Question

设y=y(x)由e^x+2y+cos(xy)=2确定,求y的导数

Answer 1

e^x+2y+cos(xy) = 2, 两边对 x 求导， 得
e^x+2y'-(y+xy')sin(xy) = 0，
解得 y' = [ysin(xy)-e^x]/[2-xsin(xy)]

Answer 2

第一步令F(x,y)=e^(x+2y)+cos(xy)-2
所以Fx=(e^(x+2y))'*(x+2y)'+(cos(xy))'*(xy)'-2'=e^(x+2y)-ysin(xy)
由偏导又得Fy=2e^(x+2y)-xsin(xy)
因为y'=dy/dx=-Fx/Fy
所以y'=e^(x+2y)-ysin(xy)/2e^(x+2y)-xsin(xy)