xy-sin(πy^2)=0 求dy/dx

xy-sin(πy^2)=0 求dy/dx
xy-sin(πy^2)=0 求dy/dx

xy-sin(πy^2)=0 求dy/dx
y+xy'-cos(πy²)2πyy'=0
y=[2πycos(πy²)-x]y'
y'=y/[2πycos(πy²)-x]
即：dy/dx=y/[2πycos(πy²)-x]

两边对x求导得
y+xy'-cos(πy^2)*(πy^2)'=0
y+xy'-πcos(πy^2)*2yy'=0