解y'=x'[x+2]-x[x+2]'/[x+2]^2=[x+2]-x]'/[x+2]^2=2/[x+2]^2 希望对你有帮助不懂追问
把原式配成Y=(X+2-2)/(x+2)=1-1/(x+2)y'=-1/(x+2)的平方。
y'=x'[x+2]-x[x+2]'/[x+2]^2=[x+2]-x]'/[x+2]^2=2/[x+2]^2