# User:StevenM

## LIMITES

Ejercicio 1

${\displaystyle \lim _{x\to -2}{\frac {5x+10}{x+2}}}$
${\displaystyle \lim _{x\to -2}{\frac {5(x+2)}{x+2}}}$
${\displaystyle \ 5}$

Ejercicio 2

${\displaystyle \lim _{x\to -4}{\frac {x^{2}-3x-4}{x^{2}+7x+6}}}$
${\displaystyle \lim _{x\to -4}{\frac {(x-4)(x+1)}{(x+6)(x+1)}}}$
${\displaystyle \lim _{x\to -4}{\frac {(x-4)}{x+6}}}$
${\displaystyle {\frac {-4-4}{-4+6}}}$
${\displaystyle {\frac {-8}{2}}}$
${\displaystyle \ -4}$

Ejercicio 3

${\displaystyle \lim _{x\to -5}{\frac {x+5}{x^{2}+8x-15}}}$
${\displaystyle \lim _{x\to -5}{\frac {-5+5}{-5^{2}+8*5-15}}}$
${\displaystyle \lim _{x\to -5}{\frac {0}{30}}}$
${\displaystyle \ 0}$

Ejercicio 4

${\displaystyle \lim _{x\to 6}{\frac {x^{2}-8x+12}{x^{2}-9x+18}}}$
${\displaystyle \lim _{x\to 6}{\frac {(x-6)(x-2)}{(x-6)(x-3)}}}$
${\displaystyle \lim _{x\to 6}{\frac {x-2}{x-3}}}$
${\displaystyle {\frac {6-2}{6-3}}}$
${\displaystyle {\frac {4}{3}}}$

Ejercicio 5

${\displaystyle \lim _{x\to \infty }{\frac {2x^{5}-x^{3}+1}{x^{5}}}}$
${\displaystyle \lim _{x\to \infty }{\frac {{\frac {2x^{5}}{x^{5}}}-{\frac {x^{3}}{x^{5}}}+{\frac {1}{x^{5}}}}{\frac {x^{5}}{x^{5}}}}}$
${\displaystyle {\frac {(2-0+0)}{(1)}}}$
${\displaystyle \ 2}$

Ejercicio 1

${\displaystyle \ y={3x^{5}-2x}{2x^{3}}}$
${\displaystyle \ y'={\frac {(15x^{4}-2)(2x^{3})-(3x^{5}-2x)(6x^{2})}{4x^{6}}}}$
${\displaystyle \ y'={\frac {(30x^{7}-4x^{3}-18x^{7}+12x^{3})}{4x^{6}}}}$
${\displaystyle \ y'={\frac {12x^{7}+8x^{3}}{4x^{6}}}}$

Ejericicio 2

${\displaystyle \ y={\frac {(3x^{2}-8x)(2x-4)}{4x^{6}}}}$
${\displaystyle \ y'={\frac {4x^{6}((3x^{2}-8x)(2)+(2x-4)(6x-8))}{(4x^{6})^{2}}}}$
${\displaystyle \ y'={\frac {4x^{6}(6x^{2}-16x+12x^{2}-16x-24x+32)}{(4x^{6})^{2}}}}$
${\displaystyle \ y'={\frac {4x^{6}(18x^{2}-56x+32)}{(4x^{6})^{2}}}}$
${\displaystyle \ y'={\frac {124x^{7}+98x^{6}}{(4x^{6})^{2}}}}$

Ejercicio 3

${\displaystyle \ f(x)=sen^{1}0({x+1})^{\frac {1}{2}}}$
${\displaystyle \ f'(x)=10sen^{9}({x+1})^{\frac {1}{2}}*cos({x+1})^{\frac {1}{2}}*{\frac {1}{2}}(x+1)^{\frac {-1}{2}}}$

Ejercicio 4

${\displaystyle \ f(x)=(cos^{2}(3X)+sen^{4}(5X))^{\frac {1}{3}}}$
${\displaystyle \ f'(x)={\frac {1}{3}}(cos^{2}(3X)+sen^{4}(5X))^{\frac {-2}{3}}*({2cos(3x)}*{-sen(3x)*(3)}{4sen^{3}(5x)}*{cos(5x)*(5)})}$

Ejercicio 5

${\displaystyle \ f(x)=cos(cos(cos(3x)))}$
${\displaystyle \ f'(x)=-sen(cos(cos(3x)))*-sen(cos(3x))*-sen(3x)*3}$

Ejercicio 6

${\displaystyle \ y^{3}tang^{2}(x^{2})+y^{2}sec(x)+(y)^{\frac {1}{3}}=0}$
${\displaystyle \ y^{3}*2tang(x^{2})*sec^{2}(x^{2})*2x+tang^{2}(x^{2})*3y*y'+y^{2}*sec(x)tang(x)+sec(x)*2y*y'+{\frac {1}{3}}(y)^{\frac {-2}{3}}*y'=0}$
${\displaystyle \ tang^{2}(x^{2})*3y*y'+sec(x)*2y*y'+{\frac {1}{3}}(y)^{\frac {-2}{3}}*y'=-(y^{3}*2tang(x^{2})*sec^{2}(x^{2})*2x+y^{2}*sec(x)tang(x))}$
${\displaystyle \ y'(tang^{2}(x^{2})*3y+sec(x)*2y+{\frac {1}{3}}(y)^{\frac {-2}{3}})=-(y^{3}*2tang(x^{2})*sec^{2}(x^{2})*2x+y^{2}*sec(x)tang(x))}$
${\displaystyle \ y'={\frac {-(y^{3}*2tang(x^{2})*sec^{2}(x^{2})*2x+y^{2}*sec(x)tang(x))}{(tang^{2}(x^{2})*3y+sec(x)*2y+{\frac {1}{3}}(y)^{\frac {-2}{3}})}}}$

## RAZON DE CAMBIO

Ejercicio 1

Dos personas "a" y "b"caminan por calles rectas, que se crusan en un angulo recto, el caminante "a" se hacerca a la intersecciòn a 2m/sg y el "b" se aleja a 1m/sg . ¿cual es la razon de cambio del angulo "x" cuando "a" este a 10m de la intersecciòn y "b" a 20m?.

${\displaystyle \ Datos:}$
${\displaystyle \ sa=10m}$
${\displaystyle \ sb=20m}$
${\displaystyle {\frac {da}{dt}}=-2m/sg}$
${\displaystyle {\frac {db}{dt}}=1m/sg}$
${\displaystyle {\frac {dx}{dt}}=?}$

${\displaystyle \ Solucion:}$
${\displaystyle \ tang(x)={\frac {sa}{sb}}}$
${\displaystyle \ tang(x)=(sa)(sb)^{-}1}$
${\displaystyle \ sec^{2}(x)*{\frac {dx}{dt}}=(sa)(-sb)^{-}2*{\frac {db}{dt}}+(sb)^{-}1*{\frac {dx}{dt}}}$
${\displaystyle \ sec^{2}(x)*{\frac {dx}{dt}}={\frac {sa}{(-sb)^{2}}}*{\frac {db}{dt}}+{\frac {1}{sb}}*{\frac {dx}{dt}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {{\frac {sa}{(-sb)^{2}}}*{\frac {db}{dt}}+{\frac {1}{sb}}*{\frac {dx}{dt}}}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {{\frac {10}{(-20)^{2}}}*1+{\frac {1}{20}}*(-2)}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {{\frac {10}{-400}}+{\frac {-1}{10}}}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {{\frac {1}{-40}}+{\frac {-1}{10}}}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {\frac {40+10}{-400}}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {\frac {50}{-400}}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {\frac {5}{-40}}{sec^{2}(x)}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {\frac {5}{-40}}{\frac {5}{4}}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {20}{-200}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {2}{-20}}}$
${\displaystyle {\frac {dx}{dt}}={\frac {1}{-10}}}$
${\displaystyle {\frac {dx}{dt}}=-0.1rad/sg}$

Ejercicio 2

Un recipiente tiene forma de cono circular con altura "h" igual a 10m y un radio "r" igual a 4m, se introduce agua en el recipiente a una velocidad constante de 5m/min , ¿con que velocidad se eleva el nivel del agua cuando la profundidad es de 5m y el vertice del cono esta hacia arriba?.

${\displaystyle \ Datos:}$
${\displaystyle \ h=10m}$
${\displaystyle \ r=4m}$
${\displaystyle {\frac {dv}{dt}}=5m^{3}/min}$
${\displaystyle {\frac {dh}{dt}}=?}$

${\displaystyle \ Solucion:}$

primero se deja todo en funciòn de una sola variable.

${\displaystyle {\frac {r}{4}}={\frac {h}{10}}}$
${\displaystyle \ r={\frac {4h}{10}}}$
${\displaystyle \ r={\frac {2h}{5}}}$

luego se deriva y se remplaza en la ecuaciòn del volumen

${\displaystyle \ v={\frac {4*phi*r^{3}*{\frac {dr}{dt}}}{3}}}$
${\displaystyle \ 5m^{3}/min=4*phi*{\frac {2h^{2}}{5^{2}}}*{\frac {dh}{dt}}}$
${\displaystyle \ 5m^{3}/min=4*phi*{\frac {2h^{2}}{25}}*{\frac {dh}{dt}}}$
${\displaystyle {\frac {dr}{dt}}={\frac {5m^{3}/min*25}{4*phi*{2h^{2}}}}}$
${\displaystyle {\frac {dr}{dt}}=75m^{3}/min}$

Ejercicio 3

un globo se eleva verticalmente desde un punto situado a 150 pies de de un observador, el globo sube a razon de 8pies/sg, ¿cual sera la rapidez de separaciòn del globo con respecto al observador, cuando la altura sea de 50 pies?.

${\displaystyle \ Datos:}$
${\displaystyle \ d=150m}$
${\displaystyle \ h=50m}$
${\displaystyle \ s=?}$
${\displaystyle {\frac {dh}{dt}}=8pies/sg}$
${\displaystyle {\frac {ds}{dt}}=?}$

${\displaystyle \ Solucion:}$

primero se despeja el valor de la que forman la altura y la distancia con el observador.

${\displaystyle \ s^{2}=h^{2}+150^{2}}$
${\displaystyle \ s=(50^{2}+150^{2})^{\frac {1}{2}}}$
${\displaystyle \ s=(2500+67500)^{\frac {1}{2}}}$
${\displaystyle \ s=70000^{\frac {1}{2}}}$
${\displaystyle \ s=264.57pies}$

luego se derivan las variables y se remplazan los valores.

${\displaystyle \ s^{2}=h^{2}+150^{2}}$
${\displaystyle \ 2s*{\frac {ds}{dt}}=2h*{\frac {dh}{dt}}}$
${\displaystyle {\frac {ds}{dt}}={\frac {2h}{2s}}*{\frac {dh}{dt}}}$
${\displaystyle {\frac {ds}{dt}}={\frac {h}{s}}*{\frac {dh}{dt}}}$
${\displaystyle {\frac {ds}{dt}}={\frac {50}{264.57}}*8pies/sg}$
${\displaystyle {\frac {ds}{dt}}=0.188*8pies/sg}$
${\displaystyle {\frac {ds}{dt}}=1.51pies/sg}$

Ejercicio 4

las aristas de un cubo variable aumentan a razon de 3pulg/sg, ¿cual sera la rapidez de variaciòn del volumen del cubo, cuando la arista tiene 10 pulgadas de largo?.

${\displaystyle \ Datos:}$
${\displaystyle \ l=10pulg}$
${\displaystyle {\frac {dl}{dt}}=3pulg/sg}$
${\displaystyle {\frac {dv}{dt}}=?}$

${\displaystyle \ Solucion:}$
${\displaystyle \ v=l^{3}}$
${\displaystyle \ v=3l^{2}*{\frac {dl}{dt}}}$
${\displaystyle \ v=3(10pulg)^{2}*3pulg/sg}$
${\displaystyle \ v=3(100pulg^{2})*3pulg/sg}$
${\displaystyle \ v=300pulg^{2}*3pulg/sg}$
${\displaystyle \ v=900pulg^{3}/sg}$

Ejercicio 5

Suponiendo que una burbuja de jabon mantenga una forma esferica constante, ¿cual sera la rapidez de cambio del radio de la burbuja, cuando el radio sea igual a 2 pulgadas, sabiendo que se le inyecta aire a razon de 4 pulgadas cubicas por segundo?.

${\displaystyle \ Datos:}$
${\displaystyle \ r=2pulg}$
${\displaystyle {\frac {dv}{dt}}=4pulg^{3}/sg}$
${\displaystyle {\frac {dr}{dt}}=?}$

${\displaystyle \ Solucion:}$
${\displaystyle \ v={\frac {4*phi*r^{3}}{3}}}$
${\displaystyle {\frac {dv}{dt}}={\frac {4*phi}{3}}*3r^{2}*{\frac {dr}{dt}}}$
${\displaystyle {\frac {dv}{dt}}=4*phi*r^{2}*{\frac {dr}{dt}}}$
${\displaystyle {\frac {dr}{dt}}={\frac {\frac {dv}{dt}}{4*phi*r^{2}}}}$
${\displaystyle {\frac {dr}{dt}}={\frac {4pulg^{3}/sg}{4*phi*(2pulg)^{2}}}}$
${\displaystyle {\frac {dr}{dt}}={\frac {4pulg^{3}/sg}{4*phi*(4pulg^{2})}}}$
${\displaystyle {\frac {dr}{dt}}={\frac {1pulg^{3}/sg}{phi*(4pulg^{2})}}}$
${\displaystyle {\frac {dr}{dt}}=0.079pulg/sg}$