# Implicit Function Theorem for $\mathbb{R}^5 \to \mathbb{R}^2$

I am trying to learn the Implicit Function Theorem and I just want to be sure I am understanding this process correctly:

Let $F=(f_1,f_2)$, where $f_1(u,v,x,y,z)=xy^5+yu^5+zv^5-1$ and $f_2(u,v,x,y,z)=x^5y+y^5u+z^5v-1$. Show $u,v$ can be solved in terms of $x,y,z$ at $(1,0,0,1,1)$ and for good measure find $\frac{\partial u}{\partial x}(0,1,1)$.

My attempt: I took the partials to get the matrix $$\begin{pmatrix} y^5 & 5xy^4-u^5 & v^5 & -5yu^4 & 5zv^4 \\ 5x^4y & x^5+5y^4u & 5z^4v & y^5 & z^5 \end{pmatrix}$$ Note that all these partials are continuous everywhere, then $F$ is $C^1$ everywhere - in particular at $(1,0,0,1,1)$. It is routine to verify $F(1,0,0,1,1)=(0,0)$. We look now at the $u,v$ partials, we have the matrix $$\begin{pmatrix} -5yu^4 & 5zv^4 \\ y^5 & z^5 \end{pmatrix}$$ so at $(1,0,0,1,1)$ we have $$\text{det} \begin{pmatrix} -5 & 0 \\ 1 & 1 \end{pmatrix} =-5\neq 0$$ so that we can solve for $u,v$ in terms of $x,y,z$ at $(1,0,0,1,1)$. Finally, we have $$\frac{\partial u}{\partial x}= -\frac{\text{det} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}}{\text{det} \begin{pmatrix} -5 & 0 \\ 1 & 1 \end{pmatrix}}= \frac{1}{5}$$ Is this correct? Or if not, where have I gone awry?

• @S.W.Cheung Thank you! I did the same thing on the rewrite but when I originally copied the problem I couldn't remember the order the problem said so I just did $(x,y,z,u,v)$. Commented Dec 30, 2015 at 6:16