$T(n) = 2T(n/2) + n \log n$ recurrence relation using master theorem Assume that 
$$T(n) = 2T\left(\frac{n}{2}\right) + \Theta(n \log n)$$
By Generic form of master theorem with $a = 2$,  $b = 2$ and $f(n) = c \, n \log n$, it can easily be proved that $T(n) = \Theta(n \log^2 n)$.
Though I am not able to understand how exactly does this derive from the master thoerem. Is there any way we can prove the generic form using the original form?

This was already answered multiple times on the site but here we go. Let $S(k)=2^{-k}T(2^k)$, then $S(k)=S(k-1)+\Theta(k)$, hence $S(k-1)+Ak\leqslant S(k)\leqslant S(k-1)+Bk$ for some finite $A$ and $B$. Iterating this yields $S(0)+A\sum\limits_{i=1}^ki\leqslant S(k)\leqslant S(0)+B\sum\limits_{i=1}^ki$, hence $S(k)=\Theta(k^2)$. 
The final step, which cannot be made rigorous without some further hypothesis (such as, that the sequence $(T(k))_k$ would be nondecreasing), is to assert that $T(2^k)=\Theta(2^kk^2)$ implies that $T(n)=\Theta(n(\log n)^2)$.
