已知x²+y²=e^y+x-2y-1；求y'.解一：2x+2yy'=(e^y)y'+1-2y'(2y-e^y+2)y'=1-2x∴y'=(1-2x)/(2y-e^y+2)解二：作F(x，y)=x²+y²-e^y-x+2y+1=0y'=dy/dx=-(∂F/∂x)/(∂F/∂y)=-(2x-1)/(2y-e^y+2)=(1-2x)/(2y-e^y+2)
供参考。
