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

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 ${\displaystyle \frac{\partial F}{\partial y}(x_0,y_0)\ne 0}$. 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 $y^{\prime }=f^{\prime }(x)={\displaystyle \frac{{\displaystyle \frac{\partial F}{\partial x}(x,y)}}{{\displaystyle \frac{\partial F}{\partial y}(x,y)}}}$, $x\in \mathcal{V}$.

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

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

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

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 ${\displaystyle \frac{\partial F}{\partial z}(x_0,y_0,z_0)\ne 0}$.
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 $$\frac{dz}{dx}(x,y)=-{\displaystyle \frac{{\displaystyle \frac{\partial F}{\partial x}(x,y)}}{{\displaystyle \frac{\partial F}{\partial z}(x,y)}}\frac{dz}{dy}(x,y)=-{\displaystyle \frac{{\displaystyle \frac{\partial F}{\partial y}(x,y)}}{{\displaystyle \frac{\partial F}{\partial z}(x,y)}}}}$$

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

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

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

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

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

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

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 $det\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\ne 0$ 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 $u={\phi }_1(x,y)$ y $v={\phi }_2(x,y)$ son funciones implícitas dadas. Se espera que ${\mathrm{\exists }}^{\prime }$n funciones $u={\phi }_1(x,y)$ y $v={\phi }_2(x,y)$ en

$F(x,y,{\phi }_1(x,y),{\phi }_2(x,y)$
$G(x,y,{\phi }_1(x,y),{\phi }_2(x,y)$

con $(x,y)$ en alguna vecindad $V$. Suponiendo que existen ${\phi }_1$ y ${\phi }_2$ 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

$det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|\ne 0$ 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 $\mathit{\text{B}}\in {\mathbb{R}}^4$ de centro $P$ las funciones $F$ y $G$ tienen (sus cuatro) derivadas parciales continuas. Si el Jacobiano ${\displaystyle \frac{\partial (F,G)}{\partial (u,v)}(P)\ne 0}$ entonces las expresiones $F(x,y,u,v)=0$ y $G(x,y,u,v)=0$ definen funciones (implícitas) $u={\phi }_1(x,y)$ y $v={\phi }_2(x,y)$ 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 $$det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|\ne 0$$ 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 ${\displaystyle \frac{\partial G}{\partial v}(p)\ne 0}$. 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
$$\frac{\partial H}{\partial u}=\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v}\frac{\partial \psi }{\partial u}=\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v}(-\frac{\frac{\partial G}{\partial u}}{\frac{\partial G}{\partial v}})=\frac{\frac{\partial F}{\partial u}\frac{\partial G}{\partial v}-\frac{\partial F}{\partial v}\frac{\partial G}{\partial u}}{\frac{\partial G}{\partial v}}\ne 0$$por lo tanto para $H(x,y,u)=0$ tenemos que existe una función $u={\phi }_1(x,y)$ y por lo tanto $v=\psi (x,y,u)=\psi (x,y,{\phi }_1(x,y,u))={\phi }_2(x,y)$ y por tanto $u,v$ se pueden expresar en términos de $x,y$ en una vecindad de $p$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/14.0.0/svg/25fb.svg">}$