求xy-sin(πy^2)=0在（0,-1）的二阶导数.
问题描述：
求xy-sin(πy^2)=0在（0,-1）的二阶导数.
答
两边对x求导得
y+xy'-cos(πy^2)*2πy*y'=0
把（0，-1）代入得
-1-2πy'=0
y'=-1/(2π)
两边再对x求导得
y'+y'+xy''+sin(πy^2)*(2πy*y')^2-cos(πy^2)*2π(y')^2-cos(πy^2)*2πy*y''=0
把x=0,y=-1,y'=-1/(2π)代入
自己代入吧
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