3.lyx

#LyX 2.0 created this file. For more info see http://www.lyx.org/
\lyxformat 413
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\begin_body

\begin_layout Part
Aplicación de algunas resultados de diferenciación en una variable.
\end_layout

\begin_layout Section
Verificar que se cumple el teorema de Rolle para las siguientes funciones
 en los intervalos indicados:
\end_layout

\begin_layout Subsection
\begin_inset Formula $f(x)=x^{2}-3x+2$
\end_inset

 en el intervalo 
\begin_inset Formula $[1,2]$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\mbox{ es continua en }\mathbb{R}\wedge f(1)=0=f(2)\wedge\left(f'(x)=2x-3=0\Leftrightarrow x=\frac{3}{2}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{3}{2}\in(1,2)\therefore\mbox{Se cumple el teorema de Rolle}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f(x)=(x-1)(x-2)(x-3)$
\end_inset

 en el intervalo 
\begin_inset Formula $[1,3]$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\mbox{ es continua en }\mathbb{R}\wedge f(1)=0=f(3)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f'(x)=D\left(\left(x^{2}-3x+2\right)\left(x-3\right)\right)=D\left(x^{3}-3x^{2}+2x-3x^{2}+9x-6\right)=D\left(x^{3}-6x^{2}+11x-6\right)=3x^{2}-12x+11
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f'(x)=0\Leftrightarrow x=2,577..\vee x=1,422...
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
2,577...\in[1,3]\therefore\mbox{Se cumple el teorema de Rolle}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f(x)=\sin^{2}x$
\end_inset

 en el intervalo 
\begin_inset Formula $[0,\pi]$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\mbox{ es continua en }\mathbb{R}\wedge f(0)=0=f(\pi)\wedge\left(f'(x)=2\cos x=0\Leftrightarrow x=\frac{\pi}{2}+k\pi\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
k=0:f'(\frac{\pi}{2})=2\cos\frac{\pi}{2}=0\wedge\frac{\pi}{2}\in(0,\pi)\therefore\mbox{Se cumple el teorema de Rolle}
\]

\end_inset


\end_layout

\begin_layout Section
Probar que la función 
\begin_inset Formula $f(x)=\sqrt[3]{\left(x-3\right)^{2}}$
\end_inset

 satisface 
\begin_inset Formula $f(1)=f(5)$
\end_inset

 pero no existe 
\begin_inset Formula $c\in\left(1,5\right)$
\end_inset

 tal que 
\begin_inset Formula $f'(c)=0$
\end_inset

 ¿Por qué no se puede aplicar el teorema de Rolle?
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f(1)=\sqrt[3]{\left(-2\right)^{2}}=\sqrt[3]{\left(2\right)^{2}}=f(5)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f'(x)=\frac{2}{3}x^{-1/3}=\frac{2}{3\sqrt[3]{x}}\neq0\forall x\in\mathbb{R}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{La función no es derivable en }3\Rightarrow\mbox{no es derivable en }\left(1,5\right)\Rightarrow\mbox{ no se cumplen las condiciones que exige el teorema}
\]

\end_inset


\end_layout

\begin_layout Section
Resolver
\end_layout

\begin_layout Subsection
Sea 
\begin_inset Formula $f$
\end_inset

 un polinomio con al menos 
\begin_inset Formula $k$
\end_inset

 raíces distintas, probar que 
\begin_inset Formula $f'$
\end_inset

 tiene al menos 
\begin_inset Formula $k-1$
\end_inset

 raíces distintas.
\end_layout

\begin_layout Standard
Es inmediato por el teorema de Rolle, si tiene 
\begin_inset Formula $k$
\end_inset

 raíces distintas puedo armar 
\begin_inset Formula $k-1$
\end_inset

 intervalos donde la derivada se anualará, entonces habrá 
\begin_inset Formula $k-1$
\end_inset

 raíces distintas.
\end_layout

\begin_layout Subsection
Probar que la ecuación 
\begin_inset Formula $3x^{5}+15x-8=0$
\end_inset

 tiene sólo una raíz real.
\end_layout

\begin_layout Standard
Es una función estrictamente creciente, entonces es inyectiva entonces no
 se puede establecer ningun intervalo para aplicar el teorema de Rolle.
\end_layout

\begin_layout Subsection
Sea 
\begin_inset Formula $a\in\mathbb{R}$
\end_inset

 y 
\begin_inset Formula $f:\mathbb{R}\rightarrow\mathbb{R}/f(x)=e^{a^{2}x}+x^{3}+x$
\end_inset

 si 
\begin_inset Formula $x\in\mathbb{R}$
\end_inset

.
 Probar que la ecuación 
\begin_inset Formula $f(x)=0$
\end_inset

 tiene exactamente una solución en el intervalo 
\begin_inset Formula $[-1,0]$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f'(x)=a^{2}e^{a^{2}x}+3x^{2}+1\neq0\rightarrow\forall x,a:a^{2}e^{a^{2}x}\geq0\wedge3x^{2}>0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f(x)\mbox{ es continua }\Rightarrow\nexists f(x')=0\Leftrightarrow x=x'\rightarrow\mbox{Sino sería aplicable el teorema de Rolle.}
\]

\end_inset


\end_layout

\begin_layout Section
Sean 
\begin_inset Formula $f$
\end_inset

 y 
\begin_inset Formula $g$
\end_inset

 funciones continuas en 
\begin_inset Formula $[a,b]$
\end_inset

 y derivables en 
\begin_inset Formula $(a,b)$
\end_inset

.
 Demostrar las siguientes afirmaciones:
\end_layout

\begin_layout Subsection
Si 
\begin_inset Formula $f'=g'$
\end_inset

 en 
\begin_inset Formula $\left(a,b\right)$
\end_inset

, entonces 
\begin_inset Formula $f(x)=g(x)+c$
\end_inset

 donde 
\begin_inset Formula $c$
\end_inset

 es una constante.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f'=g'\Leftrightarrow D(f(x)+c')=D(g(x)+c'')\Rightarrow f(x)+c'=g(x)+c''\Leftrightarrow f(x)=g(x)+c''-c'
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{sea }c=c''-c'\Rightarrow f(x)=g(x)+c
\]

\end_inset


\end_layout

\begin_layout Subsection
Si 
\begin_inset Formula $f'(x)>0\forall x\in\left(a,b\right)$
\end_inset

 , entonces 
\begin_inset Formula $f$
\end_inset

 es estrictamente creciente.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f'(x)>0\forall x\in\left(a,b\right)\Leftrightarrow\lim_{t\to0}\frac{f(x+t)-f(x)}{t}>0\Leftrightarrow f(x+t)-f(x)>t\Leftrightarrow f(x+t)-t>f(x)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f(x+t)-t>f(x)\rightarrow f(x+t)\mbox{ es mayor a }f(x)\mbox{ para cualquier }t\wedge x+t>t\Rightarrow f\mbox{ es estrictamente creciente.}
\]

\end_inset


\end_layout

\begin_layout Subsection
Si 
\begin_inset Formula $f'(x)<0\forall x\in\left(a,b\right)$
\end_inset

 , entonces 
\begin_inset Formula $f$
\end_inset

 es estrictamente decreciente.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{Análogo al anterior pero con }<\mbox{ en lugar de }>
\]

\end_inset


\end_layout

\begin_layout Section
Como consecuencia del Teorema de Lagrange, mostrar que son válidas las suguiente
s acotaciones
\end_layout

\begin_layout Subsection
\begin_inset Formula $\left|\sin x-\sin y\right|\leq\left|x-y\right|\forall x,y\in\mathbb{R}$
\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $\left|1/x-1/y\right|\leq\left|x-y\right|\forall x,y>1$
\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $\left|\arctan a-\arctan b\right|\leq\frac{1}{2}\left|a-b\right|\forall a,b\geq1$
\end_inset


\end_layout

\begin_layout Standard
¡Preguntar!
\end_layout

\begin_layout Section
Encontrar explícitamente el punto intermedio cuya existencia está garantizada
 por el teorema de Cauchy en el caso de las funciones 
\begin_inset Formula $f(x)=x^{2}$
\end_inset

 y 
\begin_inset Formula $g(x)=x^{3}$
\end_inset

 en el segmento 
\begin_inset Formula $[1,2]$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{f'(c)}{g'(c)}=\frac{f(2)-f(1)}{g(2)-g(1)}\rightarrow\mbox{Por el teorema de Cauchy}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Leftrightarrow\frac{3}{7}=\frac{2c}{3c^{2}}\Leftrightarrow9c^{2}=14c\Leftrightarrow9c^{2}-14c=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Rightarrow\left(c=0\vee9c-14=0\right)\Rightarrow c=\frac{14}{9}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
0\notin[1,2];\frac{14}{9}\in[1,2]\therefore c=\frac{14}{9}
\]

\end_inset


\end_layout

\begin_layout Section
Utilizando el teorema de Cauchy, calcular:
\end_layout

\begin_layout Subsection
\begin_inset Formula $\lim_{\left(x,y\right)\to\left(0,0\right)}f(x,y)$
\end_inset

 donde 
\begin_inset Formula $f(x,y)=\begin{cases}
\frac{\cos x-\cos y}{x^{2}-y^{2}} & \mbox{si }x\neq y\\
-\frac{1}{2} & \mbox{si }x=y
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f(x)=\cos x;g(x)=x^{2};c\in[x,y]\Rightarrow\frac{\cos x-\cos y}{x^{2}-y^{2}}=\frac{f'(c)}{g'(c)}\rightarrow\mbox{Por el teorema de Cauchy}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Leftrightarrow\lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\cos x-\cos y}{x^{2}-y^{2}}=\lim_{c\to0}\frac{-\sin c}{2c}=-\frac{1}{2}\rightarrow\mbox{Si el límite existe entonces mantiene la igualdad, por ser cociente de funciones continuas}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\cos x-\cos y}{x^{2}-y^{2}}=-\frac{1}{2}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $\sup_{x,y\in[0,2]}f(x,y)$
\end_inset

 donde 
\begin_inset Formula $f(x,y)=\begin{cases}
\frac{x^{4}-y^{4}-4(x^{3}-y^{3})+4\left(x^{2}-y^{2}\right)}{y^{2}-x^{2}} & \mbox{si }x\neq y\\
0 & \mbox{si }x=y
\end{cases}$
\end_inset

 ¿Cuánto vale 
\begin_inset Formula $\sup_{x,y\in[0,2]}f(x,y)$
\end_inset

?
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f(x)=x^{4}-4x^{3}-4x^{2};g(x)=x^{2};c\in[x,y]\Rightarrow\frac{x^{4}-y^{4}-4(x^{3}-y^{3})+4\left(x^{2}-y^{2}\right)}{y^{2}-x^{2}}=\frac{4c^{3}-12c^{2}-8c}{2c}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
=\frac{4c^{2}-12c-8}{2}=2c^{2}-6c-4=\begin{cases}
c=0 & -4\\
c=2 & 8-12-4=-8
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
??
\end_layout

\begin_layout Part
Derivadas parciales y direccionales
\end_layout

\begin_layout Section
Sea 
\begin_inset Formula $f:\mathbb{R}\to\mathbb{R}$
\end_inset

 una función continua en 
\begin_inset Formula $x=a$
\end_inset

.
 Probar que 
\begin_inset Formula $f$
\end_inset

 es derivable en 
\begin_inset Formula $x=a\Leftrightarrow\exists L(x)=m(x-a)+b$
\end_inset

 única tal que:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{x\to a}\frac{f(x)-L(x)}{x-a}=0
\]

\end_inset


\end_layout

\begin_layout Standard
Calcular el valor de 
\begin_inset Formula $m$
\end_inset

 y de 
\begin_inset Formula $b$
\end_inset

.
 Al gráfico de 
\begin_inset Formula $L(x)$
\end_inset

 se la denomina la 
\series bold
recta tangente
\series default
 a 
\begin_inset Formula $f(x)$
\end_inset

 en 
\begin_inset Formula $x=a$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{x\to a}\frac{f(x)-L(x)}{x-a}=0\Rightarrow L(x)\to f(a)\mbox{ si }x\to a
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{Sabemos que la función que mejor aproxima }f(x)\mbox{ es }f'(x)\Rightarrow L(x)=f'(x)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{x\to a}\frac{f(x)-f'(x)}{x-a}=0;f'(x)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\Rightarrow\lim_{x\to a}\frac{f(x)-\frac{f(x)-f(a)}{x-a}}{x-a}=\frac{f(x)-\frac{f(x)-f(a)}{x-a}}{x-a}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\mbox{ es derivable en }x=a\Leftrightarrow\exists\lim_{t\to0}\frac{f(x+t)-f(x)}{t}=m
\]

\end_inset


\end_layout

\begin_layout Standard
????
\end_layout

\begin_layout Section
Para cada una de las siguientes funciones 
\begin_inset Formula $f(x,y)$
\end_inset

, calcular la derivada direccional en la dirección de 
\begin_inset Formula $v$
\end_inset

 en el punto 
\begin_inset Formula $(x_{0},y_{0})$
\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=x^{2}+y^{2}$
\end_inset

, 
\begin_inset Formula $v=\left(1,0\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0}\right)=\left(1,2\right)$
\end_inset

 y 
\begin_inset Formula $v=\left(0,1\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0}\right)=\left(1,2\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{f(\left(1,2\right)+t(0,1))-f(1,2)}{t}=\lim_{t\to0}\frac{1+4+4t+t^{2}-1-4}{t}=\lim_{t\to0}\frac{2t+t^{2}}{t}=\lim_{t\to0}4+t=4
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f(x,y)=2xy-3x^{2}+y-5$
\end_inset

, 
\begin_inset Formula $v=\left(1,1\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0}\right)=\left(2,3\right)$
\end_inset

 y 
\begin_inset Formula $v=\left(1,2\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0}\right)=\left(0,1\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v'=\left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)\rightarrow\mbox{normalizo el vector}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{f\left(\left(0,1\right)+t\left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)\right)-f\left(0,1\right)}{t}=\lim_{t\to0}\frac{1+t\frac{2}{\sqrt{5}}-5-\left(1-5\right)}{t}=\lim_{t\to0}\frac{1+t\frac{2}{\sqrt{5}}-5-1+5}{t}=\lim_{t\to0}\frac{t2}{t\sqrt{5}}=\frac{2}{\sqrt{5}}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y,z\right)=e^{z}\left(xy+z^{2}\right)$
\end_inset

, 
\begin_inset Formula $v=\left(0,1,0\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0},z_{0}\right)=\left(1,1,2\right)$
\end_inset


\end_layout

\begin_layout Standard

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula 
\[
\lim_{t\to0}\frac{f\left(\left(1,1,2\right)+t\left(0,1,0\right)\right)-f\left(1,1,2\right)}{t}=\lim_{t\to0}\frac{e^{2}\left(1+t+2^{2}\right)-e^{2}\left(1+2^{2}\right)}{t}=\lim_{t\to0}\frac{e^{2}+e^{2}t+4e^{2}-e^{2}-4e^{2}}{t}=\frac{te^{2}}{t}=e^{2}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\left(x+1\right)\sin y-2$
\end_inset

, 
\begin_inset Formula $v=\left(1,0\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0}\right)$
\end_inset

cualquiera
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{f\left(\left(x_{0},y_{0}\right)+t\left(1,0\right)\right)-f\left(x_{0},y_{0}\right)}{t}=\lim_{t\to0}\frac{\left(x_{0}+t+1\right)\sin y_{0}-2-\left(\left(x_{0}+1\right)\sin y_{0}-2\right)}{t}=\lim_{t\to0}\frac{t\sin y_{0}}{t}=\sin y_{0}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\left|\left|\left(x,y\right)\right|\right|$
\end_inset

, 
\begin_inset Formula $v=\left(a,b\right)$
\end_inset

, 
\begin_inset Formula $\left(x_{0},y_{0}\right)$
\end_inset

 con 
\begin_inset Formula $\left|\left|\left(a,b\right)\right|\right|\neq0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{f\left(\left(x_{0},y_{0}\right)+t\left(a,b\right)\right)-f\left(x_{0},y_{0}\right)}{t}=\lim_{t\to0}\frac{\left|\left|\left(x_{0}+ta,y_{0}+tb\right)\right|\right|-\left|\left|\left(x_{0},y_{0}\right)\right|\right|}{t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\left|\left|\left(x_{0}+ta,y_{0}+tb\right)\right|\right|-\left|\left|\left(x_{0},y_{0}\right)\right|\right|=\lim_{t\to0}\sqrt{\left(x_{0}+ta\right)^{2}+\left(y_{0}+tb\right)^{2}}-\sqrt{\left(x_{0}\right)^{2}+\left(y_{0}\right)^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
=\lim_{t\to0}\sqrt{x_{0}^{2}+2x_{0}ta+t^{2}a^{2}+y_{0}^{2}+2y_{0}tb+t^{2}b^{2}}-\sqrt{x_{0}^{2}+y_{0}^{2}}=\lim_{t\to0}\frac{x_{0}^{2}+2x_{0}ta+t^{2}a^{2}+y_{0}^{2}+2y_{0}tb+t^{2}b^{2}-x_{0}^{2}-y_{0}^{2}}{t\left(\left|\left|\left(x_{0}+ta,y_{0}+tb\right)\right|\right|+\left|\left|\left(x_{0},y_{0}\right)\right|\right|\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{2x_{0}ta+t^{2}a^{2}+2y_{0}tb+t^{2}b^{2}}{t\left|\left|\left(x_{0}+ta,y_{0}+tb\right)\right|\right|+\left|\left|\left(x_{0},y_{0}\right)\right|\right|}=\lim_{t\to0}\frac{2x_{0}a+ta^{2}+2y_{0}b+tb^{2}}{\left|\left|\left(x_{0}+ta,y_{0}+tb\right)\right|\right|+\left|\left|\left(x_{0},y_{0}\right)\right|\right|}=\frac{x_{0}a+y_{0}b}{\left|\left|\left(x_{0},y_{0}\right)\right|\right|}
\]

\end_inset


\end_layout

\begin_layout Section
Resolver
\end_layout

\begin_layout Subsection
Sea 
\begin_inset Formula $f\left(x,y\right)=\sqrt{xy+\frac{x}{y}}$
\end_inset

 y 
\begin_inset Formula $v=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
\end_inset

.
 Calcular 
\begin_inset Formula $\frac{\partial f}{\partial y}\left(2,1\right)$
\end_inset

 y 
\begin_inset Formula $\frac{\partial f}{\partial v}\left(2,1\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}(2,1)=\lim_{t\to0}\frac{f(2,1+t)-f(2,1)}{t}=\lim_{t\to0}\frac{\sqrt{2(1+t)+\frac{2}{1+t}}-\sqrt{2+2}}{t}=\lim_{t\to0}\frac{2(1+t)+\frac{2}{1+t}-4}{t\left(\sqrt{2(1+t)+\frac{2}{1+t}}+2\right)}=\lim_{t\to0}\frac{\frac{2(1+t)^{2}+2-4-4t}{1+t}}{t\left(\sqrt{2(1+t)+\frac{2}{1+t}}+2\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
=\lim_{t\to0}\frac{2(1+t)^{2}-2-4t}{t\left(\sqrt{2(1+t)+\frac{2}{1+t}}+2\right)\left(1+t\right)}=\lim_{t\to0}\frac{2+4t+2t^{2}-2-4t}{t\left(\sqrt{2(1+t)+\frac{2}{1+t}}+2\right)\left(1+t\right)}=\lim_{t\to0}\frac{2t^{2}}{t\left(\sqrt{2(1+t)+\frac{2}{1+t}}+2\right)\left(1+t\right)}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\frac{\partial f}{\partial y}\left(2,1\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}(2,1)=\lim_{t\to0}\frac{f(2+t,1)-f(2,1)}{t}=\lim_{t\to0}\frac{\sqrt{2+t+2+t}-\sqrt{2+2}}{t}=\lim_{t\to0}\frac{\sqrt{4+2t}-2}{t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
=\lim_{t\to0}\frac{4+2t-4}{t\left(\sqrt{2(2+t)}+2\right)}=\lim_{t\to0}\frac{2t}{t\left(\sqrt{2(2+t)}+2\right)}=\lim_{t\to0}\frac{2}{\sqrt{2(2+t)}+2}=\frac{2}{\sqrt{4}+2}=\frac{1}{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Rightarrow\frac{\partial f}{\partial v}\left(2,1\right)=\left\langle \nabla f\left(2,1\right),\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right\rangle =0+\frac{1}{2}*\frac{1}{\sqrt{2}}
\]

\end_inset


\end_layout

\begin_layout Subsection
Calcular 
\begin_inset Formula $\frac{\partial f}{\partial z}(1,1,1)$
\end_inset

 para 
\begin_inset Formula $f(x,y,z)=\sqrt{x^{2}+z^{2}}+\ln(y)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial a}(1,1,1)=\lim_{t\to0}\frac{f(1,1,1+t)-f(1,1,1)}{t}=\lim_{t\to0}\frac{\sqrt{1+(1+t)^{2}}+\ln(1)-\sqrt{2}-\ln(1)}{t}=
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{1+(1+t)^{2}-2}{t\left(\sqrt{1+(1+t)^{2}}+\sqrt{2}\right)}=\lim_{t\to0}\frac{t^{2}+2t}{t\left(\sqrt{1+(1+t)^{2}}+\sqrt{2}\right)}=\lim_{t\to0}\frac{t+2}{\left(\sqrt{1+(1+t)^{2}}+\sqrt{2}\right)}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}
\]

\end_inset


\end_layout

\begin_layout Subsection
Sea 
\begin_inset Formula $f(x,y)=\begin{cases}
\frac{x^{3}}{x^{2}+y^{2}} & \mbox{si }\left(x,y\right)\neq\left(0,0\right)\\
0 & \mbox{si }\left(x,y\right)=\left(0,0\right)
\end{cases}$
\end_inset

 Calcular 
\begin_inset Formula $\frac{\partial f}{\partial v}\left(0,0\right)\forall v=(a,b)/\left|\left|v\right|\right|=1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{f(a,b)-f(0,0)}{t}=\lim_{t\to0}\frac{\frac{t^{3}a^{3}}{t^{2}\left(a^{2}+b^{2}\right)}}{t}=\lim_{t\to0}\frac{\frac{ta^{3}}{a^{2}+b^{2}}}{t}=\lim_{t\to0}\frac{ta^{3}}{\left(a^{2}+b^{2}\right)t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{a^{3}}{a^{2}+b^{2}}=\frac{a^{3}}{a^{2}+b^{2}}
\]

\end_inset


\end_layout

\begin_layout Section
Calcular las derivadas parciales de las siguientes funciones.
\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=x^{4}+2xy+y^{3}x-1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=4x^{3}+2y+y^{2}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=2x+2y^{2}x
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y,z\right)=ye^{x}+z$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=ye^{x}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=e^{x}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial z}=1
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=x^{2}\sin^{2}y$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=2x\sin^{2}y
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=2x^{2}\sin y\cos y=x^{2}\sin\left(2y\right)
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\sin x$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=\cos x
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=0
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y,z\right)=z\left(\cos\left(xy\right)+\ln\left(x^{2}+y^{2}+1\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=z\left(y\sin xy+\frac{2x}{x^{2}+y^{2}+1}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=z\left(x\sin xy+\frac{2y}{x^{2}+y^{2}+1}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial z}=\cos\left(xy\right)+\ln\left(x^{2}+y^{2}+1\right)
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=xe^{x^{2}+y^{2}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=e^{x^{2}+y^{2}}+2x^{2}e^{x^{2}+y^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=2yxe^{x^{2}+y^{2}}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\arctan\frac{y}{x}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}=-\frac{1}{\left(\frac{y}{x}\right)^{2}+1}*x^{-2}=-\frac{1}{\left(\frac{y^{2}+x^{2}}{x^{2}}\right)x^{2}}=-\frac{1}{y^{2}+x^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}=\frac{1}{\left(\frac{y}{x}\right)^{2}+1}*\frac{1}{x}=\frac{1}{\left(\frac{y^{2}+x^{2}}{x^{2}}\right)x}=\frac{x}{y^{2}+x^{2}}
\]

\end_inset


\end_layout

\begin_layout Section
Probar que al función 
\begin_inset Formula $f\left(x,y\right)=\left|x\right|+\left|y\right|$
\end_inset

 es continua pero no admite derivadas parciales en el origen.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{\left(x,y\right)\to\left(a,b\right)}\left|x\right|+\left|y\right|=\lim_{\left(x,y\right)\to\left(a,b\right)}\left|x\right|+\lim_{\left(x,y\right)\to\left(a,b\right)}\left|y\right|=\left|a\right|+\left|b\right|=f(a,b)\Rightarrow\mbox{ es continua}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}\left(0,0\right)=\lim_{t\to0}\frac{f\left(t,0\right)-f(0,0)}{t}=\lim_{t\to0}\frac{\left|t\right|}{t}\Rightarrow\nexists\lim_{t\to0}\frac{f\left(t,0\right)-f(0,0)}{t}\Rightarrow\nexists\frac{\partial f}{\partial x}\left(0,0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial y}\left(0,0\right)=\lim_{t\to0}\frac{f\left(0,t\right)-f(0,0)}{t}=\lim_{t\to0}\frac{\left|t\right|}{t}\Rightarrow\nexists\lim_{t\to0}\frac{f\left(0,t\right)-f(0,0)}{t}\Rightarrow\nexists\frac{\partial f}{\partial y}\left(0,0\right)
\]

\end_inset


\end_layout

\begin_layout Section
Consideremos la siguiente función:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\left(x,y\right)=\begin{cases}
\frac{xy}{x^{2}+y^{2}} & \mbox{si }\left(x,y\right)\neq\left(0,0\right)\\
0 & \mbox{si }\left(x,y\right)\neq\left(0,0\right)
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
Probar que las únicas direcciones 
\begin_inset Formula $v\in\mathbb{R}^{2}$
\end_inset

 para las que existe la derivada direcciones en 
\begin_inset Formula $f_{v}$
\end_inset

en el origen son 
\begin_inset Formula $v=\left(1,0\right)$
\end_inset

y 
\begin_inset Formula $v=\left(0,1\right)$
\end_inset

.
 Probar, además, que la función no es continua en el origen.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(1,0\right)\Rightarrow\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f\left(t,0\right)-f(0,0)}{t}=\lim_{t\to0}\frac{\frac{0}{t^{2}}}{t}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(0,1\right)\Rightarrow\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f\left(0,t\right)-f(0,0)}{t}=\lim_{t\to0}\frac{\frac{0}{t^{2}}}{t}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)\Rightarrow\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f\left(at,bt\right)-f(0,0)}{t}=\lim_{t\to0}\frac{\frac{abt^{2}}{t^{2}\left(a^{2}+b^{2}\right)}}{t}=\lim_{t\to0}\frac{ab}{t\left(a^{2}+b^{2}\right)}=\lim_{t\to0}\frac{ab}{t}\to\infty\forall a,b\neq0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{nota: }||v||=1\wedge a^{2}+b^{2}=||v||^{2}=1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{si }y=mx\Rightarrow\lim_{x\to0}\frac{mx^{2}}{x^{2}\left(1+m^{2}\right)}=\lim_{x\to0}\frac{m}{1+m^{2}}\neq f\left(0,0\right)\forall m\Rightarrow f\mbox{ no es continua en }\left(0,0\right)
\]

\end_inset


\end_layout

\begin_layout Section
Consideremos la siguiente función
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\left(x,y\right)=\begin{cases}
\frac{x^{3}y}{x^{6}+y^{2}} & \mbox{si }\left(x,y\right)\neq\left(0,0\right)\\
0 & \mbox{si }\left(x,y\right)=\left(0,0\right)
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
Probar que 
\begin_inset Formula $f$
\end_inset

 admite derivadas direccionales en el origen para todo vector unitario 
\begin_inset Formula $v\in R^{2}$
\end_inset

.
 Sin embargo, 
\begin_inset Formula $f$
\end_inset

 tampoco es continua en el origen.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right);||v||=1\Rightarrow\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f\left(at,bt\right)-f(0,0)}{t}=\lim_{t\to0}\frac{t^{4}a^{3}b}{t\left(t^{6}a^{6}+bt^{2}\right)}=\lim_{t\to0}\frac{t^{4}a^{3}b}{t^{3}\left(t^{4}a^{6}+b\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
=\lim_{t\to0}\frac{ta^{3}b}{t^{4}a^{6}+b}=0\forall a,b
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{si }y=x^{3}\Rightarrow\lim_{x\to0}\frac{x^{6}}{x^{6}+x^{6}}=\lim_{x\to0}\frac{1}{2}\neq f\left(0,0\right)\Rightarrow f\mbox{ no es continua en }\left(0,0\right)
\]

\end_inset


\end_layout

\begin_layout Section
Sea 
\begin_inset Formula $f\left(x,y\right)=x^{1/3}y^{1/3}$
\end_inset


\end_layout

\begin_layout Subsection
Usando la definición de derivada direccional mostrar que:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}\left(0,0\right)=\frac{\partial f}{\partial y}\left(0,0\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
y que 
\begin_inset Formula $\pm e_{1},\pm e_{2}$
\end_inset

son las únicas direcciones para las cuales existe la derivada direccional
 en el origen.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)\wedge||v||=1:\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f\left(ta,tb\right)-f\left(0,0\right)}{t}=\lim_{t\to0}\frac{\left(tatb\right)^{1/3}}{t}=\begin{cases}
a,b\neq0 & \to\infty\\
a=0\vee b=0 & 0
\end{cases}\lim_{t\to0}\frac{\left(ab\right)^{1/3}}{t^{1/3}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\left(\frac{\partial f}{\partial v}\left(0,0\right)=0\Leftrightarrow a=0\vee b=0\Leftrightarrow v=(\pm1,0)\vee v=(0,\pm1)\right)\wedge\left(\nexists\frac{\partial f}{\partial v}\left(0,0\right)\forall a,b\neq0\right)
\]

\end_inset


\end_layout

\begin_layout Subsection
Mostrar que 
\begin_inset Formula $f$
\end_inset

 es continua en 
\begin_inset Formula $\left(0,0\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
x^{1/3}\mbox{ es continua en }\mathbb{R}^{2}\wedge y^{1/3}\mbox{ es continua en }\mathbb{R}^{2}\Rightarrow x^{1/3}y^{1/3}\mbox{ es continua en }\mathbb{R}^{2}
\]

\end_inset


\end_layout

\begin_layout Part
Diferenciabilidad
\end_layout

\begin_layout Section
Estudiar la continuidad, existencia de derivadas parciales y diferenciablidad
 de la siguientes funciones en el origen:
\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y,z\right)=\sqrt{\left|xyz\right|}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
xyz\mbox{ es continua }\Rightarrow\left|xyz\right|\mbox{ es continua }\Rightarrow\sqrt{\left|xyz\right|}\mbox{ es continua}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b,c\right)/||v||=1:\frac{\partial f}{\partial v}\left(0,0,0\right)=\lim_{t\to0}\frac{f(at,bt,ct)-f\left(0,0,0\right)}{t}=\lim_{t\to0}\frac{\sqrt{\left|t^{3}abc\right|}}{t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{\sqrt{t^{2}\left|tabc\right|}}{t}=\lim_{t\to0}\sqrt{\left|tabc\right|}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f\left(x,y,z\right)\mbox{ diferenciable }\Leftrightarrow\lim_{\left(x,y,z\right)\to\left(0,0,0\right)}\frac{f(x,y,z)-f(0,0,0)-\frac{\partial f}{\partial x}\left(0,0,0\right)(x)-\frac{\partial f}{\partial y}\left(0,0,0\right)(y)-\frac{\partial f}{\partial z}\left(0,0,0\right)(z)}{||(x,y,z)||}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\sqrt{\left|xyz\right|}}{||(x,y,z)||}\leq\frac{\sqrt{\left|xyz\right|}}{||(x,y,z)||}\leq\frac{\sqrt{||(x,y,z)||^{3}}}{||(x,y,z)||}=\sqrt{||(x,y,z)||}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
0\leq\lim_{\left(x,y,z\right)\to\left(0,0,0\right)}\frac{\sqrt{\left|xyz\right|}}{||(x,y,z)||}\leq\lim_{\left(x,y,z\right)\to\left(0,0,0\right)}\sqrt{||(x,y,z)||}\leq0\rightarrow\sqrt{\left|xyz\right|}\geq0\wedge||(x,y,z)||\geq0\Rightarrow\frac{\sqrt{\left|xyz\right|}}{||(x,y,z)||}\ge0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\lim_{\left(x,y,z\right)\to\left(0,0,0\right)}\frac{\sqrt{\left|xyz\right|}}{||(x,y,z)||}=0\Rightarrow f\left(x,y,z\right)\mbox{ es diferenciable}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\begin{cases}
x\sin\left(4\arctan\frac{y}{x}\right) & \mbox{si }x\neq0\\
0 & \mbox{si }x=0
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{x\to0}x\sin\left(4\arctan\frac{y}{x}\right)=\left(\to0\right)*\sin\left(4\arctan\left(\to\infty\right)\right)=\left(\to0\right)*\sin\left(4\left(\to\frac{\pi}{2}\right)\right)=0=f\left(0,y\right)\Rightarrow\mbox{ es continua}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)/||v||=1:\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f(at,bt)-f\left(0,0\right)}{t}=\lim_{t\to0}\frac{at\sin(4\arctan\frac{bt}{at})}{t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}a\sin(4\arctan\frac{b}{a})=a\sin(4\arctan\frac{b}{a})\Rightarrow\begin{cases}
\nexists\frac{\partial f}{\partial x}\left(0,0\right):a=1,b=0 & \sin(4\arctan0)=0\\
\frac{\partial f}{\partial y}\left(0,0\right):a=0,b=1 & 0\sin(4\arctan\frac{1}{0})\mbox{ no está definida la división por 0}
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\mbox{ no es diferenciable ya que }\nexists\frac{\partial f}{\partial y}\left(0,0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
¿Puede ser que la derivada parcial en x tampoco esté definida? No veo como,
 pero wolfamalpha dice eso.
\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\begin{cases}
\frac{x^{3}-y^{3}}{x^{2}+y^{2}} & \mbox{si }\left(x,y\right)\neq\left(0,0\right)\\
0 & \mbox{si }\left(x,y\right)=\left(0,0\right)
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{si }y=\sqrt[3]{x}\Rightarrow\lim_{x\to0}\frac{x^{3}-x}{x^{2}+x^{\frac{2}{3}}}=\lim_{x\to0}\frac{x\left(x^{2}-1\right)}{x\left(x+x^{\frac{1}{3}}\right)}=\lim_{x\to0}\frac{x^{2}-1}{x+x^{\frac{1}{3}}}=\infty\Rightarrow\mbox{No es continua en }\left(0,0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)/||v||=1:\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f(at,bt)-f\left(0,0\right)}{t}=\lim_{t\to0}\frac{\frac{t^{3}\left(a^{3}-b^{3}\right)}{t^{2}\left(a^{2}+b^{2}\right)}}{t}=\lim_{t\to0}\frac{a^{3}-b^{3}}{a^{2}+b^{2}}=\begin{cases}
a=1,b=0: & \frac{\partial f}{\partial x}\left(0,0\right)=1\\
a=0,b=1: & \frac{\partial f}{\partial y}\left(0,0\right)=-1
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{No es continua por lo tanto no será diferenciable.}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\begin{cases}
\frac{xy}{\sqrt{x^{2}+y^{2}}}\sin\left(\frac{1}{\sqrt{x^{2}+y^{2}}}\right) & \mbox{si }\left(x,y\right)\neq\left(0,0\right)\\
0 & \mbox{si }\left(x,y\right)=\left(0,0\right)
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
0\leq\left|\frac{xy}{\sqrt{x^{2}+y^{2}}}\sin\left(\frac{1}{\sqrt{x^{2}+y^{2}}}\right)\right|\leq\left|\frac{xy}{\sqrt{x^{2}+y^{2}}}\right|\left|\sin\left(\frac{1}{\sqrt{x^{2}+y^{2}}}\right)\right|\leq\frac{\left|xy\right|}{\sqrt{x^{2}+y^{2}}}\leq\frac{\left|\left|(x,y)\right|\right|^{2}}{\left|\left|(x,y)\right|\right|}=\left|\left|(x,y)\right|\right|
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\left|\lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy}{\sqrt{x^{2}+y^{2}}}\sin\left(\frac{1}{\sqrt{x^{2}+y^{2}}}\right)\right|\leq\lim_{\left(x,y\right)\to\left(0,0\right)}\left|\left|(x,y)\right|\right|=0\Rightarrow\lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy}{\sqrt{x^{2}+y^{2}}}\sin\left(\frac{1}{\sqrt{x^{2}+y^{2}}}\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Rightarrow\mbox{Es continua}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)/||v||=1:\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f(at,bt)-f\left(0,0\right)}{t}=\lim_{t\to0}\frac{\frac{abt^{2}}{t\sqrt{a^{2}+b^{2}}}\sin\left(\frac{1}{t\sqrt{a^{2}+b^{2}}}\right)}{t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{ab}{\sqrt{a^{2}+b^{2}}}\sin\left(\frac{1}{t\sqrt{a^{2}+b^{2}}}\right)=\begin{cases}
a=0\vee b=0\Rightarrow & 0*\lim_{t\to0}\sin\left(\frac{1}{t\sqrt{a^{2}+b^{2}}}\right)=0\\
a,b\neq0\Rightarrow & \mbox{oscila}
\end{cases}\rightarrow\mbox{ya que }\nexists\lim_{t\to0}\sin\left(\frac{1}{t\sqrt{a^{2}+b^{2}}}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial f}{\partial x}\left(0,0\right)=\frac{\partial f}{\partial y}\left(0,0\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{No es diferenciable porque no todas sus derivadas direccionales existen.}
\]

\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $f\left(x,y\right)=\begin{cases}
\frac{x}{y}\sin\left(\frac{1}{y}\right) & \mbox{si }y\neq0\\
0 & \mbox{si }y=0
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{si }x=y:\lim_{x\to0}\frac{x}{x}\sin\left(\frac{1}{x}\right)=\lim_{x\to0}\sin\left(\frac{1}{x}\right)\rightarrow\nexists\therefore\mbox{no es continua}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)/||v||=1:\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f(at,bt)-f\left(0,0\right)}{t}=\lim_{t\to0}\frac{a}{b}\sin\left(\frac{1}{bt}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
=\frac{a}{b}\lim_{t\to0}\sin\left(\frac{1}{bt}\right)\rightarrow\nexists\therefore\mbox{No tiene derivadas direccionales \ensuremath{\Rightarrow} no es diferenciable.}
\]

\end_inset


\end_layout

\begin_layout Section
Sea 
\begin_inset Formula $f\left(x,y\right)=\begin{cases}
\frac{|x|y}{\sqrt{x^{2}+y^{2}}} & \mbox{si }\left(x,y\right)\neq\left(0,0\right)\\
0 & \mbox{si }\left(x,y\right)\neq\left(0,0\right)
\end{cases}$
\end_inset

 
\end_layout

\begin_layout Standard
Probar que en el origen 
\begin_inset Formula $f$
\end_inset

 es continua, admite todas las derivadas direccionales pero no es diferenciable.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
0\leq\left|\lim_{\left(x,y\right)\to(0,0)}\frac{|x|y}{\sqrt{x^{2}+y^{2}}}\right|\leq\lim_{\left(x,y\right)\to(0,0)}\frac{\left||x|y\right|}{\sqrt{x^{2}+y^{2}}}\leq\lim_{\left(x,y\right)\to(0,0)}\frac{\left|\left|\left(x,y\right)\right|\right|^{2}}{\left|\left|\left(x,y\right)\right|\right|}\leq\lim_{\left(x,y\right)\to(0,0)}\left|\left|\left(x,y\right)\right|\right|\leq0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\lim_{\left(x,y\right)\to(0,0)}\frac{|x|y}{\sqrt{x^{2}+y^{2}}}=0\Rightarrow\mbox{Es continua}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
v=\left(a,b\right)/||v||=1:\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f(at,bt)-f\left(0,0\right)}{t}=\lim_{t\to0}\frac{\frac{|at|bt}{t\sqrt{a^{2}+b^{2}}}}{t}=\lim_{t\to0}\frac{|at|b}{t}=\lim_{t\to0}\frac{|t||a|b}{t}\rightarrow\nexists
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\therefore\mbox{No tiene derivada direccional\Rightarrow no es diferenciable}
\]

\end_inset


\end_layout

\begin_layout Section
Sea 
\begin_inset Formula $f\left(x,y\right)=x^{2}-3xy+y^{2}$
\end_inset


\end_layout

\begin_layout Subsection
Encontrar las ecuaciones paramétricas de las rectas tangentes a las siguientes
 curvas en 
\begin_inset Formula $t_{0}=0:$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sigma_{1}\left(t\right)=\left(t+1,3,f\left(t+1,3\right)\right)\mbox{ y }\sigma_{2}\left(t\right)=\left(t+1,t+3,f\left(t+1,t+3\right)\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Resolución 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
tangente 
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\uuline default
\uwave default
\noun default
\color inherit

\begin_inset Formula $\sigma_{1}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial\sigma_{1}}{\partial t}\left(t_{0}=0\right)=\left(1,0,\frac{df}{dt}\left(1,3\right)\right);f\left(1,3\right)=1-9+9=1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{df}{dt}\left(1,3\right)=\lim_{t\to0}\frac{f\left(1+t,3\right)-f\left(1,3\right)}{t}=\lim_{t\to0}\frac{\left(1+t\right)^{2}-3\left(1+t\right)3+3^{2}-1}{t}=
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{\left(1+t\right)^{2}-9t-1}{t}=\lim_{t\to0}\frac{t^{2}+2t+1-9t-1}{t}=-7
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{tangente }\sigma_{1}(1,3,1):\left(1,3,1\right)+\lambda\left(1,0,-7\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Resolución 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
tangente 
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\uuline default
\uwave default
\noun default
\color inherit

\begin_inset Formula $\sigma_{2}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial\sigma_{2}}{\partial t}\left(t_{0}=0\right)=\left(1,1,\frac{df}{dt}\left(1,3\right)\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{df}{dt}\left(1,3\right)=\lim_{t\to0}\frac{f\left(1+t,3+t\right)-f\left(1,3\right)}{t}=\lim_{t\to0}\frac{\left(1+t\right)^{2}-3\left(3+t\right)\left(1+t\right)+\left(3+t\right)^{2}-1}{t}=\lim_{t\to0}\frac{1+2t+t^{2}-3\left(3+4t+t^{2}\right)+9+6t+t^{2}-1}{t}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{t\to0}\frac{2t+t^{2}-9-12t-3t^{2}+9+6t+t^{2}}{t}=\lim_{t\to0}2-12+6+5t=-4
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mbox{tangente }\sigma_{2}(1,3,1):\left(1,3,1\right)+\lambda\left(1,1,-4\right)
\]

\end_inset


\end_layout

\begin_layout Subsection
Encontrar la ecuación de un plano 
\begin_inset Formula $z=T\left(x,y\right)$
\end_inset

 que contenga ambas rectas y mostrar que
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\lim_{\left(x,y\right)\to\left(1,3\right)}\frac{f\left(x,y\right)-T\left(x,y\right)}{\left|\left|\left(x,y\right)-\left(1,3\right)\right|\right|}=0
\]

\end_inset


\end_layout

\end_body
\end_document