y'=1/x
y''=-1/x^2
dy=2xcosxdx
已知y=lnx,则y''= 已知y=sinx^2,则dy= 本人新手,
已知y=lnx,则y''= 已知y=sinx^2,则dy= 本人新手,
数学人气：956 ℃时间：2020-04-10 09:35:38
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