## University Calculus: Early Transcendentals (3rd Edition)

$$y'=(4x+3)^3(x+1)^{-4}(4x+7)$$
$$y=(4x+3)^4(x+1)^{-3}$$ The derivative of function $y$ is: $$y'=\Big[(4x+3)^4\Big]'(x+1)^{-3}+(4x+3)^4\Big[(x+1)^{-3}\Big]'$$ $$y'=\Big[4(4x+3)^3(4x+3)'\Big](x+1)^{-3}+(4x+3)^4\Big[-3(x+1)^{-4}(x+1)'\Big]$$ $$y'=16(4x+3)^3(x+1)^{-3}-3(4x+3)^4(x+1)^{-4}$$ $$y'=(4x+3)^3(x+1)^{-4}\Big(16(x+1)-3(4x+3)\Big)$$ $$y'=(4x+3)^3(x+1)^{-4}\Big(16x+16-12x-9\Big)$$ $$y'=(4x+3)^3(x+1)^{-4}(4x+7)$$