Introducción

Teorema de la Función Implícita (sistemas de ecuaciones

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 $\mathit{\text{B}}\in {\mathbb{R}}^4$ 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 $v$

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

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

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

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

En este caso
$$det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|=det\left|\begin{array}{cc}{\displaystyle e^u}& {\displaystyle e^v}\\ u{e}^u+e^{e^u}& v{e}^v+e^v\end{array}\right|=e^u(v{e}^v+e^v)-e^v(u{e}^u+e^u)=v{e}^{u+v}-u{e}^{v+u}\ne 0$$
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
$${\displaystyle \frac{\partial u}{\partial x}=-\frac{det\left|\begin{array}{cc}-1& -ye\\ e^v& v{e}^v+e^v\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-(v{e}^v+e^v)+e^{v}ye}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{2e-e^2}{e}=2-e}$$ $${\displaystyle \frac{\partial v}{\partial x}=-\frac{det\left|\begin{array}{cc}{e}^u& u{e}^u+e^u\\ -1& -ye\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-y{e}^{u}e+u{e}^u+e^u}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{e-1}{e}=1-e^{-1}}$$
$${\displaystyle \frac{\partial u}{\partial y}=-\frac{det\left|\begin{array}{cc}-e& -xe\\ e^v& v{e}^v+e^v\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-e(v{e}^v+e^v)+e^{v}xe}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{e^2+e^2-e^2}{e}=e}$$ $${\displaystyle \frac{\partial v}{\partial y}=-\frac{det\left|\begin{array}{cc}{e}^u& u{e}^u+e^u\\ -e& -xe\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-e^{u}xe+e(u{e}^u+e^u)}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{e-e}{e}=0}$$

Teorema de la Función Implícita (n-sistemas de ecuaciones

Considere las n-funciones
$$u_i=F_i(x_1,\dots ,x_m,y_1,\dots ,y_n),i=1,\dots ,n$$ Sea $P=(\bar{x}1,\dots ,\bar{x}m,\bar{y}1,\dots ,\bar{y}n)\in {\mathbb{R}}^{n+m}$ un punto tal que $F_i(P)=0$. Suponga que en una bola $\mathit{\text{B}}\in {\mathbb{R}}^{n+m}$ de centro $P$ las funciones $F_i$ tienen (sus $m+n$) derivadas parciales continuas. Si el Jacobiano $$\frac{\partial (F_1,F_2,\dots ,F_n)}{\partial (y_1,y_2,\dots ,y_n)}=\left|\begin{array}{cccc}\frac{\partial F_1}{\partial y_1}& \frac{\partial F_1}{\partial y_2}& \cdots & \frac{\partial F_1}{\partial y_n}\\ \frac{\partial F_2}{\partial y_1}& \frac{\partial F_2}{\partial y_2}& \cdots & \frac{\partial F_2}{\partial y_n}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{\partial F_n}{\partial y_1}& \frac{\partial F_n}{\partial y_2}& \cdots & \frac{\partial F_n}{\partial y_n}\end{array}\right|\ne 0enP$$

entonces las expresiones
$F_i(x_1,\dots ,x_m,y_1,\dots ,y_n)=0$ y $G(x,y,u,v)=0$ definen funciones (implícitas)
$y_i={\phi }_i(x_1,\dots ,x_m),i=1,\dots ,n$ definidas en una vecindad $v$ de $(\bar{x}1,\dots ,\bar{x}m)$ las cuales tienen derivadas parciales
continuas en $v$ que se pueden calcular como
$$\frac{\partial y_i}{\partial x_j}=\frac{\frac{\partial (F_1,F_2,\dots ,F_n)}{\partial (y_1,\dots ,y_{i-1},x_j,y_{i+1},\dots ,y_n)}}{\frac{\partial (F_1,F_2,\dots ,F_n)}{\partial (y_1,y_2,\dots ,y_n)}}$$

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

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
$$\frac{\partial (F,G,H)}{\partial (u,v,w)}=det{\left|\begin{array}{ccc}1& 1& 1\\ 2u& -4v& 2w\\ 4u^3& -12v^3& 32w^2\end{array}\right|}_{\begin{array}{c}u=1\\ v=-1\\ w=0\end{array}}=8\ne 0$$
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

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

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

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

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 $p$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/14.0.0/svg/25fb.svg">}$