# Thread: How did they get second derivative

1. ## How did they get second derivative

I know the answer, but I just need a little help getting it.

It's all in here:
http://i44.tinypic.com/o7r6yv.jpg

Thanks so much!

2. Originally Posted by janedoe
I know the answer, but I just need a little help getting it.

It's all in here:
http://i44.tinypic.com/o7r6yv.jpg

Thanks so much!

Differentiate $\displaystyle \frac{8-3t^2}{(8+t^2)^3} = (8-3t^2)(8+t^2)^{-3}$

The derivative is:

$\displaystyle -6t (8+t^2)^{-3} + (8-3t^2)(-3)(2t)(8+t^2)^{-4}$

Put in common denominator:

$\displaystyle \frac{-6t(8+t^2)}{(8+t^2)^4} + \frac{-6t(8-3t^2)}{(8+t^2)^4}=$
$\displaystyle \frac{-48t-6t^3-48t+18t^3}{(8+t^2)^4} =$
$\displaystyle \frac{-96t+12t^3}{(8+t^2)^4}=$
$\displaystyle \frac{-12t(8-t^2)}{(8+t^2)^4}$

Good luck!

3. $\displaystyle S''(t) = \frac {80000*(8+t^{2})^{2} - 80000t*2(8+t^{2})2t}{(8+t^{2})^{4}} = \frac {80000(8+t^{2})\Big((8+t^{2})-4t^{2}\Big)}{(8+t^{2})^{4}}$

$\displaystyle = \frac {80000(8-3t^{2})}{(8+t^{2})^{3}}$