$\textcolor{Red}{\textbf{Teorema de la Función Implícita (sistemas de ecuaciones)}}$

$\textbf{Teorema 1.}$ Considere las funciones $z_{1}=F(x,y,u,v)$ y $z_{2}=G(x,y,u,v)$. Sea $P=(x,y,u,v) \in \mathbb{R}^{4}$ un punto tal que $F(P)=G(P)=0$. Suponga que en una bola $\textit{B} \in \mathbb{R}^{4}$ de centro $P$ las funciones $F$ y $G$ tienen (sus cuatro) derivadas parciales continuas. Si el Jacobiano $\displaystyle \frac{\partial
(F,G)}{\partial (u,v)}(P)\neq0$ entonces las expresiones $F(x,y,u,v)=0$ y $G(x,y,u,v)=0$ definen funciones (implícitas) $u=\varphi_{1}(x,y)$ y $v=\varphi_{2}(x,y)$ definidas en una vecindad $v$ de $(x,y)$ las cuales tienen derivadas parciales continuas en $v$

Dadas las funciones $F$ y $G$ de las variables $u,v,x,y$ nos preguntamos cuando de las expresiones

$F(x,y,u,v)=0$
$G(x,y,u,v)=0$

podemos despejar a $u$ y $v$ en términos de $x$ y $y$ en caso de ser posible diremos que las funciones $u=\varphi_{1}(x,y)$ y $v=\varphi_{2}(x,y)$ son funciones implícitas dadas. Se espera que $\exists’$n funciones $u=\varphi_{1}(x,y)$ y
$v=\varphi_{2}(x,y)$ en

$F(x,y,\varphi_{1}(x,y),\varphi_{2}(x,y)$
$G(x,y,\varphi_{1}(x,y),\varphi_{2}(x,y)$

con $(x,y)$ en alguna vecindad $V$

Suponiendo que existen $\varphi_{1}$ y $\varphi_{2}$ veamos sus derivadas

$\displaystyle \frac{\partial F}{\partial x}\displaystyle \frac{\partial x}{\partial x}+\displaystyle \frac{\partial F}{\partial y}\displaystyle \frac{\partial y}{\partial x}+\displaystyle \frac{\partial F}{\partial u}\displaystyle \frac{\partial u}{\partial
x}+\displaystyle \frac{\partial F}{\partial v}\displaystyle \frac{\partial v}{\partial x}=0$ $~~$ $\Rightarrow$ $~~$ $\displaystyle \frac{\partial F}{\partial u}\displaystyle
\frac{\partial u}{\partial x}+\displaystyle \frac{\partial F}{\partial v}\displaystyle \frac{\partial v}{\partial x}=-\displaystyle \frac{\partial F}{\partial x}$

$\displaystyle \frac{\partial G}{\partial x}\displaystyle \frac{\partial x}{\partial x}+\displaystyle \frac{\partial G}{\partial y}\displaystyle \frac{\partial y}{\partial
x}+\displaystyle \frac{\partial G}{\partial u}\displaystyle \frac{\partial u}{\partial x}+\displaystyle \frac{\partial G}{\partial v}\displaystyle \frac{\partial v}{\partial x}=0$ $~~$ $\Rightarrow$ $~~$ $\displaystyle \frac{\partial G}{\partial u}\displaystyle \frac{\partial u}{\partial x}+\displaystyle \frac{\partial G}{\partial v}\displaystyle \frac{\partial v}{\partial x}=-\displaystyle \frac{\partial G}{\partial x}$

Lo anterior se puede ver como un sistema de 2 ecuaciones con 2 incógnitas $\displaystyle \frac{\partial u}{\partial x}$ y $\displaystyle \frac{\partial v}{\partial x}$. Aquí se ve que para que el sistema tenga solución.

$det
\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|\neq0$ en $(P)$ (el $det$ Jacobiano) y según la regla de Cramer

$\displaystyle \frac{\partial u}{\partial x}=\frac{\det
\left|\begin{array}{cc} \displaystyle -\frac{\partial F}{\partial
x}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle -\frac{\partial G}{\partial x}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}{\det
\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial(x,v)}}{\frac{\partial(F,G)}{\partial(u,v)}}$, $\displaystyle \frac{\partial
v}{\partial x}=\frac{\det \left|\begin{array}{cc} \displaystyle
\frac{\partial F}{\partial u}&\displaystyle -\frac{\partial
F}{\partial x}
\\ \displaystyle \frac{\partial G}{\partial u}&\displaystyle -\frac{\partial G}{\partial
x}\end{array}\right|}{det
\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial(u,x)}}{\frac{\partial(F,G)}{\partial(u,v)}}$.

Análogamente si derivamos con respecto a $y$ obtenemos

$\displaystyle \frac{\partial F}{\partial u}\displaystyle \frac{\partial u}{\partial y}+\displaystyle \frac{\partial F}{\partial v}\displaystyle \frac{\partial v}{\partial y}=\displaystyle \frac{\partial F}{\partial y}$

$\displaystyle \frac{\partial G}{\partial
u}\displaystyle \frac{\partial u}{\partial y}+\displaystyle
\frac{\partial G}{\partial v}\displaystyle \frac{\partial
v}{\partial y}=\displaystyle \frac{\partial G}{\partial y}$

de donde

$\displaystyle \frac{\partial u}{\partial y}=-\frac{\det
\left|\begin{array}{cc} \displaystyle -\frac{\partial F}{\partial
y}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle -\frac{\partial G}{\partial y}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}{det
\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial(y,v)}}{\frac{\partial(F,G)}{\partial(u,v)}}$, $\displaystyle \frac{\partial
v}{\partial y}=-\frac{\det \left|\begin{array}{cc} \displaystyle
\frac{\partial F}{\partial u}&\displaystyle -\frac{\partial
F}{\partial y}
\\ \displaystyle \frac{\partial G}{\partial u}&\displaystyle -\frac{\partial G}{\partial
y}\end{array}\right|}{det
\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial(u,y)}}{\frac{\partial(F,G)}{\partial(u,v)}}$.

Al determinante $det \left|\begin{array}{cc} \displaystyle
\frac{\partial F}{\partial u}&\displaystyle \frac{\partial
F}{\partial v}
\\ \displaystyle \frac{\partial G}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|$ lo llamamos Jacobiano y lo denotamos por $\displaystyle \frac{\partial (F,G)}{\partial (u,v)}$.

$\textbf{Ejemplo.}$ Analizar la solubilidad del sistema
$$e^{u}+e^{v}=x+ye$$
$$ue^{u}+ve^{v}=xye$$
$\small{Solución}$ En este caso definimos
$$F(x,y,u,v)=e^{u}+e^{v}-x-ye=0$$
$$G(x,y,u,v)=ue^{u}+ve^{v}-xye=0$$
por lo que el sistema tendra solución si $\displaystyle{\det\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|}\neq 0$

En este caso
$$\det\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|=\det\left|\begin{array}{cc} \displaystyle e^{u}&\displaystyle e^{v}\\ue^{u}+e^{e^{u}}&ve^{v}+e^{v}\end{array}\right|=e^{u}\left(ve^{v}+e^{v}\right)-e^{v}\left(ue^{u}+e^{u}\right)=ve^{u+v}-ue^{v+u}\neq 0$$
por lo tanto u y v se pueden ver en términos de x,y $\therefore$ se pueden calcular sus parciales en $u=0,~v=1,~x=1, ~y=1$ que es este caso dan
$$\displaystyle \frac{\partial u}{\partial x}=-\frac{\det
\left|\begin{matrix}-1&-ye\\e^{v}&ve^{v}+e^{v}\end{matrix}\right|}{ve^{u+v}-ue^{v+u}}=-\frac{-(ve^{v}+e^{v})+e^{v}ye}{ve^{u+v}-ue^{v+u}}\left.\right|{(0,1,1,1)}=\frac{2e-e^{2}}{e}=2-e$$ $$\displaystyle \frac{\partial v}{\partial x}=-\frac{\det \left|\begin{matrix}e^{u}&ue^{u}+e^{u}\\-1&-ye\end{matrix}\right|}{ve^{u+v}-ue^{v+u}}=-\frac{-ye^{u}e+ue^{u}+e^{u}}{ve^{u+v}-ue^{v+u}}\left.\right|{(0,1,1,1)}=\frac{e-1}{e}=1-e^{-1}$$
$$\displaystyle \frac{\partial u}{\partial y}=-\frac{\det
\left|\begin{matrix}-e&-xe\\e^{v}&ve^{v}+e^{v}\end{matrix}\right|}{ve^{u+v}-ue^{v+u}}=-\frac{-e(ve^{v}+e^{v})+e^{v}xe}{ve^{u+v}-ue^{v+u}}\left.\right|{(0,1,1,1)}=\frac{e^{2}+e^{2}-e^{2}}{e}=e$$ $$\displaystyle \frac{\partial v}{\partial y}=-\frac{\det \left|\begin{matrix}e^{u}&ue^{u}+e^{u}\\-e&-xe\end{matrix}\right|}{ve^{u+v}-ue^{v+u}}=-\frac{-e^{u}xe+e(ue^{u}+e^{u})}{ve^{u+v}-ue^{v+u}}\left.\right|{(0,1,1,1)}=\frac{e-e}{e}=0$$

$\textcolor{Red}{\textbf{Teorema de la Función Implícita (n-sistemas de ecuaciones)}}$

Considere las n-funciones
$$u_{i}=F_{i}(x_{1},…,x_{m},y_{1},…,y_{n}),~i=1,…,n$$ Sea $P=(\overline{x}{1},…,\overline{x}{m},\overline{y}{1},…,\overline{y}{n}) \in \mathbb{R}^{n+m}$ un punto tal que $F_{i}(P)=0$. Suponga que en una bola $\textit{B} \in \mathbb{R}^{n+m}$ de centro $P$ las funciones $F_{i}$ tienen (sus $m+n$) derivadas parciales continuas. Si el Jacobiano $$ \frac{\partial(F_{1},F_{2},…,F_{n})}{\partial (y_{1},y_{2},…,y_{n})}=\left|\begin{matrix}\frac{\partial F_{1}}{\partial y_{1}}&\frac{\partial F_{1}}{\partial y_{2}}&\cdots&\frac{\partial F_{1}}{\partial y_{n}}\\ \frac{\partial F_{2}}{\partial y_{1}}&\frac{\partial F_{2}}{\partial y_{2}}&\cdots&\frac{\partial F_{2}}{\partial y_{n}}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial F_{n}}{\partial y_{1}}&\frac{\partial F_{n}}{\partial y_{2}}&\cdots&\frac{\partial F_{n}}{\partial y_{n}} \end{matrix}\right|\neq0~en~~P$$

entonces las expresiones
$F_{i}(x_{1},…,x_{m},y_{1},…,y_{n})=0$ y $G(x,y,u,v)=0$ definen funciones (implícitas)
$y_{i}=\varphi_{i}(x_{1},…,x_{m}),~i=1,…,n$ definidas en una vecindad $v$ de $(\overline{x}{1},…,\overline{x}{m})$ las cuales tienen derivadas parciales
continuas en $v$ que se pueden calcular como
$$\frac{\partial y_{i}}{\partial x_{j}}=\frac{\frac{\partial(F_{1},F_{2},…,F_{n})}{\partial (y_{1},…,y_{i-1},x_{j},y_{i+1},…,y_{n})}}{\frac{\partial (F_{1},F_{2},…,F_{n})}{\partial (y_{1},y_{2},…,y_{n})}}$$

$\textbf{Ejemplo.}$ Considere las ecuaciones
$$\begin{matrix}
F(x,y,u,v,w)=x+y+u+v+w=0 \\
G(x,y,u,v,w)=x^{2}-y^{2}+u^{2}-2v^{2}+w^{2}+1=0 \\
H(x,y,u,v,w)=x^{3}+y^{3}+u^{4}-3v^{4}+8w^{4}+2=0
\end{matrix}$$

En el punto $P=(1,-1,1,-1,0)$, se tiene $F(P)=G(P)=H(P)=0$. Todas las derivadas parciales de F, G, H son continuas. Se tiene además que
$$\frac{\partial (F,G,H)}{\partial (u,v,w)}=\det\left|\begin{matrix}1&1&1\\ 2u&-4v&2w\\ 4u^{3}&-12v^{3}&32w^{2}\end{matrix}\right|_{\begin{matrix}u=1\\ v=-1\\ w=0\end{matrix}}=8\neq 0$$
Entonces el teorema asegura que en torno a P podemos despejar $u,v,w$ en términos de $x,y$ y establecer funciones
$$u=u(x,y),~v=v(x,y),~w=w(x,y)$$
las cuales tienen derivadas parciales continuas en una vecindad de $(1,-1)$ que se pueden calcular

$$\frac{\partial u}{\partial x}=-\frac{\frac{\partial (F,G,H)}{\partial (x,v,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}},\frac{\partial u}{\partial y}=-\frac{\frac{\partial (F,G,H)}{\partial (y,v,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}}$$

$$\frac{\partial v}{\partial x}=-\frac{\frac{\partial (F,G,H)}{\partial (u,x,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}},\frac{\partial v}{\partial y}=-\frac{\frac{\partial (F,G,H)}{\partial (u,y,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}}$$

$$\frac{\partial w}{\partial x}=-\frac{\frac{\partial (F,G,H)}{\partial (u,v,x)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}},~~\frac{\partial w}{\partial y}=-\frac{\frac{\partial (F,G,H)}{\partial (u,v,y)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}}$$