Answer
The Derivative is:
$y'(t)=(t\space+\space2)^{-2}$
Work Step by Step
$f(t)=\frac{t^2-1}{t^2+t-2}$
Simplifying the function
$f(t)=\frac{(t+1)\cdot (t-1)}{(t-1)\cdot(t+2)}$
$f(t)=\frac{t+1}{t+2}$
Using the Quotient Rule to find the Derivative:
$y'(t)=\frac{f'(t)\cdot g(t)\space - \space f(t)\cdot g'(t)}{g^2(t)}$
$y'(t)=\frac{((1)t^{1-1}+\space 0)(t\space +\space 2)\space -\space (t\space+\space 1)((1)t^{1-1}+\space 0)}{(t\space+\space 2)^2}$
$y'(t)=\frac{(1)(t+2)\space -\space (t+1)(1)}{(t\space+\space 2)^2}$
$y'(t)=\frac{(t+2)\space -\space (t+1)}{(t\space+\space 2)^2}$
$y'(t)=\frac{t\space+\space2\space-\space t\space-\space1}{(t\space+\space2)^2}$
$y'(t)=\frac{1}{(t\space+\space2)^2}$
OR
$y'(t)=(t\space+\space2)^{-2}$
