Calculus (3rd Edition)

$$g_u(1,2)= \frac{1}{3}+\ln 3 .$$
Since $g(u,v)=u\ln (u+v)$, then by the product rule, we have $$g_u=\ln (u+v)+u\frac{1}{u+v}=\ln (u+v)+\frac{u}{u+v}.$$ Hence, we get $$g_u(1,2)=\ln (1+3)+\frac{1}{1+2}=\frac{1}{3}+\ln 3 .$$