$\displaystyle f(x) = 2x^2+\frac{8}{x^2}$

$\displaystyle f'(x) = 4x +16x^{-1} => 4x+\frac{16}{x} $

$\displaystyle f''(x) = 4 -16 $

Is that right?

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- Sep 3rd 2009, 08:50 PMel123[SOLVED] Find first and second derivative.
$\displaystyle f(x) = 2x^2+\frac{8}{x^2}$

$\displaystyle f'(x) = 4x +16x^{-1} => 4x+\frac{16}{x} $

$\displaystyle f''(x) = 4 -16 $

Is that right? - Sep 3rd 2009, 08:55 PMRapha
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 - Sep 3rd 2009, 09:03 PMel123
is it $\displaystyle f''(x) = 4 +48x^{-4}$?

- Sep 3rd 2009, 09:06 PMRapha
- Sep 3rd 2009, 09:08 PMel123
Cheers mate.