y=ln(x+√1+x^2)+arctanx/2那么dy=[ln(x+√1+x^2)+arctanx/2]'dx显然[ln(x+√1+x^2)]'=1/(x+√1+x^2) *(x+√1+x^2)'=1/(x+√1+x^2) *(1+x/√1+x^2)=1/√1+x^2而(arctanx/2)'=1/2 *1/(1+x^2/4)=2/(4+x^2)即dy=[1/√1+x^2 +2/(4+x^2)]dx
