If f''(x) = 2*x*f'(x), and f(x) = x^(n) what is n.

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`f''(x)=2x xx f'(x)` and `f(x)=x^n`

Consider

`f(x)=x^n` (i)

Differentiate (i) w.r.t. x, then

`f'(x)=nx^(n-1)`

`f''(x)=n(n-1)x^(n-2)`

Substitute `f'(x) and f''(x)` in given equation,

`n(n-1)x^(n-2)=2xnx^(n-1)`

`n(n-1)x^(n-2)-2xnx^(n-1)=0`

`nx^(n-2)(n-1-2x^2)=0`

So either `nx^(n-2)=0` or `n-1-2x^2=0.`

`If ` `nx^(n-2)=0`

`=> n=0` or x=0

If `n-1-2x^2=0` then n will variable as x is independent variable. So this case not possible.

Thus only posible cases are either n=0 or x=0.

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