Ó 
3 −  + 2
①   =
+2
  = ℝ − {−}
< −∞, − >∪< −, +∞ >
La Función no
tiene inversa
¿Tendrá inversa?
|  =
( + 2)(−1 + 2) −(3 −  +  2 )
( + 2)2
 2 + 4 − 5
≤0
( + 2)2
 2 + 4 − 5
≥0
( + 2)2
 2 + 4 − 5 ≥ 0
( + 4)( − 1) ≥ 0
+ −4 − 1 +
< −∞, −4] ∪ [1, +∞ >
 2 + 4 − 5
=
( + 2)2
+ ≠
 ≠ −
[−4,1]
[(−4), (1)]
23
  [− , 1]
2
 −  + 
②   =
 [−,  >
+
¿Tendrá inversa?
í, á    
   [ −1 ,  1 > = [5,1 >
¿ ?
     Ó 
②   = ( − 3)2 −5    = [−3,3]
 = ( − 3)2 −5
 + 5 = ( − 3)2
¿Tendrá inversa en [1,6]?
′  = 2( − 3)
+5=−3
2( − 3) ≥ 0
−
3
+5+3=
+
[3, +∞ >
 + 5 + 3 = ()
 ⇒  
( − ) ≤ 
< −∞, 3]
¿ á   ó ?
 ⇒  
  ó 
  :
1
′  =
2 +5
ó:   = −  ; ∗ () =  +  + 
( ∗ )′  =
1
2 +5
   = (
− 3)2 −5

( ∗ )′
¿ á   ó ?
 = ( − 3)2 −5
 + 5 = ( − 3)2
+5=−3
−1 = ( − 3)2 −5
  :
′ −1 =
0 = ( − 3)2 −5 + 1
1
0 = ( − 3)2 −4
2 +5
2 −1 + 5
′  = 2( − 3)
1
1

∗
=
=
( )′  =
( − ) 2(5 − 3) 4

1
( )′ −1 =
=−
(− − )
8
 + 5 + 3 = ()
1

′()
∗
+5+3=
′  =
−1
: (∗ )′  =
0 = ( − 3 + 2)( − 3 − 2)
=
1
4
0 = ( − 1)( − 5)
=1 =5
2( − 3) ≥ 0
[3, +∞ >
       
(∗ )′  =
  = ()
 = ()

1

=
=
′()
()
 − 
′  = ()
 = ()
 = ()
() = ()
∗ () = ()
  = ()
∗ () = ()
1


1 − 2

1

=−
=−
′()
()
 − 
 = ()
(∗ )′  =
1−
2
1


  = ()
∗ () = ()
(∗ )′
1


=
=
 =
 2 ()
 + 
′()
 = ()

1 + 2
1

     
1
  = 
⇒
 = ()
  = 
⇒
 = ()
() = −
  = 
⇒
 = ()
()
Ct  = 
⇒
 = ()
  = 
⇒
  = 
⇒
 = ()
 = ()
() =
1−
2

1
2
1−

1
=

1 + 2
1
() = −

1 + 2
()
=
1
2

() = −
−1

1

2 − 1

APLICACIONES DE LA DERIVADA

∆
  = Pendiente =
∆
  + ∆ − () ()
=
∆→0

∆
lim
∆ ()
=
∆

¿  ?
  + ∆ − () ()
=
∆

  = 36 + 
∆ = 1
1
2 36 + 
Razón de cambio
Rapidez de cambio
Rapidez instantánea
  + ∆ −   = ∆ . ′()
  −   = ∆ . ′()
 = 0, ⇒  0 = 36
 = 1, ⇒  1 = 37
,  =
∆
∆
=
1
12

= ()


 =  +

 − 
 = . 
  = 49 +
¿  ?
, ()
=0
 0 = 49
=2
 2 = 51
∆ = 
1
51 = 7 +
7
1
1
−1/2 =
= (49 + )
2 49 + 
2
¿ (°) ?
 = . 
  = (30°+)
0==0

==1
180
, () = (30° + )
 0 = (30°)
 1 = (31°)
  −   = ∆ . , ()
1
51 −  =  ( )
14
∆ =
  −   = ∆ . , ()
1
3
(31°) −
= ( )
2
2
1
3 
(
)
 31° = +
2
2 180
 ° = . 
 ° = . 
DERIVADA SUPERIOR
()
= ′()

 2 ()
= ′′()
2

 ()
. 
    ó
    ó
 2 ()
. 
 ,    x    ""
 3 ()
= ′′′()
 3
    ó
   =  −  +  − ; : ′′′()
′  = 10 4 −12 3 +2
′′  = 40 3 −36 2 +2
′′′() = 120 2 −72
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