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设y=arcsinx^2则dy=
人气:410 ℃ 时间:2020-09-26 13:01:14
解答
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
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