I am having trouble finding the second derivative of this function. A step by step solution would be greatly appreciated.
$\displaystyle f(x) = 3(2-x^2)^3$
$\displaystyle y = f(x) = 3(2 - x^2)^3$.
Let $\displaystyle u = 2 - x^2$ so that $\displaystyle y = 3u^3$.
$\displaystyle \frac{du}{dx} = -2x$
$\displaystyle \frac{dy}{du} = 9u^2$
$\displaystyle = 9(2 - x^2)^2$.
Therefore $\displaystyle f'(x) = \frac{dy}{dx} = -18x(2 - x^2)^2$.
Now $\displaystyle f''(x) = \frac{d}{dx}\left(\frac{dy}{dx}\right)$
$\displaystyle = -18x\frac{d}{dx}[(2 - x^2)^2] + (2 - x^2)^2\frac{d}{dx}(-18x)$
$\displaystyle = -18x(-2x)(2)(2 - x^2) - 18(2 - x^2)^2$
$\displaystyle = 72x^2(2 - x^2) - 18(2 - x^2)^2$
$\displaystyle = 18(2-x^2)[4x^2 - (2 - x^2)]$
$\displaystyle = 18(2-x^2)(3x^2 - 2)$.
I'm grateful for your post, but I think you made a mistake. Shouldn't it be $\displaystyle f'(x) = \frac{dy}{dx} = -18x(2 - x^2)^2$?
BTW, the answer is $\displaystyle 18(2-x^2)(5x^2 - 2)$. I'm not sure where you went wrong with your method.
Here's what I did;
y' = 9(2-x^2)^2 (-2x)
y'' = 18(2-x^2)(-2x)(-2)
At this point I realized my method wasn't matching up to the given answer so I stopped. Could somebody tell me where I went wrong?