Teorema de la Función Implícita ($f:\mathbb{R}\rightarrow\mathbb{R}$

Teorema. Considere la función $y=f(x)$. Sea $(x_{0},y_{0}) \in \mathbb{R}^{2}$ un punto tal que $F(x_{0},y_{0})=0$. Suponga que la función $F$ tiene derivadas parciales continuas en alguna bola con centro $(x_{0},y_{0})$ y que $\displaystyle \frac{\partial F}{\partial y}(x_{0},y_{0})\neq 0$. Entonces $F(x,y)=0$ se puede resolver para $y$ en términos de $x$ y definir así una función $y=f(x)$ con dominio en una vecindad de $(x_{0},y_{0})$, tal que $y_{0}=f(x_{0})$, lo cual tiene derivadas continuas en $\mathcal{V}$ que pueden calcularse como $y’=f'(x)= \displaystyle \frac{\displaystyle \frac{\partial F}{\partial x}(x,y)}{\displaystyle \frac{\partial F}{\partial y}(x,y)}$, $x \in \mathcal{V}$.

Ejercicio. Si $$y’=f'(x)=-\displaystyle \frac{\displaystyle \frac{\partial F}{\partial
x}(x,y)}{\displaystyle \frac{\partial F}{\partial y}(x,y)}$$ calcular $y^{»}$

Solución. En este caso
$$y^{»}=-\frac{\left(\frac{\partial F}{\partial y}\right)\left[\frac{\partial^{2} F}{\partial x^{2}}\frac{dx}{dx}+\frac{\partial^{2} F}{\partial y\partial x}\frac{dy}{dx}\right]-\left(\frac{\partial F}{\partial x}\right)\left[\frac{\partial^{2} F}{\partial x\partial y}\frac{dx}{dx}+\frac{\partial^{2} F}{\partial y^{2}}\frac{dy}{dx}\right]}{\left(\frac{\partial F}{\partial y}\right)^{2}}$$
$$=-\frac{\left(\frac{\partial F}{\partial y}\right)\left[\frac{\partial^{2} F}{\partial x^{2}}+\frac{\partial^{2} F}{\partial y\partial x}\left(-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\right)\right]-\left(\frac{\partial F}{\partial x}\right)\left[\frac{\partial^{2} F}{\partial x\partial y}\frac{dx}{dx}+\frac{\partial^{2} F}{\partial y^{2}}\left(-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\right)\right]}{\left(\frac{\partial F}{\partial y}\right)^{2}}$$
$$=-\frac{\left(\frac{\partial F}{\partial y}\right)^{2}\left(\frac{\partial^{2} F}{\partial x^{2}}\right)-\left(\frac{\partial^{2} F}{\partial y\partial x}\right)\left(\frac{\partial F}{\partial x}\right)\left(\frac{\partial F}{\partial y}\right)-\left(\frac{\partial F}{\partial x}\right)\left(\frac{\partial F}{\partial y}\right)\left(\frac{\partial^{2} F}{\partial x\partial y}\right)+\left(\frac{\partial F}{\partial x}\right)^{2}\left(\frac{\partial^{2} F}{\partial y^{2}}\right)}{\left(\frac{\partial F}{\partial y}\right)^{3}}$$
$$=-\frac{\left(\frac{\partial F}{\partial y}\right)^{2}\left(\frac{\partial^{2} F}{\partial x^{2}}\right)-2\left(\frac{\partial^{2} F}{\partial y\partial x}\right)\left(\frac{\partial F}{\partial x}\right)\left(\frac{\partial F}{\partial y}\right)+\left(\frac{\partial F}{\partial x}\right)^{2}\left(\frac{\partial^{2} F}{\partial y^{2}}\right)}{\left(\frac{\partial F}{\partial y}\right)^{3}}$$

Teorema de la Función Implícita ($f:\mathbb{R}^{2}\rightarrow\mathbb{R}$)

Teorema. Considere la función $F(x,y,z)$. Sea $(x_{0},y_{0},z_{0}) \in
\mathbb{R}^{3}$ un punto tal que $F(x_{0},y_{0},z_{0})=0$. Suponga que la
función F tiene derivadas parciales $\displaystyle{\frac{\partial F}{\partial x},~\frac{\partial F}{\partial y},~\frac{\partial F}{\partial z}}$ continuas en alguna bola con
centro $(x_{0},y_{0},z_{0})$ y que $\displaystyle \frac{\partial
F}{\partial z}(x_{0},y_{0},z_{0})\neq 0$.
Entonces $F(x,y,z)=0$ se puede resolver para $z$ en términos de $x,y$
y definir así una función $z=f(x,y)$ con dominio en una vecindad de
$(x_{0},y_{0},z_{0})$, tal que $z_{0}=f(x_{0},y_{0})$, lo cual tiene derivadas continuas
en $\mathcal{V}$ que pueden calcularse como $$\frac{d z}{dx}(x,y)=-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
x}(x,y)}{\displaystyle \frac{\partial F}{\partial z}(x,y)}~~~\frac{d z}{dy}(x,y)=-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
y}(x,y)}{\displaystyle \frac{\partial F}{\partial z}(x,y)}$$

Ejercicio. Si
$$\frac{d z}{dx}(x,y)=-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
x}(x,y)}{\displaystyle \frac{\partial F}{\partial z}(x,y)}$$ calcular $$\frac{\partial^{2}F}{\partial x^{2}}$$

Solución. Tenemos que
$$\frac{\partial^{2}F}{\partial x^{2}}=\frac{\partial}{\partial x}\left(-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
x}(x,y)}{\displaystyle \frac{\partial F}{\partial z}(x,y)}\right)=-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial x^{2}} \frac{dx}{d x}+ \frac{\partial^{2} F}{\partial y\partial x} \frac{dy}{dx}+ \frac{\partial^{2} F}{\partial z\partial x} \frac{dz}{dx}\right]-\left( \frac{\partial F}{\partial x}\right)\left[ \frac{\partial^{2} F}{\partial x \partial z}\frac{dx}{d x}+ \frac{\partial^{2} F}{\partial y\partial z} \frac{dy}{dx}+ \frac{\partial^{2} F}{\partial z^{2}} \frac{dz}{dx}\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial x^{2}}+ \frac{\partial^{2} F}{\partial z\partial x} \frac{dz}{dx}\right]-\left( \frac{\partial F}{\partial x}\right)\left[ \frac{\partial^{2} F}{\partial x \partial z}+\frac{\partial^{2} F}{\partial z^{2}} \frac{dz}{dx}\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial x^{2}}+ \frac{\partial^{2} F}{\partial z\partial x} \left(-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}\right)\right]-\left( \frac{\partial F}{\partial x}\right)\left[ \frac{\partial^{2} F}{\partial x \partial z}+\frac{\partial^{2} F}{\partial z^{2}}\left(-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}\right)\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)^{2} \frac{\partial^{2} F}{\partial x^{2}}-2 \frac{\partial^{2} F}{\partial z\partial x} \frac{\partial F}{\partial x}\frac{\partial F}{\partial z}+\left(\frac{\partial F}{\partial x}\right)^{2}{\frac{\partial^{2} F}{\partial z^{2}}}}{\left(\frac{\partial F}{\partial z}\right)^{3}}$$

Ejercicio. Si
$$\frac{d z}{dy}(x,y)=-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
y}(x,y)}{\displaystyle \frac{\partial F}{\partial z}(x,y)}$$ calcular $$\frac{\partial^{2}F}{\partial y^{2}}$$

Solución. tenemos que
$$\frac{\partial^{2}F}{\partial y^{2}}=\frac{\partial}{\partial y}\left(-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
y}(x,y)}{\displaystyle \frac{\partial F}{\partial z}(x,y)}\right)=-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial y\partial x} \frac{dx}{d y}+ \frac{\partial^{2} F}{\partial y^{2}} \frac{dy}{dy}+ \frac{\partial^{2} F}{\partial z\partial y} \frac{dz}{dy}\right]-\left( \frac{\partial F}{\partial y}\right)\left[ \frac{\partial^{2} F}{\partial x \partial z}\frac{dx}{d y}+ \frac{\partial^{2} F}{\partial y\partial z} \frac{dy}{dy}+ \frac{\partial^{2} F}{\partial z^{2}} \frac{dz}{dy}\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial y^{2}}+ \frac{\partial^{2} F}{\partial z\partial y} \frac{dz}{dy}\right]-\left( \frac{\partial F}{\partial y}\right)\left[ \frac{\partial^{2} F}{\partial y \partial z}+\frac{\partial^{2} F}{\partial z^{2}} \frac{dz}{dy}\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial y^{2}}+ \frac{\partial^{2} F}{\partial z\partial y} \left(-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}\right)\right]-\left( \frac{\partial F}{\partial y}\right)\left[ \frac{\partial^{2} F}{\partial y \partial z}+\frac{\partial^{2} F}{\partial z^{2}}\left(-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}\right)\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)^{2} \frac{\partial^{2} F}{\partial y^{2}}-2 \frac{\partial^{2} F}{\partial z\partial y} \frac{\partial F}{\partial y}\frac{\partial F}{\partial z}+\left(\frac{\partial F}{\partial y}\right)^{2}{\frac{\partial^{2} F}{\partial z^{2}}}}{\left(\frac{\partial F}{\partial z}\right)^{3}}$$

Ejercicio. Si
$$\frac{d z}{dy}(x,y)=-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
y}}{\displaystyle \frac{\partial F}{\partial z}}$$ calcular $$\frac{\partial^{2}F}{\partial y\partial x}$$

Solución. tenemos que
$$\frac{\partial^{2}F}{\partial y\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}\right)=\frac{\partial}{\partial y}\left(-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
x}}{\displaystyle \frac{\partial F}{\partial z}}\right)=$$
$$-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial y\partial x} + \frac{\partial^{2} F}{\partial z\partial x} \frac{\partial z}{\partial y}\right]-\left(\frac{\partial F}{\partial x}\right)\left[ \frac{\partial^{2} F}{\partial y\partial z} +\frac{\partial^{2} F}{\partial z^{2}}\frac{\partial z}{\partial y}\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$-\frac{\left( \frac{\partial F}{\partial z}\right)\left[ \frac{\partial^{2} F}{\partial y\partial x} + \frac{\partial^{2} F}{\partial z\partial x} \left(-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
y}}{\displaystyle \frac{\partial F}{\partial z}}\right)\right]-\left(\frac{\partial F}{\partial x}\right)\left[ \frac{\partial^{2} F}{\partial y\partial z} +\frac{\partial^{2} F}{\partial z^{2}}\left(-\displaystyle
\frac{\displaystyle \frac{\partial F}{\partial
y}}{\displaystyle \frac{\partial F}{\partial z}}\right)\right]}{\left(\frac{\partial F}{\partial z}\right)^{2}}$$
$$=-\frac{\left( \frac{\partial F}{\partial z}\right)^{2} \frac{\partial^{2} F}{\partial y\partial x}- \frac{\partial^{2} F}{\partial z\partial x} \frac{\partial F}{\partial y}\frac{\partial F}{\partial z}-\frac{\partial^{2} F}{\partial y\partial z} \frac{\partial F}{\partial x}\frac{\partial F}{\partial z}+\frac{\partial^{2} F}{\partial z^{2}}\frac{\partial F}{\partial x}\frac{\partial F}{\partial y}}{\left(\frac{\partial F}{\partial z}\right)^{3}}$$

Teorema de la Función Implicita (version sistemas de ecuaciones)

Consideremos ahora el sistema

$au+bv-k_{1}x=0$
$cu+dv-k_{2}y=0$

con $a,b,c,d,k_{1},k_{2}$ constantes. Nos preguntamos cuando
podemos resolver el sistema para $u$ y $v$ en términos de $x$ y $y$.
Si escribimos el sistema como

$au+bv=k_{1}x$
$cu+dv=k_{2}y$

y sabemos que este sistema tiene solución si $det
\left|\begin{array}{cc} a&b\\
c&d\end{array}\right|\neq0$ en tal caso escribimos

$u=\displaystyle \frac{1}{det \left|\begin{array}{cc} a&b
\\c&d\end{array}\right|}(k_{1}dx-k_{2}by)$,~~$v=\displaystyle \frac{1}{det \left|\begin{array}{cc} a&b
\\c&d\end{array}\right|}(k_{2}ay-k_{1}cx)$.

Esta solución no cambiaría si consideramos

$au+bv=f_{1}(x,y)$
$cu+dy=f_{2}(x,y)$

donde $f_{1}$ y $f_{2}$ son funciones dadas de $x$ y $y$. La posibilidad de despejar las variables $u$ y $v$ en términos de $x$ y $y$ recae sobre los coeficientes de estas variables en las ecuaciones dadas.

Ahora si consideramos ecuaciones no lineales en $u$ y $v$ escribimos el sistema como

$g_{1}(u,v)=f_{1}(x,y)$
$g_{2}(u,v)=f_{2}(x,y)$

nos preguntamos cuando del sistema podemos despejar a $u$y $v$ en términos de $x$ y $y$. Mas generalmente, consideramos el problema siguiente, 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.

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

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

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 $\displaystyle \frac{\partial (F,G)}{\partial (u,v)}$.

Teorema de la Función Implícita (sistemas de ecuaciones)

Teorema 3. 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 $v$ que se pueden calcular como se menciona arriba.

Demostración. Dado que $$det
\left|\begin{array}{cc} \displaystyle \frac{\partial F}{\partial
u}&\displaystyle \frac{\partial F}{\partial v}
\\ \displaystyle \frac{\partial F}{\partial u}&\displaystyle \frac{\partial G}{\partial
v}\end{array}\right|\neq 0$$ entonces $\displaystyle{\frac{\partial F}{\partial u}(p)}$, $\displaystyle{\frac{\partial F}{\partial v}(p)}$, $\displaystyle{\frac{\partial G}{\partial u}(p)}$, $\displaystyle{\frac{\partial G}{\partial v}(p)}$ no son cero al mismo tiempo, podemos suponer sin perdida de generalidad que $\displaystyle{\frac{\partial G}{\partial v}(p)}\neq0$. Entonces la función $z_{1}=G(x,y,u,v)$ satisface las hipótesis del T.F.I y en una bola abierta con centro p, v se puede escribir como $v=\psi(x,y,u)$.

Hacemos ahora
$$H(x,y,u)=F(x,y,u,\psi(x,y,u))$$ y tenemos que
$$\frac{\partial H}{\partial u}=\frac{\partial F}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial F}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial F}{\partial u}\frac{\partial u}{\partial u}+\frac{\partial F}{\partial v}\frac{\partial \psi}{\partial u}=\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v}\frac{\partial \psi}{\partial u}$$

por otro lado
$$\frac{\partial \psi}{\partial u}=-\frac{\frac{\partial G}{\partial u}}{\frac{\partial G}{\partial v}}$$
por lo tanto
$$\frac{\partial H}{\partial u}=\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v}\frac{\partial \psi}{\partial u}=\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v}\left(-\frac{\frac{\partial G}{\partial u}}{\frac{\partial G}{\partial v}}\right)=\frac{\frac{\partial F}{\partial u}\frac{\partial G}{\partial v}-\frac{\partial F}{\partial v}\frac{\partial G}{\partial u}}{\frac{\partial G}{\partial v}}\neq0$$por lo tanto para $H(x,y,u)=0$ tenemos que existe una función $u=\varphi_{1}(x,y)$ y por lo tanto $v=\psi(x,y,u)=\psi(x,y,\varphi_{1}(x,y,u))=\varphi_{2}(x,y)$ y por tanto $u,v $ se pueden expresar en términos de $x,y$ en una vecindad de $p$. $\square$