T(n) = $n^{1/2}$T($n^{1/2}$) + n
put n=$2^{m}$
T($2^{m}$) = $2^{m/2}$T($2^{m/2}$) + $2^{m}$
divide by $2^{m}$
$\frac{T(2^{m})}{2^{m}} = \frac{T(2^{m/2})}{2^{m/2}} + \frac{2^{m}}{2^{m}}$
put $\frac{T(2^{m})}{2^{m}}$ = S(m)
then , S(m) = S(m/2) + 1
By master theorem S(m) = $\Theta(logm)$
T($2^{m}$) = $2^{m}$.S(m) = $2^{m}$.$\Theta(logm)$
now m = logm
T(n)= $\Theta(nloglogn)$