Question

Question #c6810

Answer 1

`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)`