# What will be $\frac {\partial F_i} {\partial x_j}$?

Suppose we are given two functional relations $y_1=f_1(x_1,x_2)$ and $y_2=f_2(x_1,x_2)$ and suppose we want to determine whether we can able to write $x_i$'s as implicit functions of $y_i$'s locally which is precisely the same argument as that was in the inverse function theorem. For doing this with the help of implicit function theorem we first write the relations in the framework of implicit function theorem. For this we introduce two functions $F_1$ and $F_2$ of four variables $x_1,x_2,y_1$ and $y_2$ defined by $F_1(x_1,x_2,y_1,y_2)=f_1(x_1,x_2)-y_1$ and $F_2(x_1,x_2,y_1,y_2)=f_2(x_1,x_2)-y_2$. Then we use implicit function theorem for system of equations to determine the condition for which we can write $x_i$'s as functions of $y_i$'s locally in an implicit way. But to do this we must have $\frac {\partial (F_1,F_2)} {\partial (x_1,x_2)} \neq 0$.

Now what is $\frac {\partial F_1} {\partial x_1}?$ My book suggests that it is $\frac {\partial f_1} {\partial x_1}$! But why is it true? Here $y_1$ is a function of $x_1$ and $x_2$. Then the partial deivative of $F_1$ w.r.t. $x_1$ should involve an additional term $-\frac {\partial y_1} {\partial x_1}$. But why is it not so? Similar problem happens for any $\frac {\partial F_i} {\partial x_j}$ for any $i,j$. Please help me in this regard.

$F_j(x_1,x_2,y_1,y_2)=f_j(x_1,x_2)-y_j$ for $j=1,2$ is to be understood as a function on $\mathbb{R}^4$ with four independent variables ( whereof three occur in the term). Of course $F_j(x_1,x_2,y_1,y_2)=0$ whenever $f_j(x_1,x_2)-y_j=0$.
By the Implicit Function Theorem for every point $(a_1,a_2,b_1,b_2)$ that satisfies $F_j(a_1,a_2,b_1,b_2)=0$ for $j=1,2$ there exists a neighborhood $U\subset\mathbb{R}^2$ of $(b_1,b_2)$ and a continuously differentiable function $g:U\rightarrow\mathbb{R}^2$ so that $F_j(g_1(y_1,y_2),g_2(y_1,y_2),y_1,y_2)=0$ for $j=1,2$ if $$D_F|_{(a,b)}=\begin{pmatrix}\frac{\partial F_1}{\partial x_1}&\frac{\partial F_1}{\partial x_2}\\\frac{\partial F_2}{\partial x_1}&\frac{\partial F_2}{\partial x_2}\end{pmatrix}|_{(a,b)}$$ is invertible. And since $F:\mathbb{R}^2\times\mathbb{R^2}\rightarrow\mathbb{R}^2,(x,y)\mapsto \begin{pmatrix}F_1(x,y)\\F_2(x,y)\end{pmatrix}$ is to be understood as a function of four independent variables one has by definition of the $F_j$ $$D_F|_{(a,b)}=\begin{pmatrix}\frac{\partial f_1}{\partial x_1}&\frac{\partial f_1}{\partial x_2}\\\frac{\partial f_2}{\partial x_1}&\frac{\partial f_2}{\partial x_2}\end{pmatrix}|_{(a,b)}$$ because $y_j$ doesn´t depend on $x$. This holds locally in a neighborhood of a point $(a,b)$ satisfying $f_j(a)=b_j$ for $j=1,2$. The point is that in the definition of $F$ the $y_j$ can take any value independently of $x_j$ but along the curves $y_j=f_j(x_1,x_2)$ there is a local description as given above possible.
• Why do we not treat $y_j$'s as functions of $x_j$'s when we take the partials of $F_j$'s? – A.Chattopadhyay Dec 16 '17 at 15:43
• @A.Chattopadhyay Because in the definition of $F:\mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}^2$ the variable $y=\begin{pmatrix}y_1\\y_2\end{pmatrix}$ can take any value in $\mathbb{R}^2$ and thus $y_j$ are independent of $x_1,x_2$ and are thus treated as constants when taking the partial derivative :$\frac{\partial y_j}{\partial x_i}=0$ for $i,j=1,2$ – Peter Melech Dec 17 '17 at 13:13
• If You would treat $y_j$ as functions of $x_1,x_2$ in the definition of $F$, the domain of $F$ would be a different, more complicated one, but this is anyway not what You want, because You are interested in an application of the IFT, which is always a local statement – Peter Melech Dec 17 '17 at 13:19