Q: Let $F$ be defined by $F(x, y, u, v) = (u^3+vx+y, uy+v^3-x).$At what points can we solve for $F(x, y, u, v)=0 $ in terms of x, y? Also compute $\frac{\partial u}{\partial x}$.
(And I found this pretty similar to Show equations determine local functions and determine $\frac{du}{dx}$) It's just that I have to find the points myself...
It seems that I can only find the representation of x and y in terms of u, v($x=\frac{-u^4+v^3}{1+uv}, y=\frac{-u^3-v^4}{1+uv}$), which is easier.
If I need to find u and v in terms of x, y, I have to solve the equations $u^3+vx+y=0$ and $uy+v^3-x =0 $, u and v are to the power of 3... seems impossible, I tried and failed. And if I can do this I believe computing $\frac{\partial u}{\partial x}$ isn't hard. Can anyone help me?
Consider the first equation and assume that $u,v$ are a function of $x,y$, so you have: $$F_1(x,y,u(x,y),v(x,y))=u^3(x,y)+v(x,y)\,x+y=0$$ Then, by the chain rule, you can derive with respect to $x$: $$\frac{d F_1}{d x}=\frac{\partial F_1}{\partial x} + \frac{\partial F_1}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F_1}{\partial v}\frac{\partial v}{\partial x}=0$$
Take now the second equation: $$F_2(x,y,u(x,y),v(x,y))=u(x,y)\, y+v^3(x,y)-x=0$$ and derive again with respect to $x$: $$\frac{d F_2}{d x}=\frac{\partial F_2}{\partial x} + \frac{\partial F_2}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F_2}{\partial v}\frac{\partial v}{\partial x}=0$$
Together, these two equations form a linear system from where you can solve for $\frac{\partial u}{\partial x},\frac{\partial v}{\partial x}$. But the theory of linear systems gives us conditions as to when a solution exists: the matrix of coefficients must be invertible. This is the Jacobian matrix: $$ J_f= \begin{pmatrix} \frac{\partial F_1}{\partial u} & \frac{\partial F_1}{\partial v} \\ \frac{\partial F_2}{\partial u} & \frac{\partial F_2}{\partial v} \end{pmatrix} $$ So at every point where the Jacobian is nonsingular, the implicit function theorem guarantees that the functions that define $u,v$ as functions of $x,y$ exist, although maybe you cannot express them in a closed formula.