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