Funciones de dos variables

Sea $T:\mathbb{R}^{n}\longrightarrow \mathbb{R}^{m}$ una función lineal dada por la $m\times n$ matriz $A=\left( a_{ij}\right) .$ Tomamos

\begin{equation*} \left\Vert A\right\Vert =\left( \sum_{i,j=1}^{n}a_{ij}^{2}\right) ^{\frac{1}{2}}= \left( \sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^{2}\right) ^{ \frac{1}{2}}=\left(\sum_{j=1}^{n}\sum_{i=1}^{m}a_{ij}^{2}\right) ^{\frac{1}{2}}. \end{equation*}

Utilizar la desigualdad de Cauchy-Schwarz para probar que \begin{equation*} \left\Vert A\overline{x}\right\Vert \leq \left\Vert A\right\Vert \left\Vert \overline{x}\right\Vert . \end{equation*}

Si $\overline{x}=\left( x_{1},x_{2},\ldots x_{n}\right) ,$ la función $T$ está definida como

\begin{equation*} T(\overline{x})=A\overline{x}=\left( \begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{m1} & \cdots & a_{mn} \end{array} \right) \left( \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right) =\left( \begin{array}{c} \sum_{j=1}^{n}a_{1j}x_{j} \\ \vdots \\ \sum_{j=1}^{n}a_{mj}x_{j} \end{array} \right) =\left( \begin{array}{c} y_{1} \\ \vdots \\ y_{m} \end{array} \right) \end{equation*}

entonces

\begin{equation*} \left\Vert A\overline{x}\right\Vert =\sqrt{\sum_{i=1}^{m}\left( y_{i}\right) ^{2}} \end{equation*}

donde

\begin{equation*} y_{i}=\sum_{j=1}^{n}a_{ij}x_{j}=\left( a_{i1},\ldots ,a_{in}\right) \cdot \left( x_{1},\ldots ,x_{n}\right) . \end{equation*}

La desigualdad de Cauchy-Schwarz afirma que

\begin{equation*} \left\vert \overline{a}\cdot \overline{b}\right\vert \leq \left\Vert \overline{a}\right\Vert \left\Vert \overline{b}\right\Vert \end{equation*}

entonces

\begin{equation*} \left\vert \overline{a}\cdot \overline{b}\right\vert ^{2}\leq \left\Vert \overline{a}\right\Vert ^{2}\left\Vert \overline{b}\right\Vert ^{2}. \end{equation*}

Así

\begin{eqnarray*} y_{i}^{2} &=&\left( \left( a_{i1},\ldots ,a_{in}\right) \cdot \left( x_{1},\ldots ,x_{n}\right) \right) ^{2} \\ \\ &\leq &\left\Vert \overline{x}\right\Vert ^{2}\left\Vert \left( a_{i1},\ldots ,a_{in}\right) \right\Vert ^{2} \\ \\ &=&\left\Vert \overline{x}\right\Vert ^{2}\sum_{j=1}^{n}a_{ij}^{2}, \end{eqnarray*}

de donde

\begin{equation*} \left\Vert A\overline{x}\right\Vert \leq \sqrt{\sum_{i=1}^{m}\left \Vert \overline{x}\right\Vert ^{2}\sum_{j=1}^{n}a_{ij}^{2}}=\sqrt{ \left\Vert \overline{x}\right\Vert ^{2}\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^{2}}=\left\Vert \overline{x}\right\Vert \sqrt{\sum_{i=1}^{m}\sum _{j=1}^{n}a_{ij}^{2}}=\left\Vert A\right\Vert \left\Vert \overline{x} \right\Vert . \end{equation*}

QED

Una función $f:U\subset \mathbb{R}^{n}\longrightarrow \mathbb{R}^{m},$ con U abierto, es derivable en $\overline{x}_{0}\in U,$ si existe una $ m\times n$ matriz $A$ que satisface

\begin{equation} \lim_{\overline{x}\rightarrow \overline{x}_{0}}\dfrac{\left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) -A\left( \overline{x}-\overline{x}_{0}\right) \right\Vert }{\left\Vert \overline{x}- \overline{x}_{0}\right\Vert }=0. \label{limdef} \tag{1} \end{equation}

Lo anterior equivale a que para cada $\varepsilon >0$ existe $\delta >0$ tal que

\begin{equation} \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) -A\left( \overline{x}-\overline{x}_{0}\right) \right\Vert \leq \varepsilon \left\Vert \overline{x}-\overline{x}_{0}\right\Vert \label{equiv} \tag{2} \end{equation}

En tal caso $A$ es la matriz jacobiana $D_{f}\left( \overline{x} _{0}\right) $ de $f$ evaluada en $\overline{x}_{0}$.

Sea $f:U\subset \mathbb{R}^{n}\longrightarrow \mathbb{R}^{m},$ con U abierto, derivable en $\overline{x}_{0}\in U,$ entonces existen $M_{1}>0,$ $ \delta _{1}>0$ tales que \[\text{ si } \left\Vert \overline{x}-\overline{x} _{0}\right\Vert <\delta _{1} \text{ entonces } \left\Vert f\left( \overline{x} \right) -f\left( \overline{x}_{0}\right) \right\Vert \leq M_{1}\left\Vert \overline{x}-\overline{x}_{0}\right\Vert .\]

Por ser $f$ derivable en $\overline{x}_{0},$ se satisface (\ref{equiv}).

Sea $M=\left\Vert D_{f}\left( \overline{x}_{0}\right) \right\Vert $. Por el Lema 1

\begin{equation*} \left\Vert D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{ x}_{0}\right) \right\Vert \leq M\left\Vert \overline{x}-\overline{x} _{0}\right\Vert . \end{equation*}

Sea $\varepsilon =1,$ por (\ref{equiv}) existe $\delta _{1}>0$ tal que si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <$ $\delta _{1}$ entonces

\begin{equation*} \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) -D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right) \right\Vert \leq \left\Vert \overline{x}-\overline{x} _{0}\right\Vert . \end{equation*}

De donde si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta _{1} $ entonces

\begin{eqnarray*} \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right\Vert & = &\left\Vert f\left( \overline{x}\right) -f\left( \overline{x} _{0}\right) -D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}- \overline{x}_{0}\right) +D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x}_{0}\right) \right\Vert \\ &\leq &\left\Vert f\left( \overline{x}\right) -f\left( \overline{x} _{0}\right) -D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}- \overline{x}_{0}\right) \right\Vert +\left\Vert D_{f}\left( \overline{x} _{0}\right) \left( \overline{x}-\overline{x}_{0}\right) \right\Vert \\ &\leq &\left\Vert \overline{x}-\overline{x}_{0}\right\Vert +M\left\Vert \overline{x}-\overline{x}_{0}\right\Vert \\ & = &\left( M+1\right) \left\Vert \overline{x}-\overline{x}_{0}\right\Vert . \end{eqnarray*}

Tomando $M_{1}=M+1,$ entonces se cumple lo que queríamos.

QED

Sean $U\subset \mathbb{R}^{n}$ y $V\subset \mathbb{R}^{m}$ abiertos, $ f:U\subset \mathbb{R}^{n}\longrightarrow \mathbb{R}^{m}$, $g:V\subset \mathbb{R}^{m}\longrightarrow \mathbb{R}^{p}$ dos funciones tales que $ f(U)\subset V.$ Supongamos que $f$ es derivable en $\overline{x}_{0}\in U$ y $g$ es derivable en $\overline{y}_{0}=f\left( \overline{x}_{0}\right) .$ Entonces $g\circ f$ es derivable en $\overline{x}_{0}$ y \begin{equation*} D_{g\circ f}\left( \overline{x}_{0}\right) =D_{g}\left( \overline{y} _{0}\right) D_{f}\left( \overline{x}_{0}\right) . \end{equation*} El lado derecho de la igualdad es un producto de matrices.

Debemos probar que la $p\times n$ matriz $D_{g}\left( \overline{y} _{0}\right) D_{f}\left( \overline{x}_{0}\right) $ satisface (\ref{equiv}), es decir que para cada $\varepsilon >0$ existe $\delta >0$ tal que

\begin{equation*} \left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right) \right\Vert \leq \varepsilon \left\Vert \overline{x}-\overline{x} _{0}\right\Vert \end{equation*}

si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta .$

Primero escribimos el lado izquierdo como

\begin{eqnarray*} & & g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x} _{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x}_{0}\right) \\ & & \\ &=& g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x} _{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) +D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right)\end{eqnarray*}

de donde

\begin{eqnarray*} & &\left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right) \right\Vert \\ & & \\ &\leq &\left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) \right\Vert +\left\Vert D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) \left( \overline{ x}-\overline{x}_{0}\right) \right\Vert \\ & & \\ & = &\left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) \right\Vert +\left\Vert D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) -D_{f}\left( \overline{x }_{0}\right) \left( \overline{x}-\overline{x}_{0}\right) \right) \right\Vert \\ & & \\ &\leq &\left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) \right\Vert +\left\Vert D_{g}\left( \overline{y}_{0}\right) \right\Vert \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) -D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right) \right\Vert , \end{eqnarray*}

ya que por el Lema 1, se satisface

\begin{equation*} \left\Vert D_{g}\left( \overline{y}_{0}\right) \overline{h}\right\Vert \leq \left\Vert D_{g}\left( \overline{y}_{0}\right) \right\Vert \left\Vert \overline{h}\right\Vert \end{equation*}

Sea $\left\Vert D_{g}\left( \overline{y}_{0}\right) \right\Vert < M.$ Entonces

Como $f$ es derivable en $\overline{x}_{0},$ dado $\varepsilon >0$ existe $ \delta _{1}>0$ tal que si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta _{1}$, entonces

\begin{equation} \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) -D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right) \right\Vert \leq \dfrac{\varepsilon }{2M}\left\Vert \overline{x}- \overline{x}_{0}\right\Vert \label{desfder}\tag{3} \end{equation}

Por ser $f$ derivable, aplicando el Teorema Auxiliar, tenemos que existen $\delta _{2}>0$ y $M_{1}>0$ tales que si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta _{2}$ entonces

\begin{equation} \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right\Vert \leq M_{1}\left\Vert \overline{x}-\overline{x}_{0}\right\Vert . \label{desfM}\tag{4} \end{equation}

Por ser $g$ derivable en $\overline{y}_{0}$ existe $\delta _{3}>0$ tal que si $\left\Vert \overline{y}-\overline{y}_{0}\right\Vert <\delta _{3}$, entonces

\begin{equation} \left\Vert g\left( \overline{y}\right) -g\left( \overline{y}_{0}\right) -D_{g}\left( \overline{y}_{0}\right) \left( \overline{y}-\overline{y} _{0}\right) \right\Vert \leq \dfrac{\varepsilon \left\Vert \overline{y}- \overline{y}_{0}\right\Vert }{2M_{1}}. \label{desgM}\tag{5} \end{equation}

Tomemos $0<\delta <\min \left( \delta _{1},\delta _{2},\dfrac{\delta _{3}}{ M_{1}}\right) $ y $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta .$ Por (\ref{desfM})

\[ \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right\Vert \leq M_{1}\left\Vert \overline{x}-\overline{x}_{0}\right\Vert < M_{1}\delta < M_{1}\dfrac{\delta _{3}}{M_{1}}=\delta _{3}, \]

Como $\left\Vert f\left( \overline{x}\right) -f\left( \overline{x} _{0}\right) \right\Vert <\delta _{3}$ se sigue de (\ref{desgM}), con $ y=f\left( \overline{x}_{0}\right) $ y $y_{0}=f\left( \overline{x}_{0}\right) ,$ que se satisface

\begin{eqnarray*} \left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) \right\Vert &\leq &\dfrac{\varepsilon \left\Vert f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right\Vert }{2M_{1}} \\ &\leq &\dfrac{\varepsilon M_{1}\left\Vert \overline{x}-\overline{x} _{0}\right\Vert }{2M_{1}} \\ & = &\dfrac{\varepsilon \left\Vert \overline{x}-\overline{x}_{0}\right\Vert }{2}. \end{eqnarray*}

En resumen si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta ,$ entonces por esa última desigualdad y la desigualdad (\ref{desfder}) tenemos.

\begin{equation*} \left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) \left( f\left( \overline{x}\right) -f\left( \overline{x}_{0}\right) \right) \right\Vert +M\left\Vert f\left( \overline{x}\right) -f\left( \overline{x} _{0}\right) -D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}- \overline{x}_{0}\right) \right\Vert \end{equation*}

es menor que

\begin{equation*} \dfrac{\varepsilon \left\Vert \overline{x}-\overline{x}_{0}\right\Vert }{2}+ \dfrac{\varepsilon \left\Vert \overline{x}-\overline{x}_{0}\right\Vert }{2} \end{equation*}

y así, si $\left\Vert \overline{x}-\overline{x}_{0}\right\Vert <\delta $ entonces

\begin{equation*} \left\Vert g\left( f\left( \overline{x}\right) \right) -g\left( f\left( \overline{x}_{0}\right) \right) -D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) \left( \overline{x}-\overline{x} _{0}\right) \right\Vert \leq \left( \dfrac{\varepsilon }{2}+\dfrac{ \varepsilon }{2}\right) \left\Vert \overline{x}-\overline{x}_{0}\right\Vert =\varepsilon \left\Vert \overline{x}-\overline{x}_{0}\right\Vert . \end{equation*}

Queda probado que $g\circ f$ es derivable en $\overline{x}_{0}$ y $ D_{g\circ f}\left( \overline{x}_{0}\right) =D_{g}\left( \overline{y} _{0}\right) D_{f}\left( \overline{x}_{0}\right) ,$ donde $\overline{y} _{0}=f\left( \overline{x}_{0}\right) .$

QED

Supongamos que se satisfacen las hipótesis del teorema anterior con $ f\left( \overline{x}\right) =\left( f_{1}\left( \overline{x}\right) ,...,f_{m}\left( \overline{x}\right) \right) $ para $\overline{x}\in \mathbb{R}^{n}$ y $g\left( \overline{y}\right) =\left( g_{1}\left( \overline{y}\right) ,...,g_{p}\left( \overline{y}\right) \right) $ para $ \overline{y}\in \mathbb{R}^{m}.$ Entonces

\begin{eqnarray} \dfrac{\partial \left( g_{i}\circ f\right) }{\partial x_{j}}\left( \overline{ x}_{0}\right) & = &\nabla g_{i}\left( f\left( \overline{x}_{0}\right) \right) \cdot \dfrac{\partial f}{\partial x_{j}}\left( \overline{x}_{0}\right) \label{DevPargral} \tag{6}\\ & = &\dfrac{\partial g_{i}}{\partial y_{1}}\left( f\left( \overline{x} _{0}\right) \right) \dfrac{\partial f_{1}}{\partial x_{j}}\left( \overline{x} _{0}\right) +\cdots +\dfrac{\partial g_{i}}{\partial y_{m}}\left( f\left( \overline{x}_{0}\right) \right) \dfrac{\partial f_{m}}{\partial x_{j}}\left( \overline{x}_{0}\right) \notag \end{eqnarray}

\begin{equation*} \dfrac{\partial f}{\partial x_{j}}\left( \overline{x}_{0}\right) =\left( \dfrac{\partial f_{1}}{\partial x_{j}}\left( \overline{x}_{0}\right) ,..., \dfrac{\partial f_{m}}{\partial x_{j}}\left( \overline{x}_{0}\right) \right) \end{equation*}

Por la regla de la cadena tenemos la siguiente igualdad para las matrices jacobianas $D_{g\circ f}\left( \overline{x}_{0}\right) ,$ $D_{g}\left( \overline{y}_{0}\right) $ y $D_{f}\left( \overline{x}_{0}\right) $ \begin{equation} D_{g\circ f}\left( \overline{x}_{0}\right) =D_{g}\left( \overline{y} _{0}\right) D_{f}\left( \overline{x}_{0}\right) \label{reg2}\tag{7} \end{equation}

donde $\overline{y}_{0}=f\left( \overline{x}_{0}\right) .$

La matriz $D_{g\circ f}\left( \overline{x}_{0}\right) $ es de tamaño $ p\times n,$ en tanto que $D_{g}\left( \overline{y}_{0}\right) $ y $ D_{f}\left( \overline{x}_{0}\right) $ tienen tamaños $p\times m$ y $ m\times n,$ respectivamente.

Para $1\leq i\leq p$ el renglón $i$ de $D_{g\circ f}\left( \overline{x} _{0}\right) $ es $\nabla \left( g_{i}\circ f\right) \left( \overline{x} _{0}\right) ,$ por lo que su entrada $(i,j)$, con $1\leq j\leq n$, es $ \dfrac{\partial \left( g_{i}\circ f\right) }{\partial x_{j}}\left( \overline{ x}_{0}\right) .$

Por otra parte, la entrada $(i,j)$ de $D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) $ es el producto escalar del renglón $i$ de $D_{g}\left( \overline{y}_{0}\right) ,$ $\nabla g_{i}\left( f\left( \overline{x}_{0}\right) \right) ,$ por la columna $j$ de $D_{f}\left( \overline{x}_{0}\right) $, \begin{equation}\dfrac{\partial f}{\partial x_{j}}\left( \overline{x}_{0}\right) =\left( \dfrac{\partial f_{1}}{\partial x_{j}}\left( \overline{x}_{0}\right) ,...,\dfrac{\partial f_{m}}{\partial x_{j}}\left( \overline{x}_{0}\right) \right) , \end{equation} es decir \begin{equation}\nabla g_{i}\left( f\left( \overline{x}_{0}\right) \right) \cdot \dfrac{\partial f}{\partial x_{j}} \left( \overline{x}_{0}\right) . \end{equation}

Debido a la igualdad (\ref{reg2}), la entrada $\left( i,j\right) $ de $ D_{g\circ f}\left( \overline{x}_{0}\right) $ coincide con la de $D_{g}\left( \overline{y}_{0}\right) D_{f}\left( \overline{x}_{0}\right) ,$ con lo que obtenemos la igualdad (\ref{DevPargral}).

QED

Dos casos particulares del corolario anterior son los que se tratan en los siguientes dos resultados en los que $g$ tiene una sola función componente, es decir es una función real.

Sean $U\subset \mathbb{R}^{2}$ y $V\subset \mathbb{R}^{2}$ abiertos. Supongamos que $f:U\subset \mathbb{R}^{2}\longrightarrow \mathbb{R}^{2}$ y $ g:V\subset \mathbb{R}^{2}\longrightarrow \mathbb{R}$ son derivables en sus dominios. Si $f(U)\subset V$ y $f\left( x,y\right) =\left( u\left( x,y\right) ,v\left( x,y\right) \right) $ , entonces la función

\begin{equation*} h\left( x,y\right) =g\left( f\left( x,y\right) \right) =g\left( u\left( x,y\right) ,v\left( x,y\right) \right) \end{equation*}

\begin{eqnarray} \dfrac{\partial h}{\partial x}\left( x,y\right) & = &\dfrac{\partial g}{ \partial u}\left( f\left( x,y\right) \right) \dfrac{\partial u}{\partial x} \left( x,y\right) +\dfrac{\partial g}{\partial v}\left( f\left( x,y\right) \right) \dfrac{\partial v}{\partial x}\left( x,y\right) \label{ForCor1}\tag{8} \\ & = &\nabla g\left( f\left( x,y\right) \right) \cdot \dfrac{\partial f}{ \partial x}\left( x,y\right) \notag \\ & & \notag \\ \dfrac{\partial h}{\partial y}\left( x,y\right) & = &\dfrac{\partial g}{ \partial u}\left( f\left( x,y\right) \right) \dfrac{\partial u}{\partial y} \left( x,y\right) +\dfrac{\partial g}{\partial v}\left( f\left( x,y\right) \right) \dfrac{\partial v}{\partial y}\left( x,y\right) \notag \\ & = &\nabla g\left( f\left( x,y\right) \right) \cdot \dfrac{\partial f}{ \partial y}\left( x,y\right) \notag \end{eqnarray}

donde $\dfrac{\partial f}{\partial x}\left( x,y\right) =\left( \dfrac{ \partial u}{\partial x}\left( x,y\right) ,\dfrac{\partial v}{\partial x} \left( x,y\right) \right) $ y $\dfrac{\partial f}{\partial y}\left( x,y\right) =\left( \dfrac{\partial u}{\partial y}\left( x,y\right) ,\dfrac{ \partial v}{\partial y}\left( x,y\right) \right) .$

La función $f$ tiene dos funciones componentes $u\left( x,y\right) $ y $ v\left( x,y\right) $ y $g$ sólo una, ella misma. El resultado se sigue de (\ref{DevPargral}) tomando $g_{1}\left( f\left( x,y\right) \right) =g\left( u\left( x,y\right) ,v\left( x,y\right) \right) ,f_{1}\left( x,y\right) =u\left( x,y\right) $ y $f_{2}\left( x,y\right) =v\left( x,y\right) .$

\begin{equation*} h\left( x,y\right) =g\left( u\left( x,y\right) ,v\left( x,y\right) \right) \end{equation*}

entonces una variación en $x$ implica una variación en $u\left( x,y\right) $ y $v\left( x,y\right) $ lo que a su vez implica que $g$ varíe. Así, resulta natural que para calcular $\dfrac{\partial h}{ \partial x}$ debamos tomar en cuenta las parciales $\dfrac{\partial u}{ \partial x}$ y $\dfrac{\partial v}{\partial x}$ así como $\dfrac{ \partial g}{\partial u}$ y $\dfrac{\partial g}{\partial v}.$ La regla de la cadena nos dice cómo se relacionan todos estos valores.

Podemos considerar que $\dfrac{\partial g}{\partial u}\dfrac{\partial u}{ \partial x}$ nos da la variación de $h$ debido al cambio de $u$ por variar $x$; en tanto que, $\dfrac{\partial g}{\partial v}\dfrac{\partial v}{ \partial x}$ nos da la variación de $h$ debido al cambio de $v$ por variar $x.$ Y la suma de las dos nos da la medida de cómo cambia $h$ al variar $x;$ es decir \begin{equation*} \dfrac{\partial h}{\partial x}=\dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{\partial x}. \end{equation*}

Una discusión similar se puede hacer para $\dfrac{\partial h}{\partial y}.$

QED

Observamos que, por ejemplo, para encontrar $\dfrac{\partial h}{\partial x},$ identificamos en la parte inferior, todos los símbolos $x.$ Seguimos desde $g$, los caminos que lleven a dichos símbolos, calculamos las derivadas indicadas siguiendo cada uno de los caminos multiplicando las derivadas obtenidas y sumamos el resultado de cada uno de ellos. Así, observamos que hay dos caminos que terminan en $x,$ en los que vemos

\begin{equation*} \dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial x}\qquad \qquad \text{y}\qquad \qquad \dfrac{\partial g}{\partial v}\dfrac{\partial v}{ \partial x} \end{equation*}

\begin{equation*} \dfrac{\partial h}{\partial x}=\dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{\partial x}. \end{equation*}

Sean $f:U\subset \mathbb{R}^{3}\longrightarrow \mathbb{R}^{3}$ una función derivable en un conjunto abierto $U,$ dada por $f\left(x,y,z\right) =\left( u\left( x,y,z\right) ,v\left( x,y,z\right) ,w\left( x,y,z\right) \right) $ y $g:V\subset \mathbb{R}^{3}\longrightarrow \mathbb{R}$ una función derivable en su dominio. Si \begin{equation*} h\left( x,y,z\right) =g\left( f\left( x,y,z\right) \right) =g\left( u\left( x,y,z\right) ,v\left( x,y,z\right) ,w\left( x,y,z\right) \right) \end{equation*}

\begin{eqnarray*} \dfrac{\partial h}{\partial x}\left( x,y,z\right) & = &\dfrac{\partial g}{ \partial u}\dfrac{\partial u}{\partial x}\left( x,y,z\right) +\dfrac{ \partial g}{\partial v}\dfrac{\partial v}{\partial x}\left( x,y,z\right) + \dfrac{\partial g}{\partial w}\dfrac{\partial w}{\partial x}\left( x,y,z\right) \\ & & \\ \dfrac{\partial h}{\partial y}\left( x,y,z\right) & = &\dfrac{\partial g}{ \partial u}\dfrac{\partial u}{\partial y}\left( x,y,z\right) +\dfrac{ \partial g}{\partial v}\dfrac{\partial v}{\partial y}\left( x,y,z\right) + \dfrac{\partial g}{\partial w}\dfrac{\partial w}{\partial y}\left( x,y,z\right) \\ & & \\ \dfrac{\partial h}{\partial z}\left( x,y,z\right) & = &\dfrac{\partial g}{ \partial u}\dfrac{\partial u}{\partial z}\left( x,y,z\right) +\dfrac{ \partial g}{\partial v}\dfrac{\partial v}{\partial z}\left( x,y,z\right) + \dfrac{\partial g}{\partial w}\dfrac{\partial w}{\partial z}\left( x,y,z\right) \end{eqnarray*}

Donde las parciales de $g$ están evaluadas en $f\left( x,y,z\right) =\left( u\left( x,y,z\right) ,v\left( x,y,z\right) ,w\left( x,y,z\right) \right) .$

La función $f$ tiene tres funciones componentes $u,v,$ $w$ y $g$ sólo una, ella misma. El resultado se sigue de (\ref{DevPargral}) tomando \[ g_{1}\left( f\left( x,y,z\right) \right) =g\left( f\left( x,y,z\right) \right),\ f_{1}\left( x,y,z\right) =u\left( x,y,z\right),\ f_{2}\left( x,y,z\right) =v\left( x,y,z\right)\ \text{y}\ f_{3}\left( x,y,z\right) =w\left( x,y,z\right) . \]

QED

Sea $h\left( x,y\right) =\left( x+y\right) \ \text{sen}\ \left( x^{2}+y^{3}\right) .$ Encuentra $\dfrac{\partial h}{\partial x}$ y $\dfrac{ \partial h}{\partial y}.$

Observamos que $h\left( x,y\right) $ es el producto de las funciones $ f\left( x,y\right) =x+y$ y $g\left( x,y\right) =\ \text{sen}\ \left( x^{2}+y^{3}\right) ,$ donde a su vez $g$ es una composición de funciones.

\begin{eqnarray*} \dfrac{\partial h}{\partial x} & = &\dfrac{\partial }{\partial x}\left( \left( x+y\right) \ \text{sen}\ \left( x^{2}+y^{3}\right) \right) \\ & = &\left( \dfrac{\partial }{\partial x}\left( x+y\right) \right) \ \text{sen}\ \left( x^{2}+y^{3}\right) +\left( x+y\right) \left( \dfrac{\partial }{ \partial x}\ \text{sen}\ \left( x^{2}+y^{3}\right) \right) \\ & = &\ \text{sen}\ \left( x^{2}+y^{3}\right) +\left( x+y\right) \left( \dfrac{ \partial }{\partial x}\ \text{sen}\ \left( x^{2}+y^{3}\right) \right) \end{eqnarray*}

\begin{equation*} \dfrac{\partial h}{\partial x}=\ \text{sen}\ \left( x^{2}+y^{3}\right) +\left( x+y\right) \left( \cos \left( x^{2}+y^{3}\right) \right) 2x. \end{equation*}

\begin{eqnarray*} \dfrac{\partial h}{\partial y} & = &\dfrac{\partial }{\partial y}\left( \left( x+y\right) \ \text{sen}\ \left( x^{2}+y^{3}\right) \right) \\ & = &\left( \dfrac{\partial }{\partial y}\left( x+y\right) \right) \ \text{sen}\ \left( x^{2}+y^{3}\right) +\left( x+y\right) \left( \dfrac{\partial }{ \partial y}\ \text{sen}\ \left( x^{2}+y^{3}\right) \right) \\ & = &\ \text{sen}\ \left( x^{2}+y^{3}\right) +\left( x+y\right) \left( \cos \left( x^{2}+y^{3}\right) \right) 3y^{2}. \end{eqnarray*}

Recordamos que si una función $f:U\subset \mathbb{R}^{n}\longrightarrow \mathbb{R}^{n}$ tiene derivadas parciales continuas en $U$, entonces $f$ es derivable en $U$. Así, cuando $f$ es de clase $C^{2},$ entonces ella y sus derivadas parciales son funciones derivables.

Sean $f:U\subset \mathbb{R}^{2}\longrightarrow \mathbb{R}^{2}$ una transformación de clase $C^{2}$ en el conjunto abierto $U$ dada por $f\left( x,y\right) =\left( u\left( x,y\right) ,v\left( x,y\right) \right) $ y $g:\mathbb{R}^{2}\longrightarrow \mathbb{R}$ una función de clase $C^{2}$. Si

\begin{equation*} h\left( x,y\right) =g\left( f\left( x,y\right) \right) =g\left( u\left( x,y\right) ,v\left( x,y\right) \right) \end{equation*}

\begin{eqnarray*} \dfrac{\partial ^{2}h}{\partial x^{2}} & = &\dfrac{\partial ^{2}g}{\partial u^{2}}\left( \dfrac{\partial u}{\partial x}\right) ^{2}+\dfrac{\partial ^{2}g }{\partial v\partial u}\dfrac{\partial v}{\partial x}\dfrac{\partial u}{ \partial x}+\dfrac{\partial g}{\partial u}\dfrac{\partial ^{2}u}{\partial x^{2}}+\dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{\partial u}{ \partial x}\dfrac{\partial v}{\partial x}+\dfrac{\partial ^{2}g}{\partial v^{2}}\left( \dfrac{\partial v}{\partial x}\right) ^{2}+\dfrac{\partial g}{ \partial v}\dfrac{\partial ^{2}v}{\partial x^{2}} \\ & & \\ \dfrac{\partial ^{2}h}{\partial y\partial x} & = &\dfrac{\partial ^{2}g}{ \partial u^{2}}\dfrac{\partial u}{\partial y}\dfrac{\partial u}{\partial x}+ \dfrac{\partial ^{2}g}{\partial v\partial u}\dfrac{\partial v}{\partial y} \dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial u}\dfrac{\partial ^{2}u}{\partial y\partial x}+\dfrac{\partial ^{2}g}{\partial u\partial v} \dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial x}+\dfrac{\partial ^{2}g}{\partial v^{2}}\dfrac{\partial v}{\partial y}\dfrac{\partial v}{ \partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial ^{2}v}{\partial y\partial x} \\ & & \\ \dfrac{\partial ^{2}h}{\partial y^{2}} & = &\dfrac{\partial ^{2}g}{\partial u^{2}}\left( \dfrac{\partial u}{\partial y}\right) ^{2}+\dfrac{\partial ^{2}g }{\partial v\partial u}\dfrac{\partial v}{\partial y}\dfrac{\partial u}{ \partial y}+\dfrac{\partial g}{\partial u}\dfrac{\partial ^{2}u}{\partial y^{2}}+\dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{\partial u}{ \partial y}\dfrac{\partial v}{\partial y}+\dfrac{\partial ^{2}g}{\partial v^{2}}\left( \dfrac{\partial v}{\partial y}\right) ^{2}+\dfrac{\partial g}{ \partial v}\dfrac{\partial ^{2}v}{\partial y^{2}} \end{eqnarray*}

Por el corolario 1 sabemos que

\begin{equation*} \dfrac{\partial h}{\partial x}=\dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{\partial x} \quad \quad \quad \text{y}\quad \quad \quad \dfrac{\partial h}{\partial y}= \dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{\partial y} \end{equation*}

entonces

\begin{eqnarray} \dfrac{\partial }{\partial x}\left( \dfrac{\partial h}{\partial x}\right) & = & \dfrac{\partial }{\partial x}\left( \dfrac{\partial g}{\partial u}\dfrac{ \partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{ \partial x}\right) \label{cadena1}\tag{9} \\ & = &\dfrac{\partial }{\partial x}\left( \dfrac{\partial g}{\partial u}\dfrac{ \partial u}{\partial x}\right) +\dfrac{\partial }{\partial x}\left( \dfrac{ \partial g}{\partial v}\dfrac{\partial u}{\partial x}\right) \quad \quad \quad \text{derivada de una suma} \notag \\ & = &\dfrac{\partial }{\partial x}\left( \dfrac{\partial g}{\partial u}\right) \dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial u}\dfrac{\partial }{\partial x}\left( \dfrac{\partial u}{\partial x}\right) +\dfrac{\partial }{ \partial x}\left( \dfrac{\partial g}{\partial v}\right) \dfrac{\partial u}{ \partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial }{\partial x} \left( \dfrac{\partial u}{\partial x}\right) \quad \quad \quad \text{ derivada de un producto} \notag \end{eqnarray}

Observamos que $\dfrac{\partial g}{\partial u}$ y $\dfrac{\partial g}{ \partial v}$ son funciones de $u$ y $v$ que a su vez dependen de $x$ y $y$ por lo que para calcular sus derivadas, con respecto a estas variables, debemos usar nuevamente la regla de la cadena (Ver Ejemplo)

Por lo tanto,

\begin{eqnarray} \dfrac{\partial ^{2}h}{\partial x^{2}} & = &\dfrac{\partial u}{\partial x} \left( \dfrac{\partial ^{2}g}{\partial u^{2}}\dfrac{\partial u}{\partial x}+ \dfrac{\partial ^{2}g}{\partial v\partial u}\dfrac{\partial v}{\partial x} \right) +\dfrac{\partial g}{\partial u}\dfrac{\partial ^{2}u}{\partial x^{2}} +\dfrac{\partial v}{\partial x}\left( \dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{\partial u}{\partial x}+\dfrac{\partial ^{2}g}{\partial v^{2}}\dfrac{\partial v}{\partial x}\right) +\dfrac{\partial g}{\partial v} \dfrac{\partial ^{2}v}{\partial x^{2}} \notag \\ & = &\dfrac{\partial ^{2}g}{\partial u^{2}}\left( \dfrac{\partial u}{\partial x }\right) ^{2}+\dfrac{\partial ^{2}g}{\partial v\partial u}\dfrac{\partial v}{ \partial x}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial u} \dfrac{\partial ^{2}u}{\partial x^{2}}+\dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial x}+ \dfrac{\partial ^{2}g}{\partial v^{2}}\left( \dfrac{\partial v}{\partial x} \right) ^{2}+\dfrac{\partial g}{\partial v}\dfrac{\partial ^{2}v}{\partial x^{2}} \notag \end{eqnarray}

Veamos el caso de la derivada mixta.

\begin{eqnarray*} \dfrac{\partial }{\partial y}\left( \dfrac{\partial h}{\partial x}\right) & = & \dfrac{\partial }{\partial y}\left( \dfrac{\partial g}{\partial u}\dfrac{ \partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{ \partial x}\right) \\ & = &\dfrac{\partial u}{\partial x}\dfrac{\partial }{\partial y}\left( \dfrac{ \partial g}{\partial u}\right) +\dfrac{\partial g}{\partial u}\dfrac{ \partial }{\partial y}\left( \dfrac{\partial u}{\partial x}\right) +\dfrac{ \partial v}{\partial x}\dfrac{\partial }{\partial y}\left( \dfrac{\partial g }{\partial v}\right) +\dfrac{\partial g}{\partial v}\dfrac{\partial }{ \partial y}\left( \dfrac{\partial v}{\partial x}\right) \\ & = &\dfrac{\partial u}{\partial x}\left( \dfrac{\partial ^{2}g}{\partial u^{2} }\dfrac{\partial u}{\partial y}+\dfrac{\partial ^{2}g}{\partial v\partial u} \dfrac{\partial v}{\partial y}\right) +\dfrac{\partial g}{\partial u}\dfrac{ \partial ^{2}u}{\partial y\partial x}+\dfrac{\partial v}{\partial x}\left( \dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{\partial u}{\partial y}+ \dfrac{\partial ^{2}g}{\partial v^{2}}\dfrac{\partial v}{\partial y}\right) + \dfrac{\partial g}{\partial v}\dfrac{\partial ^{2}v}{\partial y\partial x} \\ & = &\dfrac{\partial ^{2}g}{\partial u^{2}}\dfrac{\partial u}{\partial y} \dfrac{\partial u}{\partial x}+\dfrac{\partial ^{2}g}{\partial v\partial u} \dfrac{\partial v}{\partial y}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial u}\dfrac{\partial ^{2}u}{\partial y\partial x}+\dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{\partial u}{\partial y}\dfrac{\partial v}{ \partial x}+\dfrac{\partial ^{2}g}{\partial v^{2}}\dfrac{\partial v}{ \partial y}\dfrac{\partial v}{\partial x}+\dfrac{\partial g}{\partial v} \dfrac{\partial ^{2}v}{\partial y\partial x} \end{eqnarray*}

Por lo tanto,

\begin{equation*} \dfrac{\partial ^{2}h}{\partial y\partial x}=\dfrac{\partial ^{2}g}{\partial u^{2}}\dfrac{\partial u}{\partial y}\dfrac{\partial u}{\partial x}+\dfrac{ \partial ^{2}g}{\partial v\partial u}\dfrac{\partial v}{\partial y}\dfrac{ \partial u}{\partial x}+\dfrac{\partial g}{\partial u}\dfrac{\partial ^{2}u}{ \partial y\partial x}+\dfrac{\partial ^{2}g}{\partial u\partial v}\dfrac{ \partial u}{\partial y}\dfrac{\partial v}{\partial x}+\dfrac{\partial ^{2}g}{ \partial v^{2}}\dfrac{\partial v}{\partial y}\dfrac{\partial v}{\partial x}+ \dfrac{\partial g}{\partial v}\dfrac{\partial ^{2}v}{\partial y\partial x} \end{equation*}

QED

Para recordar las fórmulas anteriores observamos, por ejemplo, que en el segundo renglón de (\ref{cadena1}) aparecen $\dfrac{\partial }{\partial x }\left( \dfrac{\partial g}{\partial u}\right) $ y $\dfrac{\partial }{ \partial x}\left( \dfrac{\partial g}{\partial v}\right) ,$ las cuales podemos calcular con el uso del esquema siguiente

Regla de la Cadena de Funciones de Varias Variables

Angel Carrillo Hoyo, Elena de Oteyza de Oteyza\(^2\), Carlos Hernández Garciadiego\(^1\), Emma Lam Osnaya\(^2\)

Resultados sobre la Regla de la Cadena de Funciones de Varias Variables