Question #c6810

1 Answer
Oct 1, 2017

y_n=(d^ny)/dx^n=2n(2n-1)(2n-2)...(n+1)x^(n)

Explanation:

y=x^(2n)
First Derivative
y_1=dy/dx=2nx^(2n-1)

Second Derivative
y_2=(d^2y)/dx^2=2n(2n-1)x^(2n-2)

Hence
n^"th" derivative
y_n=(d^ny)/dx^n=2n(2n-1)(2n-2)...(2n-(n-1))x^(2n-(n)
y_n=(d^ny)/dx^n=2n(2n-1)(2n-2)...(n+1)x^(n)