how to find n-th derivative of a function f(x)=[x^(n-1)]logx what will be n-th derivate of f(x)

Dear pritam

F_n-1(x)=x^(n-1)logx

differentiate

F¹_n-1  = x^n-2 + (n-1)x^n-2logx

          =  x^n-2 + (n-1)F_n-2

again differentiate

F²_n-1 = (n-2)x^n-3  + (n-1) x^n-3 +(n-1)(n-2)F_n-3

similerly

Fⁿ_n-1 = (n-1)(n-2)(n-3) ........3.2.1 x^n-n

           =(n-1)!

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed
solution very  quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin

this answer is wrong. The correct answer is (n-1)!/x You can get the answer by successively differentiating and finally reaching a step where (n-1)(n-2)....(n-(n-1))D^(n-(n-1){(x^(n-n)*logx} = (n-1)! d/dx(logx) = (n-1)!/x