Answer
$$y'=\dfrac{1}{1+t^2}$$
Work Step by Step
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:
$$y'=\dfrac{2}{1^2-2t+t^2+1+2t+t^2}\\
y'=\dfrac{2}{2+2t^2}\\
y'=\dfrac{1}{1+t^2}$$
