## Calculus (3rd Edition)

$$\frac{d}{dx}\frac{y}{y+1}= \frac{1}{(y+1)^2}\frac{dy}{dx}.$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Using the quotient and chain rules, we have $$\frac{d}{dx}\frac{y}{y+1}=\frac{(y+1)\frac{dy}{dx}-y\frac{dy}{dx}}{(y+1)^2}= \frac{1}{(y+1)^2}\frac{dy}{dx}.$$