Question

Find the nth differential cofficent of x^x-1logx

Answer 1

F1n-1  = xn-2 + (n-1)xn-2logx=  xn-2 + (n-1)Fn-2differentiate again to get:F2n-1 = (n-2)xn-3  + (n-1) xn-3 +(n-1)(n-2)Fn-3similarly we can say that,Fnn-1 = (n-1)(n-2)(n-3) ...3.2.1 xn-n=(n-1)!