※ 引述《TECO2oo (麻雀)》之銘言：
: 算了很久，還是毫無頭緒......？
: 求詳解，麻煩各位了！！
: x^2y-3y^2+2xy^3-sin(xy)=5 求 dx/dy?
※ Method 1
Let F(x,y) = x^2y-3y^2+2xy^3-sin(xy)-5
Then dx/dy = -(F_y)/(F_x)
= [ 6y - x^2 - 6xy^2 + xcos(xy) ] / [ 2xy + 2y^3 - ycos(xy) ]
※ Method 2
2xydx + x^2dy - 6ydy + 2y^3dx + 6xy^2dy - cos(xy)(ydx+xdy) = 0
Then dx/dy = [ 6y - x^2 - 6xy^2 + xcos(xy) ] / [ 2xy + 2y^3 - ycos(xy) ]