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.

f(x) = x^(n)

Hence, f'(x) = d/dx [x^(n)] = n x^(n-1) .... (1)

Hence, f''(x) = d/dx [f'(x)] = d/dx [n x^(n-1)] = n (n - 1) x^(n-2) .... (2)

As f''(x) = 2*x*f'(x)

So plugging in the values from (1) and (2) we get,

n (n - 1) x^(n-2) = 2*x*n x^(n-1)

or (n - 1) x^(n-2) = 2*x* x^(n-1) assuming n ≠ 0

or (n -1) x^(n-2) = 2* x^(n-1+1)

or (n-1) x^(n-2) = 2*x^(n)

or (n-1)/2 = x^(n) / x^(n-2)

or (n-1)/2 = x^(n - n + 2)

or n-1 = 2*x^2

or **n =** **1 + 2*x^2**

If n is not a variable function of x, we are left with the other option of **n = 0**

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