Question

Hey guys, How to find the nth derivative of this equation:- y=x^2n

Answer 1

It is simple and easy. differentiate once, second time. Observe the way sequence develops. Identify how to group factors.

See document enclosed for better answer

d/dx   X^2n  =  2n X^2n-1

d^2/dx^2  X^2n  =  2n  (2n-1) X^2n-2 

d^3/dx^3  X^2n  =  2n  (2n-1) (2n -2) X^2n-3

d^n/dx^n  X^2n  =  2n  (2n-1) (2n-2) ....  [2n – (n -1) ] X^2n-n    
                 =  P(2n , n)  X^n 

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Answer 2

so
$y= e^{2nlogx} \\ \\ dy/dx = e^{2nlogx}d[(2nlogx)]/dx \\ \\ dy/dx = x^{2n}[2n(d[logx)]/dx) + logx (d(2n)/dx) \\ \\ dy/dx = x^{2n}[(2n/x) + 2logx) ]$