## Calculus (3rd Edition)

$$y'=(x+1)^2(x+4)^3(7x+16) .$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(x^n)'=nx^{n-1}$ Since $y=(x+1)^3(x+4)^4$, then by the product and chain rules, the derivative $y'$ is given by $$y'=3(x+1)^2(1)(x+4)^4+4(x+1)^3(x+4)^3(1)=3(x+1)^2(x+4)^4+4(x+1)^3(x+4)^3 =(x+1)^2(x+4)^3(3(x+4)+4(x+1))=(x+1)^2(x+4)^3(7x+16) .$$