Answer
$ g’(-2) = \frac{8}{9} $
Work Step by Step
$ g(t) = \frac{t^{2}+1}{t^{2}-1}$
$ g’(t) = \frac{(t^{2}-1)(2t)-(t^{2}+1)(2t)}{(t^{2}-1)^2} $
$ g’(t) = \frac{-4t}{(t^{2}-1)^2} $
$ g’(-2) = \frac{-4(-2)}{((-2)^2-1)^2} $
$ g’(-2) = \frac{8}{9} $