Answer
The Derivative is:
$v'=\frac{t^2-2t-1}{(1+t^2)^2}$
Work Step by Step
$v=\left(1-t\right)\space (1+t^2)^{-1}$
$v=\frac{1\space-\space t}{1\space+\space t^2}$
Using the Quotient Rule to find the Derivative:
$v'=\frac{f'(t)\cdot g(t)\space-\space f(t)\cdot g'(t)}{g^2(t)}$
$v'=\frac{(0\space-\space(1)t^{1-1})(1\space+\space t^2)\space-\space(1\space-\space t)(0\space+\space(2)t^{2-1})}{(1\space+\space t^2)^2}$
$v'=\frac{(-1)(1+t^2)-(1-t)(2t)}{(1+t^2)^2}$
$v'=\frac{-1-t^2-2t+2t^2}{(1+t^2)^2}$
$v'=\frac{t^2-2t-1}{(1+t^2)^2}$
