## Calculus (3rd Edition)

$$y'= -\frac{8}{(x+1)^{3}} .$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Given $y=\frac{4x}{x+1}$, then, by using the quotient rule, we have $$y'=\frac{(x+1)(4)-4x(1)}{(x+1)^2}=\frac{4}{(x+1)^2}=4(x+1)^{-2},$$ and the seond derivative $y''$ is given by $$y'=(-2)4(x+1)^{-3}=-8(x+1)^{-3}=-\frac{8}{(x+1)^{3}} .$$