=1/√(1-(√(1-x^2)^2)) * (-x)/√(1-x^2)
=1/|x| * (-x)/√(1-x^2)
=-x/|x| * √(1-x^2)-1
求y=arcsin√(1-x^2)的微分,根据arcsinx'=1/√(1-x^2)
求y=arcsin√(1-x^2)的微分,根据arcsinx'=1/√(1-x^2)
根据arcsinx'=1/√(1-x^2),我算得是[-1/√(1-x^2)]dx
答案却是dy=[1/√(1-x^2)]dx,当-1
根据arcsinx'=1/√(1-x^2),我算得是[-1/√(1-x^2)]dx
答案却是dy=[1/√(1-x^2)]dx,当-1
数学人气：743 ℃时间：2019-08-17 22:27:18
优质解答
dy/dx
=1/√(1-(√(1-x^2)^2)) * (-x)/√(1-x^2)
=1/|x| * (-x)/√(1-x^2)
=-x/|x| * √(1-x^2)-1
=1/√(1-(√(1-x^2)^2)) * (-x)/√(1-x^2)
=1/|x| * (-x)/√(1-x^2)
=-x/|x| * √(1-x^2)-1
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