## University Calculus: Early Transcendentals (3rd Edition)

$$v'=\frac{t^2-2t-1}{(1+t^2)^2}$$
$$v=(1-t)(1+t^2)^{-1}=\frac{1-t}{1+t^2}$$ Apply the Derivative Quotient Rule, we have $$v'=\frac{(1-t)'(1+t^2)-(1-t)(1+t^2)'}{(1+t^2)^2}$$ $$v'=\frac{(0-1)(1+t^2)-(1-t)(0+2t)}{(1+t^2)^2}$$ $$v'=\frac{-(1+t^2)-(1-t)\times2t}{(1+t^2)^2}$$ $$v'=\frac{-1-t^2-(2t-2t^2)}{(1+t^2)^2}$$ $$v'=\frac{-1-t^2-2t+2t^2}{(1+t^2)^2}$$ $$v'=\frac{t^2-2t-1}{(1+t^2)^2}$$