Ejercicio 1
lim
x
→
−
2
5
x
+
10
x
+
2
{\displaystyle \lim _{x\to -2}{\frac {5x+10}{x+2}}}
lim
x
→
−
2
5
(
x
+
2
)
x
+
2
{\displaystyle \lim _{x\to -2}{\frac {5(x+2)}{x+2}}}
5
{\displaystyle \ 5}
lim
x
→
−
4
x
2
−
3
x
−
4
x
2
+
7
x
+
6
{\displaystyle \lim _{x\to -4}{\frac {x^{2}-3x-4}{x^{2}+7x+6}}}
lim
x
→
−
4
(
x
−
4
)
(
x
+
1
)
(
x
+
6
)
(
x
+
1
)
{\displaystyle \lim _{x\to -4}{\frac {(x-4)(x+1)}{(x+6)(x+1)}}}
lim
x
→
−
4
(
x
−
4
)
x
+
6
{\displaystyle \lim _{x\to -4}{\frac {(x-4)}{x+6}}}
−
4
−
4
−
4
+
6
{\displaystyle {\frac {-4-4}{-4+6}}}
−
8
2
{\displaystyle {\frac {-8}{2}}}
−
4
{\displaystyle \ -4}
lim
x
→
−
5
x
+
5
x
2
+
8
x
−
15
{\displaystyle \lim _{x\to -5}{\frac {x+5}{x^{2}+8x-15}}}
lim
x
→
−
5
−
5
+
5
−
5
2
+
8
∗
5
−
15
{\displaystyle \lim _{x\to -5}{\frac {-5+5}{-5^{2}+8*5-15}}}
lim
x
→
−
5
0
30
{\displaystyle \lim _{x\to -5}{\frac {0}{30}}}
0
{\displaystyle \ 0}
Ejercicio 4
lim
x
→
6
x
2
−
8
x
+
12
x
2
−
9
x
+
18
{\displaystyle \lim _{x\to 6}{\frac {x^{2}-8x+12}{x^{2}-9x+18}}}
lim
x
→
6
(
x
−
6
)
(
x
−
2
)
(
x
−
6
)
(
x
−
3
)
{\displaystyle \lim _{x\to 6}{\frac {(x-6)(x-2)}{(x-6)(x-3)}}}
lim
x
→
6
x
−
2
x
−
3
{\displaystyle \lim _{x\to 6}{\frac {x-2}{x-3}}}
6
−
2
6
−
3
{\displaystyle {\frac {6-2}{6-3}}}
4
3
{\displaystyle {\frac {4}{3}}}
lim
x
→
∞
2
x
5
−
x
3
+
1
x
5
{\displaystyle \lim _{x\to \infty }{\frac {2x^{5}-x^{3}+1}{x^{5}}}}
lim
x
→
∞
2
x
5
x
5
−
x
3
x
5
+
1
x
5
x
5
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}}}}}
(
2
−
0
+
0
)
(
1
)
{\displaystyle {\frac {(2-0+0)}{(1)}}}
2
{\displaystyle \ 2}
Ejercicio 1
y
=
3
x
5
−
2
x
2
x
3
{\displaystyle \ y={3x^{5}-2x}{2x^{3}}}
y
′
=
(
15
x
4
−
2
)
(
2
x
3
)
−
(
3
x
5
−
2
x
)
(
6
x
2
)
4
x
6
{\displaystyle \ y'={\frac {(15x^{4}-2)(2x^{3})-(3x^{5}-2x)(6x^{2})}{4x^{6}}}}
y
′
=
(
30
x
7
−
4
x
3
−
18
x
7
+
12
x
3
)
4
x
6
{\displaystyle \ y'={\frac {(30x^{7}-4x^{3}-18x^{7}+12x^{3})}{4x^{6}}}}
y
′
=
12
x
7
+
8
x
3
4
x
6
{\displaystyle \ y'={\frac {12x^{7}+8x^{3}}{4x^{6}}}}
y
=
(
3
x
2
−
8
x
)
(
2
x
−
4
)
4
x
6
{\displaystyle \ y={\frac {(3x^{2}-8x)(2x-4)}{4x^{6}}}}
y
′
=
4
x
6
(
(
3
x
2
−
8
x
)
(
2
)
+
(
2
x
−
4
)
(
6
x
−
8
)
)
(
4
x
6
)
2
{\displaystyle \ y'={\frac {4x^{6}((3x^{2}-8x)(2)+(2x-4)(6x-8))}{(4x^{6})^{2}}}}
y
′
=
4
x
6
(
6
x
2
−
16
x
+
12
x
2
−
16
x
−
24
x
+
32
)
(
4
x
6
)
2
{\displaystyle \ y'={\frac {4x^{6}(6x^{2}-16x+12x^{2}-16x-24x+32)}{(4x^{6})^{2}}}}
y
′
=
4
x
6
(
18
x
2
−
56
x
+
32
)
(
4
x
6
)
2
{\displaystyle \ y'={\frac {4x^{6}(18x^{2}-56x+32)}{(4x^{6})^{2}}}}
y
′
=
124
x
7
+
98
x
6
(
4
x
6
)
2
{\displaystyle \ y'={\frac {124x^{7}+98x^{6}}{(4x^{6})^{2}}}}
f
(
x
)
=
s
e
n
1
0
(
x
+
1
)
1
2
{\displaystyle \ f(x)=sen^{1}0({x+1})^{\frac {1}{2}}}
f
′
(
x
)
=
10
s
e
n
9
(
x
+
1
)
1
2
∗
c
o
s
(
x
+
1
)
1
2
∗
1
2
(
x
+
1
)
−
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}}}
f
(
x
)
=
(
c
o
s
2
(
3
X
)
+
s
e
n
4
(
5
X
)
)
1
3
{\displaystyle \ f(x)=(cos^{2}(3X)+sen^{4}(5X))^{\frac {1}{3}}}
f
′
(
x
)
=
1
3
(
c
o
s
2
(
3
X
)
+
s
e
n
4
(
5
X
)
)
−
2
3
∗
(
2
c
o
s
(
3
x
)
∗
−
s
e
n
(
3
x
)
∗
(
3
)
4
s
e
n
3
(
5
x
)
∗
c
o
s
(
5
x
)
∗
(
5
)
)
{\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)})}
f
(
x
)
=
c
o
s
(
c
o
s
(
c
o
s
(
3
x
)
)
)
{\displaystyle \ f(x)=cos(cos(cos(3x)))}
f
′
(
x
)
=
−
s
e
n
(
c
o
s
(
c
o
s
(
3
x
)
)
)
∗
−
s
e
n
(
c
o
s
(
3
x
)
)
∗
−
s
e
n
(
3
x
)
∗
3
{\displaystyle \ f'(x)=-sen(cos(cos(3x)))*-sen(cos(3x))*-sen(3x)*3}
y
3
t
a
n
g
2
(
x
2
)
+
y
2
s
e
c
(
x
)
+
(
y
)
1
3
=
0
{\displaystyle \ y^{3}tang^{2}(x^{2})+y^{2}sec(x)+(y)^{\frac {1}{3}}=0}
y
3
∗
2
t
a
n
g
(
x
2
)
∗
s
e
c
2
(
x
2
)
∗
2
x
+
t
a
n
g
2
(
x
2
)
∗
3
y
∗
y
′
+
y
2
∗
s
e
c
(
x
)
t
a
n
g
(
x
)
+
s
e
c
(
x
)
∗
2
y
∗
y
′
+
1
3
(
y
)
−
2
3
∗
y
′
=
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}
t
a
n
g
2
(
x
2
)
∗
3
y
∗
y
′
+
s
e
c
(
x
)
∗
2
y
∗
y
′
+
1
3
(
y
)
−
2
3
∗
y
′
=
−
(
y
3
∗
2
t
a
n
g
(
x
2
)
∗
s
e
c
2
(
x
2
)
∗
2
x
+
y
2
∗
s
e
c
(
x
)
t
a
n
g
(
x
)
)
{\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))}
y
′
(
t
a
n
g
2
(
x
2
)
∗
3
y
+
s
e
c
(
x
)
∗
2
y
+
1
3
(
y
)
−
2
3
)
=
−
(
y
3
∗
2
t
a
n
g
(
x
2
)
∗
s
e
c
2
(
x
2
)
∗
2
x
+
y
2
∗
s
e
c
(
x
)
t
a
n
g
(
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))}
y
′
=
−
(
y
3
∗
2
t
a
n
g
(
x
2
)
∗
s
e
c
2
(
x
2
)
∗
2
x
+
y
2
∗
s
e
c
(
x
)
t
a
n
g
(
x
)
)
(
t
a
n
g
2
(
x
2
)
∗
3
y
+
s
e
c
(
x
)
∗
2
y
+
1
3
(
y
)
−
2
3
)
{\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}})}}}
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?.
D
a
t
o
s
:
{\displaystyle \ Datos:}
s
a
=
10
m
{\displaystyle \ sa=10m}
s
b
=
20
m
{\displaystyle \ sb=20m}
d
a
d
t
=
−
2
m
/
s
g
{\displaystyle {\frac {da}{dt}}=-2m/sg}
d
b
d
t
=
1
m
/
s
g
{\displaystyle {\frac {db}{dt}}=1m/sg}
d
x
d
t
=
?
{\displaystyle {\frac {dx}{dt}}=?}
S
o
l
u
c
i
o
n
:
{\displaystyle \ Solucion:}
t
a
n
g
(
x
)
=
s
a
s
b
{\displaystyle \ tang(x)={\frac {sa}{sb}}}
t
a
n
g
(
x
)
=
(
s
a
)
(
s
b
)
−
1
{\displaystyle \ tang(x)=(sa)(sb)^{-}1}
s
e
c
2
(
x
)
∗
d
x
d
t
=
(
s
a
)
(
−
s
b
)
−
2
∗
d
b
d
t
+
(
s
b
)
−
1
∗
d
x
d
t
{\displaystyle \ sec^{2}(x)*{\frac {dx}{dt}}=(sa)(-sb)^{-}2*{\frac {db}{dt}}+(sb)^{-}1*{\frac {dx}{dt}}}
s
e
c
2
(
x
)
∗
d
x
d
t
=
s
a
(
−
s
b
)
2
∗
d
b
d
t
+
1
s
b
∗
d
x
d
t
{\displaystyle \ sec^{2}(x)*{\frac {dx}{dt}}={\frac {sa}{(-sb)^{2}}}*{\frac {db}{dt}}+{\frac {1}{sb}}*{\frac {dx}{dt}}}
d
x
d
t
=
s
a
(
−
s
b
)
2
∗
d
b
d
t
+
1
s
b
∗
d
x
d
t
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {{\frac {sa}{(-sb)^{2}}}*{\frac {db}{dt}}+{\frac {1}{sb}}*{\frac {dx}{dt}}}{sec^{2}(x)}}}
d
x
d
t
=
10
(
−
20
)
2
∗
1
+
1
20
∗
(
−
2
)
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {{\frac {10}{(-20)^{2}}}*1+{\frac {1}{20}}*(-2)}{sec^{2}(x)}}}
d
x
d
t
=
10
−
400
+
−
1
10
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {{\frac {10}{-400}}+{\frac {-1}{10}}}{sec^{2}(x)}}}
d
x
d
t
=
1
−
40
+
−
1
10
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {{\frac {1}{-40}}+{\frac {-1}{10}}}{sec^{2}(x)}}}
d
x
d
t
=
40
+
10
−
400
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {\frac {40+10}{-400}}{sec^{2}(x)}}}
d
x
d
t
=
50
−
400
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {\frac {50}{-400}}{sec^{2}(x)}}}
d
x
d
t
=
5
−
40
s
e
c
2
(
x
)
{\displaystyle {\frac {dx}{dt}}={\frac {\frac {5}{-40}}{sec^{2}(x)}}}
d
x
d
t
=
5
−
40
5
4
{\displaystyle {\frac {dx}{dt}}={\frac {\frac {5}{-40}}{\frac {5}{4}}}}
d
x
d
t
=
20
−
200
{\displaystyle {\frac {dx}{dt}}={\frac {20}{-200}}}
d
x
d
t
=
2
−
20
{\displaystyle {\frac {dx}{dt}}={\frac {2}{-20}}}
d
x
d
t
=
1
−
10
{\displaystyle {\frac {dx}{dt}}={\frac {1}{-10}}}
d
x
d
t
=
−
0.1
r
a
d
/
s
g
{\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?.
D
a
t
o
s
:
{\displaystyle \ Datos:}
h
=
10
m
{\displaystyle \ h=10m}
r
=
4
m
{\displaystyle \ r=4m}
d
v
d
t
=
5
m
3
/
m
i
n
{\displaystyle {\frac {dv}{dt}}=5m^{3}/min}
d
h
d
t
=
?
{\displaystyle {\frac {dh}{dt}}=?}
S
o
l
u
c
i
o
n
:
{\displaystyle \ Solucion:}
primero se deja todo en funciòn de una sola variable.
r
4
=
h
10
{\displaystyle {\frac {r}{4}}={\frac {h}{10}}}
r
=
4
h
10
{\displaystyle \ r={\frac {4h}{10}}}
r
=
2
h
5
{\displaystyle \ r={\frac {2h}{5}}}
luego se deriva y se remplaza en la ecuaciòn del volumen
v
=
4
∗
p
h
i
∗
r
3
∗
d
r
d
t
3
{\displaystyle \ v={\frac {4*phi*r^{3}*{\frac {dr}{dt}}}{3}}}
5
m
3
/
m
i
n
=
4
∗
p
h
i
∗
2
h
2
5
2
∗
d
h
d
t
{\displaystyle \ 5m^{3}/min=4*phi*{\frac {2h^{2}}{5^{2}}}*{\frac {dh}{dt}}}
5
m
3
/
m
i
n
=
4
∗
p
h
i
∗
2
h
2
25
∗
d
h
d
t
{\displaystyle \ 5m^{3}/min=4*phi*{\frac {2h^{2}}{25}}*{\frac {dh}{dt}}}
d
r
d
t
=
5
m
3
/
m
i
n
∗
25
4
∗
p
h
i
∗
2
h
2
{\displaystyle {\frac {dr}{dt}}={\frac {5m^{3}/min*25}{4*phi*{2h^{2}}}}}
d
r
d
t
=
75
m
3
/
m
i
n
{\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?.
D
a
t
o
s
:
{\displaystyle \ Datos:}
d
=
150
m
{\displaystyle \ d=150m}
h
=
50
m
{\displaystyle \ h=50m}
s
=
?
{\displaystyle \ s=?}
d
h
d
t
=
8
p
i
e
s
/
s
g
{\displaystyle {\frac {dh}{dt}}=8pies/sg}
d
s
d
t
=
?
{\displaystyle {\frac {ds}{dt}}=?}
S
o
l
u
c
i
o
n
:
{\displaystyle \ Solucion:}
primero se despeja el valor de la que forman la altura
y la distancia con el observador.
s
2
=
h
2
+
150
2
{\displaystyle \ s^{2}=h^{2}+150^{2}}
s
=
(
50
2
+
150
2
)
1
2
{\displaystyle \ s=(50^{2}+150^{2})^{\frac {1}{2}}}
s
=
(
2500
+
67500
)
1
2
{\displaystyle \ s=(2500+67500)^{\frac {1}{2}}}
s
=
70000
1
2
{\displaystyle \ s=70000^{\frac {1}{2}}}
s
=
264.57
p
i
e
s
{\displaystyle \ s=264.57pies}
luego se derivan las variables y se remplazan los valores.
s
2
=
h
2
+
150
2
{\displaystyle \ s^{2}=h^{2}+150^{2}}
2
s
∗
d
s
d
t
=
2
h
∗
d
h
d
t
{\displaystyle \ 2s*{\frac {ds}{dt}}=2h*{\frac {dh}{dt}}}
d
s
d
t
=
2
h
2
s
∗
d
h
d
t
{\displaystyle {\frac {ds}{dt}}={\frac {2h}{2s}}*{\frac {dh}{dt}}}
d
s
d
t
=
h
s
∗
d
h
d
t
{\displaystyle {\frac {ds}{dt}}={\frac {h}{s}}*{\frac {dh}{dt}}}
d
s
d
t
=
50
264.57
∗
8
p
i
e
s
/
s
g
{\displaystyle {\frac {ds}{dt}}={\frac {50}{264.57}}*8pies/sg}
d
s
d
t
=
0.188
∗
8
p
i
e
s
/
s
g
{\displaystyle {\frac {ds}{dt}}=0.188*8pies/sg}
d
s
d
t
=
1.51
p
i
e
s
/
s
g
{\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?.
D
a
t
o
s
:
{\displaystyle \ Datos:}
l
=
10
p
u
l
g
{\displaystyle \ l=10pulg}
d
l
d
t
=
3
p
u
l
g
/
s
g
{\displaystyle {\frac {dl}{dt}}=3pulg/sg}
d
v
d
t
=
?
{\displaystyle {\frac {dv}{dt}}=?}
S
o
l
u
c
i
o
n
:
{\displaystyle \ Solucion:}
v
=
l
3
{\displaystyle \ v=l^{3}}
v
=
3
l
2
∗
d
l
d
t
{\displaystyle \ v=3l^{2}*{\frac {dl}{dt}}}
v
=
3
(
10
p
u
l
g
)
2
∗
3
p
u
l
g
/
s
g
{\displaystyle \ v=3(10pulg)^{2}*3pulg/sg}
v
=
3
(
100
p
u
l
g
2
)
∗
3
p
u
l
g
/
s
g
{\displaystyle \ v=3(100pulg^{2})*3pulg/sg}
v
=
300
p
u
l
g
2
∗
3
p
u
l
g
/
s
g
{\displaystyle \ v=300pulg^{2}*3pulg/sg}
v
=
900
p
u
l
g
3
/
s
g
{\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?.
D
a
t
o
s
:
{\displaystyle \ Datos:}
r
=
2
p
u
l
g
{\displaystyle \ r=2pulg}
d
v
d
t
=
4
p
u
l
g
3
/
s
g
{\displaystyle {\frac {dv}{dt}}=4pulg^{3}/sg}
d
r
d
t
=
?
{\displaystyle {\frac {dr}{dt}}=?}
S
o
l
u
c
i
o
n
:
{\displaystyle \ Solucion:}
v
=
4
∗
p
h
i
∗
r
3
3
{\displaystyle \ v={\frac {4*phi*r^{3}}{3}}}
d
v
d
t
=
4
∗
p
h
i
3
∗
3
r
2
∗
d
r
d
t
{\displaystyle {\frac {dv}{dt}}={\frac {4*phi}{3}}*3r^{2}*{\frac {dr}{dt}}}
d
v
d
t
=
4
∗
p
h
i
∗
r
2
∗
d
r
d
t
{\displaystyle {\frac {dv}{dt}}=4*phi*r^{2}*{\frac {dr}{dt}}}
d
r
d
t
=
d
v
d
t
4
∗
p
h
i
∗
r
2
{\displaystyle {\frac {dr}{dt}}={\frac {\frac {dv}{dt}}{4*phi*r^{2}}}}
d
r
d
t
=
4
p
u
l
g
3
/
s
g
4
∗
p
h
i
∗
(
2
p
u
l
g
)
2
{\displaystyle {\frac {dr}{dt}}={\frac {4pulg^{3}/sg}{4*phi*(2pulg)^{2}}}}
d
r
d
t
=
4
p
u
l
g
3
/
s
g
4
∗
p
h
i
∗
(
4
p
u
l
g
2
)
{\displaystyle {\frac {dr}{dt}}={\frac {4pulg^{3}/sg}{4*phi*(4pulg^{2})}}}
d
r
d
t
=
1
p
u
l
g
3
/
s
g
p
h
i
∗
(
4
p
u
l
g
2
)
{\displaystyle {\frac {dr}{dt}}={\frac {1pulg^{3}/sg}{phi*(4pulg^{2})}}}
d
r
d
t
=
0.079
p
u
l
g
/
s
g
{\displaystyle {\frac {dr}{dt}}=0.079pulg/sg}
