consider the equation
e^2x-y+cos(x^2+xy)-2-2y=0 (x,y) in R^2
Does the set of solutions of this equation in a neighborhood of the solution (0,0) implicitly define one of the components of the point (x,y) as a function of the other component?
If so,compute the derivative of this function (these functions?) at the point 0.
- mathmanliuLv 71 decade agoFavorite Answer
設F(x,y)= exp(2x-y)+cos(x^2+xy)-2-2y, 則
∂F/∂x = 2 exp(2x-y)- sin(x^2+xy) *(2x+y)
∂F/∂y = - exp(2x-y) - x sin(x^2+xy) - 2
(x,y)=(0,0)處, ∂F/∂x=2, ∂F/∂y= -3
兩項均不為0, 故y可視為x之函數, x亦可視為y之函數, 且
dy/dx = -[∂F/∂x] / [∂F/∂y] = 2/3
dx/dy = - [∂F/∂y] / [∂F/∂x] = 3/2