Answer
$$ y'= \frac{(1+\sec t)-t(\sec t\tan t)}{(1+\sec t)^2}.$$
Work Step by Step
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
$$ y'= \frac{(1+\sec t)-t(\sec t\tan t)}{(1+\sec t)^2}.$$
