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Y=x√(1-x^2 )+arcsinx,dy=?
人气:431 ℃ 时间:2020-09-25 13:27:00
解答
y'=(√(1-x^2)+x*(1/2)(1-x^2)^(-1/2)*(-2x)+1/√(1-x^2)
=(√(1-x^2)-x^2/√(1-x^2)+1/√(1-x^2)
=(1-x^2-x^2)/√(1-x^2)+1/√(1-x^2)
=(2-2x^2)/√(1-x^2)
=2√(1-x^2),
∴dy=2√(1-x^2)dx.
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