设y=y(x)是由下列方程确定的隐函数,求dy/dx.
(1)ex=xy2-siny;
(2)y=2x+cos(x-y);
【正确答案】:(1)令F(x,y)ex-xy2+siny,则
∂F/∂x=ex-y2,∂F/∂y=-2xy+cosy,
dy/dx=-[(∂F/∂x)/( ∂F/∂y)]
所以 =-[(ex-y2)/(-2xy+cosy)]
=(ex-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)
设y=y(x)是由下列方程确定的隐函数,求dy/dx. (1)ex=xy2-siny; (2)y=2x+cos(x-y);
- 2024-11-07 09:08:48
- 高等数学(经管类)(13125)