I am working in Economic and get some trouble with the following problem. My text book did not give any definition or hint, could you please help me to verify or give me some hint?
Consider the system $u^3 + xu - 2y -z = 0 $ and $u - zv^3 + xy + xv =0$.
Can you define $u$ and $v$ as differentiable functions of $x$, $y$, $z$ in a neighborhood of $(x, y, z, u, v) = (0, 1, -1, 1, -1)$. If the answer is yes, then compute the partial derivatives of $u,v$ w.r.t $x, y, z$.
For the part of computing partial derivative, I can do it well.
In order to prove the differentiability, I have no idea.
Thank you for your help.
Let $F:\mathbb{R^5} \rightarrow \mathbb{R^2}$, $F(x,y,z,u,v)=(u^3+xu-2y-z, u-zv^3+xy+xv)$;
Note that:
$F$ is $C^{\infty}(\mathbb{R^5},\mathbb{R^2)}$ since its components are polynomials, so it is differentiable.
Our function is equal to zero at the given point, because $F(0,1,-1,1,-1)=(1^3+0-2-(-1),1-(-1)(-1)+0+0)=(0,0);$
The last two columns of the jacobian at the given point are linearly indepentent. To see this, note that $Jac(F(x,y,z,u,v))=\begin{bmatrix}u & -2 & -1 & x+3u^2 & 0\\y+v & x & -v^3 & 1 & -3zv^2+x\end{bmatrix}$, so $Jac(F(0,1,-1,1,-1))=\begin{bmatrix}1 & -2 & -1 & 1 & 0\\0 & 0 & -1 & 1 & 3\end{bmatrix}.$ Finally, $det\begin{bmatrix}1 & 0\\1 & 3\end{bmatrix}=3 \neq 0$.
Applying the implicit function theorem there exists an open set $V \subset \mathbb{R^3}$ such that $(0,1,-1) \in V$ and a $C^{\infty}$ (and therefore differentiable) function $\phi:V \rightarrow \mathbb{R^2}$ verifying: