#### Answer

$f' = \frac{1}{(t+2)^2}$

#### Work Step by Step

$f = \frac{t^{2} - 1}{t^{2}+t-2}$
$f' = \frac{(t^{2}+t-2)(2t) - (t^{2}-1)(2t+1)}{(t^{2}+t-2)^{2}}$
$f' = \frac{2t^{3}+2t^{2}-4t-2t^{3}-t^{2}+2t+1}{(t^{2}+t-2)^{2}}$
$f' = \frac{t^{2}-2t+1}{(t^{2}+t-2)^{2}}$
$f' = \frac{(t-1)(t-1)}{((t-1)(t+2))^2}$
$f' = \frac{1}{(t+2)^2}$