Calculus (3rd Edition)

$$y'= \frac{(1+\sec t)-t(\sec t\tan t)}{(1+\sec t)^2}.$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(\sec x)'=\sec x\tan x$. Since $y=\frac{t}{1+\sec t}$, then by the quotient rule, the derivative $y'$ is given by $$y'= \frac{(1+\sec t)-t(\sec t\tan t)}{(1+\sec t)^2}.$$