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