Answer
$$ y'= \frac{-x^2 - 2x+1}{(x^2+1)^2}.$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\frac{x+1}{x^2+1}$, then using the quotient rule, the derivative $ y'$ is given by
$$ y'= \frac{(x^2+1)(1)-(x+1)(2x)}{(x^2+1)^2}= \frac{-x^2 - 2x+1}{(x^2+1)^2}.$$
