Derivar
1)
(
x
2
)
(
c
o
s
(
x
)
)
+
(
x
2
)
(
c
o
s
(
x
)
)
2
{\displaystyle \ (x^{2})(cos(x))+(x^{2})(cos(x))^{2}}
=
(
2
x
c
o
s
x
)
−
(
x
2
s
e
n
x
)
{\displaystyle \ (2xcosx)-(x^{2}senx)}
2)
c
o
s
5
(
2
x
2
−
1
)
{\displaystyle \ cos^{5}(2x^{2}-1)}
=
5
c
o
s
4
(
−
s
e
n
(
2
x
2
−
1
)
)
(
4
x
)
{\displaystyle \ 5cos^{4}(-sen(2x^{2}-1))(4x)}
3)
(
3
x
−
2
x
2
)
(
5
−
4
x
)
{\displaystyle \ (3x-2x^{2})(5-4x)}
=
(
3
x
−
2
)
(
4
)
+
(
5
+
4
x
)
(
3
−
4
x
)
p
{\displaystyle \ (3x-2)(4)+(5+4x)(3-4x)p}
(
12
x
−
8
x
2
)
(
15
−
8
x
−
16
x
2
)
{\displaystyle \ (12x-8x^{2})(15-8x-16x^{2})}
−
24
x
2
+
4
x
+
15
{\displaystyle \ -24x^{2}+4x+15}
4)
2
x
2
−
4
x
+
3
2
−
3
x
{\displaystyle {\frac {2x^{2}-4x+3}{2-3x}}}
=
(
4
x
−
4
)
(
2
−
3
x
)
−
(
2
x
2
−
4
x
+
3
)
(
−
3
)
(
2
−
3
x
)
2
{\displaystyle {\frac {(4x-4)(2-3x)-(2x^{2}-4x+3)(-3)}{(2-3x)^{2}}}}
(
−
12
x
2
+
20
x
−
8
)
−
(
−
6
x
2
+
12
x
−
9
)
(
2
−
3
)
2
{\displaystyle {\frac {(-12x^{2}+20x-8)-(-6x^{2}+12x-9)}{(2-3)^{2}}}}
(
−
6
x
2
+
8
x
+
1
)
(
2
−
3
x
)
2
{\displaystyle {\frac {(-6x^{2}+8x+1)}{(2-3x)^{2}}}}
5)
3
x
2
−
1
4
x
+
3
{\displaystyle {\frac {3x^{2}-1}{4x+3}}}
=
6
x
(
4
x
+
3
)
−
(
3
x
2
−
1
)
4
(
4
x
+
3
)
2
{\displaystyle {\frac {6x(4x+3)-(3x^{2}-1)4}{(4x+3)^{2}}}}
12
x
2
+
18
x
+
4
(
4
x
+
3
)
2
{\displaystyle {\frac {12x^{2}+18x+4}{(4x+3)^{2}}}}
6)
(
x
3
−
3
x
2
+
10
x
)
1
0
{\displaystyle \ (x^{3}-3x^{2}+10x)^{1}0}
=
10
(
x
3
−
3
x
2
+
10
x
)
9
.
(
3
x
2
−
6
x
+
10
)
{\displaystyle \ 10(x^{3}-3x^{2}+10x)^{9}.(3x^{2}-6x+10)}
3
(
4
x
3
+
11
x
)
3
{\displaystyle {\frac {3}{(4x^{3}+11x)^{3}}}}
=
−
3
(
4
x
3
+
11
x
)
3
(
(
4
x
3
+
11
x
)
2
)
2
{\displaystyle {\frac {-3(4x^{3}+11x)^{3}}{((4x^{3}+11x)^{2})^{2}}}}
−
9
(
12
x
2
+
11
)
(
4
x
3
+
11
)
4
{\displaystyle {\frac {-9(12x^{2}+11)}{(4x^{3}+11)^{4}}}}
s
e
n
(
3
x
2
+
11
x
)
{\displaystyle \ sen(3x^{2}+11x)}
=
c
o
s
(
3
x
2
+
11
x
)
.
(
6
x
+
11
)
{\displaystyle \ cos(3x^{2}+11x).(6x+11)}
s
e
n
3
x
{\displaystyle \ sen^{3}x}
=
3
s
e
n
2
x
.
c
o
s
x
{\displaystyle \ 3sen^{2}x.cosx}
3
x
−
1
2
x
+
5
{\displaystyle {\frac {3x-1}{2x+5}}}
=sen
3
x
−
1
2
x
+
5
{\displaystyle {\frac {3x-1}{2x+5}}}
3
x
−
1
2
x
+
5
{\displaystyle {\frac {3x-1}{2x+5}}}
(
2
x
−
1
)
3
.
(
x
2
−
3
)
2
{\displaystyle \ (2x-1)^{3}.(x^{2}-3)^{2}}
=
(
2
x
−
1
)
3
.
(
x
2
+
3
)
2
+
(
2
x
−
1
)
.
(
x
2
−
3
)
2
{\displaystyle \ (2x-1)^{3}.(x^{2}+3)^{2}+(2x-1).(x^{2}-3)^{2}}
(
2
x
−
1
)
2
.
(
2
)
.
(
x
2
−
3
)
2
+
(
2
x
−
1
)
.2
(
x
2
−
3
)
.
(
2
x
)
{\displaystyle \ (2x-1)^{2}.(2).(x^{2}-3)^{2}+(2x-1).2(x^{2}-3).(2x)}
s
e
n
x
c
o
s
2
x
{\displaystyle {\frac {senx}{cos2x}}}
= 3
s
e
n
x
c
o
s
2
x
{\displaystyle {\frac {senx}{cos2x}}}
(
c
o
s
x
)
.
(
c
o
s
2
x
)
−
(
s
e
n
x
)
.
(
−
s
e
n
2
x
)
(
c
o
s
2
x
)
2
{\displaystyle {\frac {(cosx).(cos2x)-(senx).(-sen2x)}{(cos2x)^{2}}}}
13)Suponga que:
f
(
2
)
=
4
{\displaystyle \ f(2)=4}
f
∗
(
4
)
=
6
{\displaystyle \ f*(4)=6}
f
∗
(
2
)
=
−
2
{\displaystyle \ f*(2)=-2}
HALLAR:
13.1)
(
f
(
x
)
)
3
{\displaystyle \ (f(x))^{3}}
=
(
f
(
2
)
)
3
{\displaystyle \ (f(2))^{3}}
(
4
)
3
{\displaystyle \ (4)^{3}}
3
(
4
)
2
{\displaystyle \ 3(4)^{2}}
48
{\displaystyle \ 48}
13.2)
3
f
(
x
)
{\displaystyle {\frac {3}{f(x)}}}
=
3
f
(
2
)
{\displaystyle {\frac {3}{f(2)}}}
3
4
{\displaystyle {\frac {3}{4}}}
RAZON DE CAMBIO 1- Con que rapidez baja el nivel de un fluido contenido en un tanque cilìndrico, si bombeamos hacia afuera el fluido, a razón de : 3.000 l/min?
R: DATOS:
M
3
{\displaystyle \ M^{3}}
r
{\displaystyle \ r}
H
(
t
)
{\displaystyle \ H(t)}
D
v
D
t
{\displaystyle {\frac {Dv}{Dt}}}
D
h
D
t
{\displaystyle {\frac {Dh}{Dt}}}
TENIENDO LOS DATOS CLAROS , PODEMOS HALLAR ECUALCIONES:
V
{\displaystyle \ V}
r
2
{\displaystyle \ r^{2}}
h
{\displaystyle \ h}
=
D
v
D
t
{\displaystyle {\frac {Dv}{Dt}}}
r
2
{\displaystyle \ r^{2}}
= -3000 :1000&pi .
r
2
{\displaystyle \ r^{2}}
D
h
D
t
{\displaystyle {\frac {Dh}{Dt}}}
=
−
3
p
i
(
r
2
)
{\displaystyle {\frac {-3}{pi(r^{2})}}}
D
h
D
t
:
−
3
p
i
(
r
2
)
m
/
m
i
n
{\displaystyle {\frac {Dh}{Dt}}:{\frac {-3}{pi(r^{2})}}m/min}
R: DATOS:
X(T) :0.6KM
Y(T) :0.8KM
D
z
D
t
{\displaystyle {\frac {Dz}{Dt}}}
D
x
D
t
{\displaystyle {\frac {Dx}{Dt}}}
D
y
D
t
{\displaystyle {\frac {Dy}{Dt}}}
sabiendo esto, entonces:
Z:(0.6)^2+(0.8)^2 : 1
= 2Z.
D
x
D
z
{\displaystyle {\frac {Dx}{Dz}}}
D
x
D
t
{\displaystyle {\frac {Dx}{Dt}}}
D
y
D
t
{\displaystyle {\frac {Dy}{Dt}}}
=
2
(
20
)
−
2
(
0.6
)
(
−
60
)
2
(
0.8
)
{\displaystyle {\frac {2(20)-2(0.6)(-60)}{2(0.8)}}}
D
y
D
t
{\displaystyle {\frac {Dy}{Dt}}}
=
D
y
D
t
{\displaystyle {\frac {Dy}{Dt}}}
V:1/3PI.r^2.h
Dv/Dt: 9M3/MIN
Dh/Dt:? Pregunta del problema
H:10m
R:5m3
H:6F
SABIENDO ESTO, ENTONCES:
=V:
1
3
.
P
I
.
r
2
.
h
{\displaystyle {\frac {1}{3}}.PI.r^{2}.h}
1
12
.
P
I
.
h
3
{\displaystyle {\frac {1}{12}}.PI.h^{3}}
1
12
.
P
I
.3
h
2
.
{\displaystyle {\frac {1}{12}}.PI.3h^{2}.}
1
P
I
{\displaystyle {\frac {1}{PI}}}
