(a) \[\frac{dy}{dx}=\frac{\frac2{2x+1}(x+3)-\ln(2x+1)}{(x+3)^2}.\]

(b) Setting the numerator to zero gives \(2(x+3)=(2x+1)\ln(2x+1)\). Rearranging gives \(x=(x+3)/\ln(2x+1)-1/2\).

(c) Define \(g(x)=x-(x+3)/\ln(2x+1)+1/2\). Then \(g(2.5)=-0.0696084\), \(g(3)=0.416610\). Continuity and the sign change show a root in (2.5,3).

(d) Starting with \(x_0=2.5\), use \(x_{n+1}=(x_n+3)/\ln(2x_n+1)-0.5\). Values to 6 significant figures: 2.56961, 2.56917, 2.56917, 2.56917, 2.56917, 2.56917. Hence \(x=2.569\) to 4 significant figures.