T(\,)$, we then have, with $n=2^m$ (and so $m=\lg n$),
$$
\frac{T(n)}{n} = \Theta(\lg\,\lg n)
$$
So $T(n) =\Theta(n\,\lg\,\lg n)$.