Question

解答该题，写出具体过程

Answer 1

方程两边对x求导，其中y看作x的函数y=(x)，
-sin(xy)(y+xy')=2xy(y+xy')
将y'解出来，
y'= - (y·sin(xy)+2xy^2)/(2x^2y+x·sin(xy))

追答

dy=y'dx

Answer 2

dcos(xy)=d(x²y²)
-sin(xy)d(xy)=y²dx²+x²dy²
-sin(xy)(ydx+xdy)=2xy²dx+2x²ydy
-ysin(xy)dx-xsin(xy)dy=2xy²dx+2x²ydy
xsin(xy)dy+2x²ydy=-dx
所以dy=-[ysin(xy)dx+2xy²]dx/[xsin(xy)+2x²y]