-2x／3³√﹙1－x²﹚²的二阶导数,求 具体过程和用什么公式
问题描述：
-2x／3³√﹙1－x²﹚²的二阶导数,求 具体过程和用什么公式
答
y=-2x/[3* (1-x^2)^(2/3)]=(-2/3) x*[1/(1-x^2)^(2/3)]
y'=(-2/3) *(1/(1-x^2)^(2/3)) +(-2/3) [1/(1-x^2)^(2/3)]'(1-x^2)'=-2x
=(-2/3)*(1/(1-x^2)^(2/3))+(-2/3) [(-2x) *(-2/3)*(1/(1-x^2)^(5/3) )]
=(-2/3)(1/(1-x^2)^(2/3)-2x/(1-x^2)^(5/3)