IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

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f(x)=(x^n-1)^n
1st derivative=(n^2)((x^n-1)^n-1)*x^n-1
now basically we need to bother only abt the terms in which differentiation of (x-1)^n goes on as in the other terms when u later substitue x=1 they become 0 anyway..in the nth derivative the (x-1)^n term which was there initially becomes(x-1)^0..so this term alond with its coeff and x power sometihng is the only 1 which does not become 0 for x=1..the coeff of this will be n^n*n! as on each diff 'n' is contributed due to the x^n inside the bracket and overall n! is produced due to continuous diff of the initial (...)^n..

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