因为ln‘t=1/t=t^-1
(t^-1)‘=-t^-2=-1/t^2
y‘ =ln‘ (1+x'2)=2x/(1+x'2)
y’’=[2x(1+x‘2)^-1]‘
=2(1+x‘2)^-1-2x(1+x‘2)^-2*2x
=2(1-x‘2)/(1+x‘2)^2
因为ln‘t=1/t=t^-1
(t^-1)‘=-t^-2=-1/t^2
y‘ =ln‘ (1+x'2)=2x/(1+x'2)
y’’=[2x(1+x‘2)^-1]‘
=2(1+x‘2)^-1-2x(1+x‘2)^-2*2x
=2(1-x‘2)/(1+x‘2)^2