Introduccion

Diferenciales de funciones $f:A\subset {\mathbb{R}}^2\to \mathbb{R}$

Tenemos que $f:A\subset {\mathbb{R}}^2\to \mathbb{R}$ es diferenciable si
$$f(x_o+h_1,y_0+h_2)=f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)h_1+\frac{\partial f}{\partial y}(x_0,y_0)h_2+r(h_1,h_2)$$
cumple
$$\underset{(h_1,h_2)\to (0,0)}{lim}\frac{r(h_1,h_2)}{|(h_1,h_2)|}=0$$
Esto se puede escribir como
$$f(x_o+h_1,y_0+h_2)-f(x_0,y_0)=\frac{\partial f}{\partial x}(x_0,y_0)h_1+\frac{\partial f}{\partial y}(x_0,y_0)h_2+r(h_1,h_2)$$

tomando
$$f(x_o+h_1,y_0+h_2)-f(x_0,y_0)=\mathrm{△}z$$
$$\frac{\partial f}{\partial x}(x_0,y_0)h_1=\frac{\partial f}{\partial x}(x_0,y_0)\mathrm{△}x$$
$$\frac{\partial f}{\partial y}(x_0,y_0)h_2=\frac{\partial f}{\partial y}(x_0,y_0)\mathrm{△}y$$
tenemos que
$$\mathrm{△}z=\frac{\partial f}{\partial x}(x_0,y_0)\mathrm{△}x+\frac{\partial f}{\partial y}(x_0,y_0)\mathrm{△}y+r(\mathrm{△}x,\mathrm{△}y)$$
haciendo $\mathrm{△}x,\mathrm{△}y\to 0$ tenemos
$$dz=\frac{\partial f}{\partial x}(x_0,y_0)dx+\frac{\partial f}{\partial y}(x_0,y_0)dy$$
$\mathbf{\text{Definición.}}$Si $z=f(x,y)$ es una función diferenciable, la diferencial de f denotada $dz$ se define
$$dz=\frac{\partial f}{\partial x}(x_0,y_0)dx+\frac{\partial f}{\partial y}(x_0,y_0)dy$$

$\mathbf{\text{Ejemplo.}}$ Calcular la diferencial de $z=4x^2-xy$\En este caso
$$dz=\frac{\partial (4x^2-xy)}{\partial x}dx+\frac{\partial (4x^2-xy)}{\partial y}dy=(8x-y)dx-xdy$$

Ahora bien
$$f(x_o+h_1,y_0+h_2)-f(x_0,y_0)=\mathrm{△}z\approx \frac{\partial f}{\partial x}(x_0,y_0)\mathrm{△}x+\frac{\partial f}{\partial y}(x_0,y_0)\mathrm{△}y$$
expresa el cambio aproximado de $z=f(x,y)$ cuando $(x,y)$ pasa a $(x+\mathrm{△}x,y+\mathrm{△}y)$

Ejemplo. Aproximar el cambio de $z=4x^2-xy$ cuando $(x,y)$ pasa de $(2,1)$ a $(2.1,1.5)$\
En este caso tomamos $x_0=2$, $y_0=1$, $\mathrm{△}x=0.1$ y $\mathrm{△}y=.5$ y el valor de cambio será
$$\frac{\partial f}{\partial x}(2,1)\mathrm{△}x+\frac{\partial f}{\partial y}(2,1)\mathrm{△}y=(15)(0.1)-2(0.5)=1.5$$
mientras que
$$f(2.1,1.5)-f(2,1)=14.49-14=0.49$$
por lo tanto en la aproximacion se cometió un error de $0.01$

Ejemplo. Usando diferenciales se quiere calcular aproximadamente
$$A=\frac{0.97}{\sqrt{15.05}+\sqrt[3]{0.98}}$$

Solución. Considerando la función
$$f(x,y,z)=\frac{x}{\sqrt{y+\sqrt[3]{z}}}$$
con $x=1$, $y=15$, $z=1$, $dx=-0.03$, $dy=0.05$ y $dz=-0.02$ se tiene
$$f(x+dx,y+dy,z+dz)=f(x,y,z)+df(x,y,z)$$
en este caso
$$f(x,y,z)=f(1,15,1)=\frac{1}{4}$$
$$\frac{\partial f}{\partial x}=\frac{1}{\sqrt{y}+\sqrt[3]{z}},\frac{\partial f}{\partial y}=-\frac{x}{2}{(y+\sqrt[3]{z})}^{\frac{-3}{2}},\frac{\partial f}{\partial z}=-\frac{x}{2}(y+\sqrt[3]{z}{)}^{\frac{-3}{2}}\frac{1}{3}{z}^{\frac{-2}{3}}$$
evaluando en $(1,15,1)$ se tiene
$$\frac{\partial f}{\partial x}(1,15,1)=\frac{1}{4},\frac{\partial f}{\partial y}(1,15,1)=-\frac{1}{128},\frac{\partial f}{\partial z}(1,15,1)=-\frac{1}{384}$$
de modo que
$$df(1,15,1)=\frac{1}{4}(-0.03)-\frac{1}{128}(0.05)-\frac{1}{384}(-0.02)=-\frac{3.01}{384}$$
por lo que
$$A=\frac{1}{4}-\frac{3.01}{384}=0.242161$$
(el valor es $0.2421726$)

Diferencial de orden 2

Si ${\displaystyle df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy}$ entonces una diferencial de orden 2 seria:
$$d^{2}f=d(df)=d(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy)=\frac{\partial }{\partial x}(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy)dx+\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy)dy$$
$$=(\frac{{\partial }^{2}f}{\partial x^2}dx+\frac{{\partial }^{2}f}{\partial x\partial y}dy)dx+(\frac{{\partial }^{2}f}{\partial y\partial x}dx+\frac{{\partial }^{2}f}{\partial y^2}dy)dy=\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y\partial x}dydx+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2$$
$$=\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2$$
Por lo tanto

$$d^{2}f=d(df)=\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2$$

Ejemplo. Hallar la diferencial de orden 2 para $f(x,y)=e^{x^2+y^y}$

Solución. En este caso tenemos la fórmula
$$d^{2}f=d(df)=\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2$$
vamos a calcular las derivadas parciales correspondientes
$$\frac{\partial (e^{x^2+y^2})}{\partial x}=2x{e}^{x^2+y^2}$$
$$\frac{\partial (e^{x^2+y^2})}{\partial y}=2y{e}^{x^2+y^2}$$
$$\frac{{\partial }^2(e^{x^2+y^2})}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial (e^{x^2+y^2})}{\partial x}\right)=\frac{\partial (2x{e}^{x^2+y^2})}{\partial x}=4x^2{e}^{x^2+y^2}+2e^{x^2+y^2}$$
$$\frac{{\partial }^2(e^{x^2+y^2})}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial (e^{x^2+y^2})}{\partial y}\right)=\frac{\partial (2y{e}^{x^2+y^2})}{\partial y}=4y^2{e}^{x^2+y^2}+2e^{x^2+y^2}$$
$$\frac{{\partial }^2(e^{x^2+y^2})}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial (e^{x^2+y^2})}{\partial x}\right)=\frac{\partial (2x{e}^{x^2+y^2})}{\partial y}=4xy{e}^{x^2+y^2}$$
$$\frac{{\partial }^2(e^{x^2+y^2})}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial (e^{x^2+y^2})}{\partial y}\right)=\frac{\partial (2y{e}^{x^2+y^2})}{\partial x}=4xy{e}^{x^2+y^2}$$
y la diferencial de orden 2 sería:
$$d^{2}f=(4x^2{e}^{x^2+y^2}+2e^{x^2+y^2})d{x}^2+8xy{e}^{x^2+y^2}dxdy+(4y^2{e}^{x^2+y^2}+2e^{x^2+y^2})d{y}^2$$

Diferencial de orden 3

Si ${\displaystyle d^{2}f=\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2}$ entonces una diferencial de orden 3 seria:
$$d^{3}f=d(d^{2}f)=d(\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2)=$$
$$\frac{\partial }{\partial x}(\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2)dx+\frac{\partial }{\partial y}(\frac{{\partial }^{2}f}{\partial x^2}d{x}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}dxdy+\frac{{\partial }^{2}f}{\partial y^2}d{y}^2)dy=$$

$$(\frac{{\partial }^{3}f}{\partial x^3}d{x}^2+2\frac{{\partial }^{3}f}{\partial x^2\partial y}dxdy+\frac{{\partial }^{3}f}{\partial x\partial y^2}d{y}^2)dx+(\frac{{\partial }^{3}f}{\partial x^2\partial y}d{x}^2+2\frac{{\partial }^{3}f}{\partial x\partial y^2}dxdy+\frac{{\partial }^{3}f}{\partial y^3}d{y}^2)dy=$$

$$\frac{{\partial }^{3}f}{\partial x^3}d{x}^3+2\frac{{\partial }^{3}f}{\partial x^2\partial y}d{x}^{2}dy+\frac{{\partial }^{3}f}{\partial x\partial y^2}dxd{y}^2+\frac{{\partial }^{3}f}{\partial x^2\partial y}dyd{x}^2+2\frac{{\partial }^{3}f}{\partial x\partial y^2}dxd{y}^2+\frac{{\partial }^{3}f}{\partial y^3}d{y}^3=$$

$$\frac{{\partial }^{3}f}{\partial x^3}d{x}^3+3\frac{{\partial }^{3}f}{\partial x^2\partial y}d{x}^{2}dy+3\frac{{\partial }^{3}f}{\partial x\partial y^2}dxd{y}^2+\frac{{\partial }^{3}f}{\partial y^3}d{y}^3$$
Por lo tanto
$$d^{3}f=d(d^{2}f)=\frac{{\partial }^{3}f}{\partial x^3}d{x}^3+3\frac{{\partial }^{3}f}{\partial x^2\partial y}d{x}^{2}dy+3\frac{{\partial }^{3}f}{\partial x\partial y^2}dxd{y}^2+\frac{{\partial }^{3}f}{\partial y^3}d{y}^3$$

Diferencial de orden 3

Si ${\displaystyle d^{3}f=\frac{{\partial }^{3}f}{\partial x^3}d{x}^3+3\frac{{\partial }^{3}f}{\partial x^2\partial y}d{x}^{2}dy+3\frac{{\partial }^{3}f}{\partial x\partial y^2}dxd{y}^2+\frac{{\partial }^{3}f}{\partial y^3}d{y}^3}$ entonces una diferencial de orden 4 seria:
$$d^{4}f=d(d^{3}f)=\frac{{\partial }^{4}f}{\partial x^4}d{x}^4+4\frac{{\partial }^{4}f}{\partial x^3\partial y}d{x}^{3}dy+6\frac{{\partial }^{4}f}{\partial x^2\partial y^2}d{x}^{2}d{y}^2+4\frac{{\partial }^{4}f}{\partial x\partial y^3}dxd{y}^3+\frac{{\partial }^{4}f}{\partial y^4}d{y}^4$$

Diferencial de orden n

$$d^{n}f=\frac{{\partial }^{n}f}{\partial x^n}d{x}^n+\left(\begin{array}{c}n\\ 1\end{array}\right)\frac{{\partial }^{n-1}f}{\partial x^{n-1}\partial y}d{x}^{n-1}dy+\left(\begin{array}{c}n\\ 2\end{array}\right)\frac{{\partial }^{n-2}f}{\partial x^{n-2}\partial y^2}d{x}^{n-2}d{y}^2+\cdots +\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\partial }^{n-k}f}{\partial x^{n-k}\partial y^k}d{x}^{n-k}d{y}^k+\cdots +\frac{{\partial }^{n}f}{\partial y^n}d{y}^n$$

que se puede escribir
$$d^{n}f=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j$$

Mas adelante

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