## Calculus (3rd Edition)

$$y'=\dfrac{1}{1+t^2}$$
Rewrite $y$ as follows $$y=\tan^{-1}u, \quad u:=\frac{1+t}{1-t}.$$ Now, uisng the chain rule, the derivative $y'$ is given by $$y'=\frac{1}{1+u^2}\frac{du}{dt}$$ where $\frac{du}{dt}$, by the qoutient rule, is given by $$\frac{du}{dt}=\frac{(1-t)+(1+t)}{(1-t)^2}=\frac{2}{(1-t)^2}.$$ Hence, we get $$y'=\frac{2}{(1-t)^2+(1+t)^2} .$$ This can be simplified to: $$y'=\dfrac{2}{1^2-2t+t^2+1+2t+t^2}\\ y'=\dfrac{2}{2+2t^2}\\ y'=\dfrac{1}{1+t^2}$$