$\displaystyle f(x) = 2x^2+\frac{8}{x^2}$
$\displaystyle f'(x) = 4x +16x^{-1} => 4x+\frac{16}{x} $
$\displaystyle f''(x) = 4 -16 $
Is that right?
Hi el123!
No, that is not correct. You tried to integrate or something.
If
$\displaystyle g(x) := 8*\frac{1}{x^2} = 8*x^{-2} $
then
$\displaystyle g'(x) = -2*8*x^{-2-1} =-2*8*x^{-3} = -16*\frac{1}{x^3}$
Thus $\displaystyle f'(x) = 4x -16*\frac{1}{x^3}$
Do you think you find the second derivative on your own?
Yours
Rapha