设y=arcsinx^2则dy=?

2025-06-26 10:54:45
推荐回答(1个)
回答1:

y= (arcsinx)^2 or y = arcsin(x^2)
case 1:
y= (arcsinx)^2
dy = 2 [(arcsinx) /√(1-x^2)] dx
case 2:
y= arsin(x^2)
dy = [2x/√(1-x^4)] dx,1,