$\begin{align}  & y={{(2x+5)}^{2}}(x-4) \\ & y=uv\text{   (where }u\text{ and }v\text{ are function of }x) \\ & \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx} \\ & u={{(2x+5)}^{2}},\text{    }\frac{du}{dx}=2(2x+5)\times 2=4(2x+5) \\ & v=(x-4),\text{         }\frac{dv}{dx}=1 \\ & \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx} \\ & \frac{dy}{dx}={{(2x+5)}^{2}}\cdot 1+(x-4)[4(2x+5)] \\ & \frac{dy}{dx}={{(2x+5)}^{2}}+4(x-4)(2x+5) \\ & \frac{dy}{dx}=(2x+5)\left[ 2x+5+4(x-4) \right] \\ & \frac{dy}{dx}=(2x+5)\left[ 2x+5+4x-16 \right] \\ & \frac{dy}{dx}=(2x+5)(6x-11) \\\end{align}$