二阶导数,y=ln((x^2)+1)
发布网友
共1个回答
热心网友
y=ln(x^2+1)
y'=1/(x^2+1)×2x
=2x/(x^2+1)
y"=[2(x^2+1)-2x×2x]/(x^2+1)^2
=(2x^2+2-4x^2)/(x^2+1)^2
=(2-2x^2)/(x^2+1)^2
