$\begin{align}  & y=\frac{x(x+1)}{(x+2)(x+3)} \\ & \text{Let} \\ & \text{ }u=x(x+1)={{x}^{2}}+x,\text{  } \\ & \frac{du}{dx}=2x+1 \\ & v=(x+2)(x+3)={{x}^{2}}+5x+6 \\ & \frac{dv}{dx}=2x+5 \\ & \text{Using quotient rule} \\ & \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{{v}^{2}}} \\ & \frac{dy}{dx}=\frac{({{x}^{2}}+5x+6)(2x+1)-({{x}^{2}}+x)(2x+5)}{{{({{x}^{2}}+5x+6)}^{2}}} \\ & \frac{dy}{dx}=\frac{2{{x}^{3}}+10{{x}^{2}}+12x+{{x}^{2}}+5x+6-2{{x}^{3}}-5{{x}^{2}}-2{{x}^{2}}-5x}{{{({{x}^{2}}+5x+6)}^{2}}} \\ & \frac{dy}{dx}=\frac{4{{x}^{2}}+12x+6}{{{({{x}^{2}}+5x+6)}^{2}}} \\\end{align}$