为自己加油

设y=y(x)是由下列方程确定的隐函数，求dy/dx．
(1)e^x=xy²-siny；
(2)y=2x+cos(x-y)；

(1)令F(x，y)e^x-xy²+siny，则 ∂F/∂x=e^x-y²，∂F/∂y=-2xy+cosy， dy/dx=-[(∂F/∂x)/( ∂F/∂y)] 所以 =-[(e^x-y²)/(-2xy+cosy)] =(e^x-y²)/(2xy-cosy) (2)令F(x，y)=2x+cos(x-y)-y，则 ∂F/∂x=2-sin(x-y)，∂F/∂y=sin(x-y)-1， dy/dx=-[(∂F/∂x)/( ∂F/∂y)] 所以 =-[2-sin(x-y)]/[sin(x-y)-1] =[2-sin(x-y)]/[1-sin(x-y)