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求xy-sin(πy^2)=0在(0,-1)的二阶导数.
人气:238 ℃ 时间:2020-06-05 21:15:39
解答
y''=-(y'(x+2πycos(πy^2))-y(1+2πy'cos(πy^2)-2πysin(πy^2)*2πyy'))/(x+2πycos(πy^2))^2
=-(1/(2π)*(-2π)-(1-2π*1/(2π)-0))/(-2π)^2
=-(-1)/((2π)^2)
=1/4π^2
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