积法则+链式
y'=x'[arcsin (x/2)]+x[arcsin (x/2)]'
=arcsin (x/2)+x*[1/根号(1-(x/2)^2)]*(x/2)'
=arcsin (x/2)+x/[2*根号(1-(x/2)^2)]那如果再加上一个平方呢？y=x[arcsin (x/2)]²链式多一步而已y'=x'[arcsin (x/2)]^2+x{[arcsin (x/2)]^2}'=[arcsin (x/2)]^2+x*2arcsin(x/2)*[arcsin (x/2)]'=[arcsin (x/2)]^2+x*2arcsin(x/2)[1/根号(1-(x/2)^2)]*(x/2)'=[arcsin (x/2)]^2+xarcsin(x/2)/[根号(1-(x/2)^2)]