两边对x求导:y+xy'-cos(πy2) (2πyy')=0 y'[x-2πycos(πy^2)]=-y
y' = 2πcos(πy^2) - x/y (1)
2xy+x^2 y'+ 2yy'cos(y^2)=0 y'[x^2+2ycos(y^2)]=-2xy
y' = -2xy / [x^2+2ycos(y^2)] (2)
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