UNIVERSIDAD CENTRAL DE ECUADOR ´
FACULTAD DE CIENCIAS ADMINISTRATIVAS CARRERA CARRER A DE CONTABILIDAD CONTABILIDAD Y AUDITORÍA PORTA PORTAFOLIO FOLIO VIRTUAL VIRTUAL DE MATEMÁTICA
ÍNDICE DATOS DATOS PERSONALES:................ PERSONALES:........................................ .................................. .................... ............ ..3 3 SÍLABO..................................................................................5 FILOSOFÍA CORPORATIVA DE LA FACULTAD.....................................................6 3.- COMPETENCIA DE LA ASIGNATURA. RESULTADOS DE APRENDIZAJE..........8 UNIDAD UNIDAD II.- DIFERENCIACIÓN DIFERENCIACIÓN UNO................. UNO............................ ............................................... .................................... 15 UNIDAD UNIDAD III.- DIFERENCIACIÓN DIFERENCIACIÓN DOS.................. DOS............................ ..................... ...................... .........................18 ..............18 UNIDAD IV.- INTEGRALES...............................................................................21
Unidad I...............................................................................30 LÍMITES Y CONTINUIDAD........ CONTINUIDAD.................. ..................... ..................... ..................... ........................................ ............................. 3 1.1 D!"#$%$&#............... D!"#$%$&#.......................... ..................... ..................... ..................... ..................... ....................................... ............................ 3 PROPIEDADES........ PROPIEDADES................... ...................... ..................... ..................... ....................................................... ............................................ 33
HIPERBOLA...........................................................................33 UNIDAD II.......................................... II............................................................................. ................................... 45 DERIVACIONES. DERIVACIONES............ ..................... ..................... ..................... ..................... ..................... ........................................ .............................. '5 DEFINICIÓN(........... DEFINICIÓN(..................... ..................... ..................... ..................... ..................... ............................................. ................................... '5 REGLAS.................. REGLAS............................ ..................... ..................... ..................... ..................... ............................................. ................................... '5 E)PONENTE............ E)PONENTE....................... ..................... ..................... ...................... ..................... ..................... ..................... ....................'5 ..........'5 CONSTANTE................................................................................................'5 COCIENTE ..................... ............................... ..................... ..................... ..................... ......................................... .............................. '6
*DE LA CADENA+............ CADENA+....................... ..................... ..................... ..................... ..................... ...................... ..................... ............. ...'8 '8
DERIVACIÓN DERIVACIÓN IMPLÍCITA................... IMPLÍCITA........................................... ....................................... ............... 59 DERIVACIÓN DERIVACIÓN LOARÍTMICA.......................................... LOARÍTMICA.................................................... ..........!" !" #UNCIÓN E$PONENCIAL E$PONENCIAL % LOARÍTMICA....... LOARÍTMICA................................!3 .........................!3 COSTOS...............................................................................!& #UNCIÓN DEL COSTO............................................................&0 ...........................................................................................&" UNIDAD III....................................... III............................................................................ ..................................... &' M($IMOS % MÍNIMOS............................... MÍNIMOS................................................. ............................ .......... &' M)TODO DE LA SEUNDA DERIVADA......................... DERIVADA.................................... ............&' .&' UNIDAD IV INTERALES........................ INTERALES................................................ ................................. .........9" 9" PRO,LEMA DE APLICACIÓN....... APLICACIÓN.................. ..................... ..................... ..................... .................................... ..........................
INTERAL INTERAL DE#INIDA..................................... DE#INIDA.............................................. .................... .............. ..."0" "0" DEBERES............................................................................""9 DE,ER N 1.................... 1.............................. ..................... ..................... ..................... .............................................. ................................... 11 1
ÍNDICE DATOS DATOS PERSONALES:................ PERSONALES:........................................ .................................. .................... ............ ..3 3 SÍLABO..................................................................................5 FILOSOFÍA CORPORATIVA DE LA FACULTAD.....................................................6 3.- COMPETENCIA DE LA ASIGNATURA. RESULTADOS DE APRENDIZAJE..........8 UNIDAD UNIDAD II.- DIFERENCIACIÓN DIFERENCIACIÓN UNO................. UNO............................ ............................................... .................................... 15 UNIDAD UNIDAD III.- DIFERENCIACIÓN DIFERENCIACIÓN DOS.................. DOS............................ ..................... ...................... .........................18 ..............18 UNIDAD IV.- INTEGRALES...............................................................................21
Unidad I...............................................................................30 LÍMITES Y CONTINUIDAD........ CONTINUIDAD.................. ..................... ..................... ..................... ........................................ ............................. 3 1.1 D!"#$%$&#............... D!"#$%$&#.......................... ..................... ..................... ..................... ..................... ....................................... ............................ 3 PROPIEDADES........ PROPIEDADES................... ...................... ..................... ..................... ....................................................... ............................................ 33
HIPERBOLA...........................................................................33 UNIDAD II.......................................... II............................................................................. ................................... 45 DERIVACIONES. DERIVACIONES............ ..................... ..................... ..................... ..................... ..................... ........................................ .............................. '5 DEFINICIÓN(........... DEFINICIÓN(..................... ..................... ..................... ..................... ..................... ............................................. ................................... '5 REGLAS.................. REGLAS............................ ..................... ..................... ..................... ..................... ............................................. ................................... '5 E)PONENTE............ E)PONENTE....................... ..................... ..................... ...................... ..................... ..................... ..................... ....................'5 ..........'5 CONSTANTE................................................................................................'5 COCIENTE ..................... ............................... ..................... ..................... ..................... ......................................... .............................. '6
*DE LA CADENA+............ CADENA+....................... ..................... ..................... ..................... ..................... ...................... ..................... ............. ...'8 '8
DERIVACIÓN DERIVACIÓN IMPLÍCITA................... IMPLÍCITA........................................... ....................................... ............... 59 DERIVACIÓN DERIVACIÓN LOARÍTMICA.......................................... LOARÍTMICA.................................................... ..........!" !" #UNCIÓN E$PONENCIAL E$PONENCIAL % LOARÍTMICA....... LOARÍTMICA................................!3 .........................!3 COSTOS...............................................................................!& #UNCIÓN DEL COSTO............................................................&0 ...........................................................................................&" UNIDAD III....................................... III............................................................................ ..................................... &' M($IMOS % MÍNIMOS............................... MÍNIMOS................................................. ............................ .......... &' M)TODO DE LA SEUNDA DERIVADA......................... DERIVADA.................................... ............&' .&' UNIDAD IV INTERALES........................ INTERALES................................................ ................................. .........9" 9" PRO,LEMA DE APLICACIÓN....... APLICACIÓN.................. ..................... ..................... ..................... .................................... ..........................
INTERAL INTERAL DE#INIDA..................................... DE#INIDA.............................................. .................... .............. ..."0" "0" DEBERES............................................................................""9 DE,ER N 1.................... 1.............................. ..................... ..................... ..................... .............................................. ................................... 11 1
DE,ER N 2..................... 2............................... ..................... ..................... ..................... ............................................ ................................. 123 DE,ER N 3.................... 3.............................. ..................... ..................... ..................... .............................................. ................................... 125 DE,ER N '.................... '.............................. ..................... ..................... ..................... .............................................. ................................... 1'5 DE,ER N 5.................... 5.............................. ..................... ..................... ..................... .............................................. ................................... 15 DE,ER N 6.................... 6.............................. ..................... ..................... ..................... .............................................. ................................... 153 DE,ER N /.................... /.............................. ..................... ..................... ..................... .............................................. ................................... 158 DE,ER N 8.................... 8.............................. ..................... ..................... ..................... .............................................. ................................... 168 DE,ER N .................... .............................. ..................... ..................... ..................... .............................................. ................................... 1/1
DEBERES SEUNDO HEMISEMESTRE........... HEMISEMESTRE..................................... .........................."&* "&* DE,ER N 1.................... 1.............................. ..................... ..................... ..................... .............................................. ................................... 1/8 DE,ER N 2.................... 2.............................. ..................... ..................... ..................... .............................................. ................................... 1/ DE,ER N 3.................... 3.............................. ..................... ..................... ..................... .............................................. ................................... 2/ DE,ER N '.................... '.............................. ..................... ..................... ..................... .............................................. ................................... 223
CORRECCIÓNES CORRECCIÓNES DE LAS PRUEBAS PRIMER HEMISEMESTRE.....'40 HEMISEMESTRE.....'40 CORRECCIÓN DE LA PRUE,A N 1.................... 1.............................. ..................... ..................... ...................... ............2' 2' CORRECCIÓN DE LA PRUE,A N 2.................... 2.............................. ..................... ..................... ...................... ............251 251 CORRECCIÓN DE LA MICRO PRUE,A..................... PRUE,A................................ ...................................... ...........................258 258
CORRECCIÓNES CORRECCIÓNES DE LAS PRUEBAS SEUNDO SEUNDO HEMISEMESTRE. HEMISEMESTRE. '!0 CORRECCIÓN DE LA PRUE,A N 1.................... 1.............................. ..................... ..................... ...................... ............26 26 CORRECCIÓN DE LA PRUE,A N 2.................... 2.............................. ..................... ..................... ...................... ............2/2 2/2
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12.- EVALUACIÓN INSTRUMENTOS EVALUACIÓN
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13.- BIBLIORA#ÍA TE)TO
AEUSSLER= J.= E#!: F=K R$%0> S. P0 @
13.- BIBLIORA#ÍA TE)TO
AEUSSLER= J.= E#!: F=K R$%0> S. P0 @ R$%0> J. <<>= Matemáticas para Administración RESULTADOS DEL CONTRI,UCIÓ EL ESTUDIANTE DE,E( = D!%$?<:!#>0 !>$%$&#= E>. P!0:<#= y Economía APRENDIZAJE A7< 28=NM4$%<. *ALTA= MEDIA= ,AJA+ 0+ R!:<%$&# >! ,:$%<: OFFMANN L0!#%!= ! 0=*21'+= Matemáticas, MEDIA S! %009 >! !:<! <!?0:= $$90#>< Aplicadas a la Administración y los Negocios, <!?0: 0$%0#>< [?$!:= 0$%0><:México, 0 0 Mc Graw Hill Edcation. ?$!: @ %<#$#$>0> !?!:0. TAN= S<< R= Matemáticas aplicadas a los negocios, S! %009 >! !:<! + R!:<%$&#las ciencias >! MEDIA = ;$#0 !>$%$&#= E>. sociales y la !ida <!?0: >! D!$0>0: <!?0:= %<# CENGAGE L!0#$#= A7< 211= M4$%<. $$90%$&# >! >!$0>0:= MEDIA S! %009 >! !:<! 0$%0><: 0 0 !?!:0. ARYA= L0#>!= I00= Matemáticas a la @ <!?0:Aplicadas >! M$?<: K '0 !>$%$&#= E>. Administración y a la Economía ?#$?<:. %+ R!:<%$&#P!0:<#= >! A7<. 22 <!?0:= %<# $$90%$&# >! >!$0>0:= L!%0: (QQ<<\:.<<!.%0: $>^,'>P>DVD8C$#:!%^H<#%<!>^?0 ?$?0 @ ?#$?0= !?0$%0_0$%0>0_0_0_0>?$#$:0%$<#_@_!%<# 0$%0><: 0 0 !?!:0. !_0@0^!#:0^)!$^#5U$H,!'A OP1IAY!>^CC';6AEWAA
MATERIA 2/
PRIMER HEMISEMESTRE
28
MATERIA
Unidad I LÍMITES Y CONTINUIDAD 1.1 Definii!n
L0 >$$:$&# ! ?0%0 #0 :!00%$&# !#! ><: !$<#!: :! %<#<%! %
k 0
0
=∞
=0
k
∞ = Indeterminación ∞ ∞
∞ = Indeterminación
2
E/8i8i,1: 3 2 2 t + 19 t −8 t −9 − 1. S ( t )= −t 2+ 8 t − 7
S F
lim s ( t )=4
0.9 ! S 3.8 6
0.9 * 3. 1
t→2
G"%<( $ % " I ' & 3 "0.5 4 "5 5 '' >" > >' ".'5 >3 >3.! >4 >! >5 > *.'G > "0.5
3
0.9 9 3. /
0.99 " 9 3. '
".00 " '. 2
".0 " '. 3
"." ".'
'.2 8
'.5 /
2.lim x → 1
lim x→ 1
√ x −1 x −1
√ x −1 = √ 1−1 = 0 x −1
1− 1
0
( x − 1) 1 = x −1 √ x + 1 ( x −1 )( √ x + 1) √ x + 1
√ x −1 = √ x −1 · √ x +1 = x −1
lim x→ 1
1
=
1
√ x +1 √ 1+ 1
1 2
= =0.5
G"%<( $ " ' 3 4 5 ! & *
% I 0.4" 0.3! 0.33 0.30 0.'* 0.'& 0.'!
31
PROPIEDADES
lim ( f ( x ) ± g( x ) ) =lim f ( x ) ± lim g( x )
1.-
lim k f ( x )= k lim f ( x )
2.-
f ( x )
3.-$?
g( x )
=
lim f ( X ) lim g( x )
HIPERBOLA 1. 9 y 9 y
2
2
−4 x 2−54 y −16 x + 29=0
− 4 x 2−54 y −16 x + 29= 0
( 9 y 2−54 y ) – ( 4 x 2+ 16 x )=−29 9 ( y
2
−6 y + 9 )−4 ( x 2 +4 x + 4 )=−29 + 81−16
32
9
( y −3 )2 36
−4
( x −2 )2 36
=
36 36
9
( y −3 )2 36
−4
( x −2 )2 36
=
36 36
( y −3 )2 ( x − 2 )2 − =1 4
9
Centr ( ! "k ) • •
^ %<<>!#0>0 X^ %<<>!#0>0 @
Centr (−2 " 3 )
a2 = 4 → a = 2 #2= 9 →a =3 c 2= 4 + 9 c 2=13 c =3.6
G"%<( $ " ' 4 3 5 >" >' >4 >3 >5
% 0."& >0.33 >0.** >".4& >'.0& 0.*9 " 0.*9 0.59 0."&
33
2.−5 x
5 x 2
2
2
−2 y 2=2
2 y − 2
2
2
=1
2
x y − 2 =1 →!ri$nta% 2 a #
a2= 2 → a =1.41 #2= 2 → a =1.41 c 2 =2 + 2 c 2= 4 c =2
3'
G"%<( $
%
G"%<( $ " ' 4 3 5 >" >' >4 >3 >5
% ".'' 3.0 0 !.' 4 4.! 3 &.* 4 ".' ' 3.0 0 !.' 4 4.! 3 *.! 3
E!%$%$<:( 2
x −1 1.lim 2 x → 1 x −3 x + 2
x 2−1 1−1 0 lim 2 = = = I 1− 3 + 2 0 x→ 1 x −3 x + 2 2 ( x + 1 ) ( x −1 ) ( x + 1 ) x −1 = = 2 x −3 x + 2 ( x −2 ) ( x −1 ) ( x −2 )
lim ( x + 1 ) x → 1
( x −2 )
=
1 +1 2 = =−2 1− 2 − 1
G"%<( 35
$ " '
% I I
$ " ' 3 4 5 >" >' >3 >4 >5
% I I 4 '.5 ' 0 0.'5 0.40 0.50 0.5&
2.lim x → 1
2
+ 14 x −3 2 x + 3 x
5 x
x 2+ 14 x −3 5 (−3) + 14 (−3 )− 3 45− 42−3 0 lim = = = I 9 −9 0 x→ 1 x 2+ 3 x (−3 )2+ ( 3 ) (−3) 2
X =
x =
(5 x + 15 )( 5 x −1 ) 5
5 ( x + 3)( 5 x −1 ) 5
x =( x + 3 )( 5 x −1 )
+ 14 x −3 = ( x + 3 )( 5 x −1 )= (5 x −1 ) x ( x + 3) x x 2 + 3 x
5 x
2
36
lim
(5 x −1 ) 58 −39− 1 16 = =
(5 x −1 ) 58 −39− 1 16 = = 3 −3 x x→ 1 lim
P#< C$%< P *-3K 5.33+ G"%<( $ " ' 3 4 5 >" >' >3 >4 >5
% 4 4.5 4.! 4.& 4.* ! 5.5 I 5.'5 5."*
3. lim x →∞
2
+5 2 x + 1 4 x
2
+ 5 4 ∞ 2 + 5 4 ∞ 2+ 5 ∞ lim = 2 = 2 = = I 2 x → ∞ 2 x + 1 2 ∞ + 1 2 ∞ +1 ∞ 4 x
2
2
4 x 2
+
5
x
5
4+
5
x
lim x → ∞
2+
1 2
x
x
x
4+
2
2
4+
2 x + 5 x x = 2 = 2 1 2 x + 1 2 x 1 2+ 2 + 2 2
4 x
2
= 2+
5
∞2 1
=
4+ 0 =2 2+0
∞2
G"%<( 3/
$ " '
% 4 4.5
$ " ' 3 4 5 >" >' >3 >4 >5
% 4 4.5 4.! 4.& 4.* ! 5.5 I 5.'5 5."*
EJERCICIOS 2
x + x −6 x −2 X → 3 lim
x 2 + x −6 22 + 2−6 0 = = = I 2 −2 0 X → 3 x − 2 lim
x 2 + x −6 ( x +3 ) ( x −2 ) = = x + 3 ( x −2 ) x −2 lim X + 3 =2 + 3=5 X → 3
$ " ' 3 4 5 >" >' >3 >4
% 4 & ! & * ' " 0 >"
G"%<(
38
2.lim x → 9
√ x −3 x − 9
√ x −3 = √ 9−3 = 3 −3 = 0 = I
lim
x −9
x→ 9
9−9
0
0
1 x −9 √ x −3 · √ x + 3 = = x − 9 √ x + 3 ( x −9 )( √ x + 3 ) √ x + 3
lim 1 x → 9
√ x + 3
=
1 1 1 = = √ 3 + 3 3 + 3 6
G"%<( $ " ' 3 4 5 ! &
% 0.'5 0.'' 0.'" 0.'0 0."9 0."* 0."&
3
4
2
x + 13 x −2 x +7 3. lim 3 x →∞ x + x −1
x 4 + 13 x 2−2 x + 7 ∞ lim = = I ∞ x → ∞ x 3+ x −1 x 4 + 13 x 2−2 x + 7 x 4 3 x 2 2 x 7 3 2 7 + 4 − 4 + 4 1 + 2 − 3 + 4 4 4 x x x x x x x x = = 3 3 1 1 1 x + x −1 x x 1 − 3+ 4 − + 4 4 4 4 x x x x x x x ∞=¿4
¿ ¿ ¿
1
∞ 3
2
7
x
x
x 4
1+ lim
¿
n →∞
−
1
− 2
−
1
+ 3
+
1
1
1
3
(∞ )
+¿
1+
=
3
2
7
− 3+ 4 2 (∞ ) ( ∞) (∞) ¿
x x 3 x 4
'
G"%<( $
%
G"%<( $ " ' 3 4 5 >" >' >3 >4 >5
% 3 '.* 3.5 4.4 5.3 >"3 >4.3 >4." >4.& >5.5
3
4. lim
x
n →∞
3
lim
−
2 x
2
2
x + 1
2 x
−
2
x x 2+ 1
x → ∞
3
lim
x
x → ∞
∞
2 x
lim
3
= = 2
2
x + 1
x → ∞
'1
2 x 2
2 x 2
=
x + 1 x
2
x 2 2
1
=
2 1
2 x 2
2 x
2
x 2
=
x 2 + 1 x 2 1 + 2 2 x x lim 2 x → ∞
1+
1
1+
x 2
lim n →∞
3
2 1
=
−
2
=
=
1+
1
2
=2
x 2
1+ 0
∞
2 x
2
x x2 + 1
=0 −2=−2
G"%<( $ " ' 3 5 & >" >' >3 >4
% ' >0." >0.* >".3 >".5 >4 >3." >' >'.5
REPASO PARA LA PRUEBA 3
+ 3 x −5 n → ∞ − x + 2 x + 7
1.lim
2 x
2
'2
∞
¿ ¿ ¿ 3 +3 ( ∞ ) 5
∞
¿ ¿ ¿ 3 + 3 ( ∞ ) −5 ¿ ∞
¿ ¿ ¿ 2 + 2 ( ∞ )+ 7 2¿ 3 2 x + 3 x −5 lim =¿ 2 n →∞ − x + 2 x + 7 2
2 x
+ 3 x −5
3
2 x
3 x
− 3
5
2+
3
− 2
5
+ 3 x −5 x 3 x 3 x x3 x x 3 = = = − x 2 + 2 x + 7 − x 2+ 2 x + 7 − x 3 2 x 7 −1 + 2 + 7 + 3 + 3 2 3 2 x
3
+
3
3
5
x
x 3
lim 2 + n→∞
− 2
−1 2 + 2 +7 x
x
x
2+
=
3
x
x
x
x
5
− 3 2 (∞ ) ( ∞)
−1 2 + +7 ( ∞) ( ∞)2
=
2 + 0 −0 2 = =∞ −0 + 0 −0 0
G"%<( $ % " 0 ' '.4 3 "4.5 4 >"35 5 > >" 3'.5 >' >'.5 >3 '& >4 *.5 *.5
'3
2. lim x →∞
2 x + 1 3 x
2
+ 2 x −7
∞
¿ ¿ ¿ 2 + 2 ( ∞ ) −7 3¿ 2 ( ∞ )+ 1 2 x + 1 = 2 ¿ 3 x + 2 x −7 lim ¿ x →∞
2 x + 1
2 x
− 2
2
1
2
2
x x x 2 x + 1 = 2 = 2 2 3 x + 2 x −7 3 x + 2 x −7 3 x 2 x 7 x
lim x →∞
3+
2
−
x x 2 2
+
7
x x
$ " ' 3 4 5 >" >' >3 >4 >5
2
1
2
% >".5 '.5 0.' 0." 0.4 0."! >3 >0.3 >0.' >0."
=
2
−
x
2
+
x
− 2
x
2
1
x x 2 3+
2 7
+
x x 2
1
( ∞) ( ∞ )2 3+
2
+
=
−
7
0 3
= =0
( ∞) (∞ )2
G"%<(
''
UNIDAD II DERIVACIONES DEFINICIÓN" R09&# >! C0?$< %<# !:!%< 0 <<= $#!!0%$&# !
REGLAS & y dy → & x dx
E$PONENTE
' = x n
dy d n = x dx dx
dy =n xn−1 dx
'5
E-,: ' = x 5
E-,: ' = x 5 dy d 5 = x =5 x 5−1=5 x 4 dx dx y =C
CONSTANTE dy d = c dx dx
E!?<( ' =5
dy d = 5 =0 dx dx
( COCIENTE y = ( * ) → f ( x ) )
d d ( −( ) dy dx dx = 2 dx ) )
E!?<(
x 5 x 5 = √ x x 1 /2
% F
M4<>< 1 x
1/ 2
dy = dx
d 5 5 d 1 /2 x − x x dx dx
( x )
1 2 2
'6
1
1 2 −1 ¿ x 2
¿
1
1 2 −1 ¿ x 2
¿
1/ 2
5− 1
x + x − x 5 ¿ dy =¿ dx 1/ 2
dy = dx
dy = dx
5 x x
4
1 2
− x −1 /2 x 5 x
5 x
9/ 2
1 2
− x 9/ 2 x 1 2
9/ 2
X ( 5− )
dy = dx
X 9 2
9/ 2
X ( )
dy = dx
X 9
−1 9 dy = X 2 ( ) 2 dx
dy 9 7 /2 = X dx 2
M4<>< 2 x5 y = 1/ 2 x 9
dy d 2 = x dx dx 9
dy 9 2 −1 = x dx 2
dy 9 7 /2 = X dx 2
'/
".> y =( ( ± ) ) "()→f ( x )
".> y =( ( ± ) ) "()→f ( x ) dy d d = ( ± ) dx dx dx
2.−
dy d d = cx= c d dx dx dx
' = cx
DE LA CADENA y =( x n−c )n dy d n = ( x −c )n dx dx
n ( x n− c )n−1 ·
d n ( x − c ) dx
'8
E-,:
E-,: 4
2 3
' =( x + 5 x )
3 dy d 4 = ( x + 5 x 2 ) dx dx
dy d =3 ( x 4 + 5 x 2)3−1 · ( x 4 + 5 x 2 ) dx dx dy dx
3 ( x
^
4
+ 5 x 2)2 ·
(
d 4 d x − 5 x 2 dx dx
)
x 4−5 x 2 dy x 4−1 -1 x 2−1 ¿ ¿ ' dx ^ * 2
¿¿ ¿
PRODUCTO
y =( ) → f ( x )
dy dx d d = () =( ) + ) ( dx dx dx dx
E!%$%$<:(
2
y =( x 5− 4 ) ( x −3 )2
'
2 dy d 5 = ( x −4 ) ( x −3 )2 dx dx
2 dy d 5 = ( x −4 ) ( x −3 )2 dx dx
2 dy ( 5 )2 d ( 2 2 d x −3 ) + ( x − 3 ) ( x 5− 4 ) = x −4 dx dx dx
2 2− 1 dy d =( x5 −4 ) · 2 ( x −3 )2−1+ ( x −3 ) +( x −3 )2 · 2 ( x 5− 4 ) dx dx
(
)
(
2 dy d d d d =2 ( x 5− 4 ) ( x −3 ) x − 3 + 2 ( x −3 )2 x 5− 4 dx dx dx dx dx
d 5 ( x − 4 ) dx
)
2 dy =2 ( x 5− 4 ) ( x −3 ) ( 1− 0 )+ 2 ( x −3 )2 ( 5 x 4− 0 ) dx
2 dy =2 ( x −3 ) ( x 5− 4 ) + 10 x 4 ( x −3 )2 ( x 5−4 ) dx
dy = 2 ( x −3 ) ( x 5−4 ) [ ( x 5− 4 ) + 5 x 4 ( x −3 ) ] dx dy =2 ( x −3 ) ( x 5− 4 ) [ x 5− 4 + 5 x 5−15 x 4 ] dx
dy =2 ( x −3 ) ( x 5− 4 ) [ 6 x , 5−15 x 4− 4 ] dx
E/8i8i,1
5
x − 4 x 1.− y = 3 x 5
x 4 x y = 3 − 3 x x
2
2
5
4
y = x 2− x
4
y = x 2− x y = x 2− 4 x−1 dy d 2 d = x − 4 x−1 dx dx dx dy =2 x −(−1 ) x −1−1 dx dy =2 x + 4 x−2 dx dy 4 =2 x + 2 dx x
2.− y =
−1
1 1
1
t t
√ t
+ 2 −
−2
−1
y =t + t − t 2 −1
dy d −1 −2 d 2 = ( t + t ) − t dt dt dt
−3
1 dy = (−1 t −2+ 2 t −1) − t 2 dt 2
dy 1 1 1 = − + 3 dt t 2 t 2 t 2
51
3 1 x x 2 2 − 3.− y = + − x 2 + 2 +
16
x
3 x
3
3 1 x x 2 2 − 3.− y = + − x 2 + 2 +
16
y =
−1
x
3 x
3 2
1 3
1 3
x + 2 x − x + x−2 + x 2
16
−1
3
3
dy −1 d 2 d −1 d 1 d −2 = x −2 x − x 2 + x dx 16 dx dx dx 3 dx
2 ¿ x
2− 1
−2 ( 1 ) x
3
3 −1 1 1 − x 2 + (−2 ) x−2−1 + 2 3 3 dy −1 = ¿ dx 16
− 1− 1
−1
dy −1 3 2 1 = 2 x −2 x −2− x 2 + x−3 + dx 8 2 3 3
( x )=¿ 2 x 4 + 3 x + 1 " x =−1 4.− f ¿ d f x d = (2 x 4 + 3 x + 1) dx dx d f x dx
=2
d 4 d d x + 3 x + 1 dx dx dx
d f x =8 x 3+ 3 dx d f x = 8 (−1 )3 + 3 dx → 1
d f x =5 dx → 1
52
E/8i8i,1
D!!?$#0 0 !%0%$&# >!0 !%0 0#!#!= G0"%0( 1.− y =− x
3
− 5 x 2 + 3 x − 1 " (−1,−8 )
/8,: RECTA DE LA
$ >" >' >3 >4 >5 >! " '
% >* >"9 >'* >'9 >"! "& >4 >'3
$ % 0 ' >"J5 0
53
3
2
y =− x −5 x + 3 x −1 d f x dx d f x dx
=
=
d (− x 3−5 x 2+ 3 x −1 ) dx
−d dx
3
x −5
d 2 d d x + 3 x − 1 dx dx dx
d f x =−3 x2− 10 x + 3 ( 1 )−0 dx d f x dx
=−3 x2− 10 x + 3
d f x 2 =−3 (−1 ) −10 (−1)+ 3 dx →−1
d f x =10 dx → 1
m=10 y − y 0 =m ( x − x 0) y −(−8 ) =10 [ x −(−1) ] y + 8 = 10 x + 0
y =10 x + 2
5'
1.− y = x
5
−3 x 3− 5 x + 2 " ( 1,−5 )
/8,: RECTA DE LA
$ " ' 3 4 >" >' >3 >4
% >5 0 "49 *"4 9 4 >"45 >*"0
$ 0 4J9
% 4 0
55
y = x 5−3 x 3−5 x + 2 d f x d 5 = ( x −3 x 3−5 x + 2 ) dx dx d f x d 5 d d d = x −3 x3 −5 x + 2 dx dx dx dx dx d f x =5 x 4− 9 x 2−5 ( 1 )− 0 dx d f x =5 x 4− 9 x 2−5 dx d f x dx →−1
d f x dx → 1
=5 (1)4 −9 (1 )2−5
=−9
m=−9 y − y 0 =m ( x − x 0) y −(−5 )=−9 x −(−5 ) y + 5=−9 ( x −1)
y =−9 x + 9−5
56
y =−9 x + 4
y =−9 x + 4
E?ERCICIOS 2
1.−( x + x )( 3 + 2 x ) " (−1,2)
y =3 x 2+ 2 x 3−3 x −2 x 2 2
3
y = x + 2 x − 3 x
y =2 x 3 + x 2−3 x
/8,: RECTA DE LA
$ " ' 3 4 >" >' >3 >4
% >5 0 "49 *"4 9 4 >"45 >*"0
$ 0 4J9
% 4 0
5/
y =2 x 3 + x 2−3 x d f x d = (2 x 3 + x 2−3 x ) dx dx d f x d d d =2 x 3 + x 2−3 x dx dx dx dx d f x =3 ( 2 ) x 3−1 + 2 x2−1−3 dx d f x =6 x 2+ 2 x −3 dx d f x =6 (−1)2 + 2 (−1)−3 dx →−1
58
d f x dx → 1
=1
d f x dx → 1
=1
m =1 y − y 0 =m ( x − x 0) y −( 2 ) =−9 [ x −(−1) ] y −2 = x + 1
y = x + 3
'. Una =/i8a d/-ina K6 n T -11 d16<1 d in/,d68i/ a -/8ad, 6n 2 /,d68, n6, 1 6d d8i/ x ( t )=t + 3 t 6nidad1 2 67, nd/,1 a 6n 2 3
. /8i, - (t )=−2 t + 30 ( ( )
". E/1 in7/1, a/a 1 /,d68, 8,-, 6na =6n8i;n d i-,
I#!:< %! $!?< I#!:<^ N >! #$>0>!: !%$< I = X ( t ) + / t t 2 3
(¿¿ 2 + 3 t )(−2 t + 30 ) I =¿
5
7 2
5 2
I =−2 t + 30 t − 6 t + 90 t ¿
7 2
5 2
I =−2 t + 30 t − 6 t + 90 t ¿
I =( +
. (
'. C6 1 a /a;n d 8ai, d in7/1, /18, a i-, d16<1 d 86a/, -11 −2 t 5
(¿ ¿ 7 + 30 t −6 t 2 + 90 t ) 2 dI d = ¿ dt dt 7
5
dI d −1 d d −1 d =−2 t 2 + 30 t 2−1−6 t 2 + 90 dt dt dt dt dt
()
()
7
5
7 −1 5 −1 dI =−2 t 2 + 30 ( 2 ) t 2−1− 6 t 2 + 90 dt 2 2 5
3
dI =7 t 2 + 60 t −15 t 2 +90 dt 5
3
dI =7 ( 4 ) 2 + 60 ( 4 )− 15 ( 4 ) 2 + 90 dt → 4
dI =−14 dt → 4
DERIVACIÓN IMPLÍCITA
y = f x x 2+ y 2 =4
E$PLÍCITA IMPLÍCITA
6
dy d = f =f 0 ( x) = y 0 dx dx x
dy d = f =f 0 ( x) = y 0 dx dx x dy d = y 0 = y 0 0 dx dx dy d = y 0 0 = y 0 0 0 dx dx dy d n = y = y n 0 dx dx
E!%$%$<: 1.− x
2
+ y 2=4
dy d 2 2 = ( x + y =4 ) dx dx dy d 2 d d = x + 3 y 2+ 4 dx dx dx dx 2 x + 2 yy 0 =0 2 yy 0 =−2 x
y 0 =
y 0 =
−2 x 2 y
− x y
2
3
2.− x y
−6 =5 y 2 + x
61
x d dx
(¿ ¿ 2 y 3−6 )= ( 5 y2 + x )
x d dx
(¿ ¿ 2 y 3−6 )= ( 5 y2 + x ) d ¿ dx
d 2 3 d d d x y − 6 = 5 y2 + x dx dx dx dx x
2
d 3 3 d 2 y + y x −0 =10 yy0 + 1 dx dx
x 2 ( 3 y 2 y 0 ) + y 3 ( 2 x )=10 yy 0 + 1 2
2
2
2
3 x y y 0 + 2 xy
3
=10 yy 0 + 1
3 x y y 0 −10 yy 0 =1−2 xy 2
2
y 0 ( 3 x y −10 y )=1−2 xy
y 0 =
1 −2 xy 2
3 x y
2
3
3
3
−10 y
2
3.− x y + 2 y
3
=3 x + 2 y
d 2 ( x y +2 y 3 )= d ( 3 x + 2 y ) dx dx
d 2 d d d x y + 2 y 3 =3 x + 2 y dx dx dx dx x
2
d d 2 2 y + y x + 6 y y 0 =3 + 2 y 0 dx dx
2
2
x y 0 + 2 xy+ 6 y y 0 =3 + 2 y 0
x 2 y 0 + 6 y 2 y 0 −2 y 0 =3−2 xy 2
2
y 0 ( x + 6 y −2 )=3 −2 xy
62
3− 2 xy y 0 = 2 x + 6 y 2−2
3− 2 xy y 0 = 2 x + 6 y 2−2
DERIVACIÓN LOARÍTMICA Vulgar
(Base 10)
Natura
base e
E/8i8i,1: 1.− y =ln x
n
y =n ln x
dy d = n ln x dx dx dy d =n ln x dx dx
dy 1 =n dx x dy n = dx x x (¿¿ m+ a )n 2.− y = ln ¿
x (¿¿ m + a ) y =n ln ¿ x (¿¿ m+ a ) dy d =n ln ¿ dx dx
63
m
x + a dy
¿
=n
1
d
¿
m
x + a
¿
1 dy d =n m + ¿ dx x + a dx
1 dy d d =n m + x m + a dx x + a dx dx
x
m− 1
+0 m¿
dy n = m ¿ dx x + a m−1
dy n mx = dx x m + a
dy n = dx x dy = n
dy =
dx x
dx y
x
3.− y =e
/8,: $ 3 ' " 0 > " > ' > 3
% '0 0* &3 * '& " " 03 0." 0.0 5 6'
y = e x x
ln y = ln e ln y = x ln e
ln y = x 1
y
· y 0 =1 y 0 = y x
y 0 =e
#UNCIÓN E$PONENCIAL % LOARÍTMICA E/8i8i,1: x
1.− y =e
65
dy d x = e dx x
dy d x = e dx x dy x d =e x dx x x
y = e
dy d 2 x = e dx dx
dy 2 x d =e + ( 2 x ) dx dx dy 2 x =e ( 2) dx dy =2 e 2 x dx
#ORMA E$PONENCIAL '
2
x + 2
+− y =e
dy d x +2 = e dx dx 2
dy x +2 d 2 =e ( x + 2 ) dx dx 2
dy x +2 =e (2 x + 0) dx 2
dy x +2 =e (2 x ) dx 2
66
dy =2 e x +2+ 2 dx 2
dy =2 e x +2+ 2 dx 2
F<?0 L<0?$%0 2
x + 2
y = e
2
x + 2
ln y = ln e
x y =(¿¿ 2 + 2) ln e ln ¿
ln y =
1
y
d 2 x + 2 dx
· y 0 =
1
y
d 2 d x + 2 dx dx
· y 0 =2 x
y 0 =2 x + y 2
y 0 =2 e x +2 + 2
3 +− y =4 x ln y = ln 4 x ln y = x ln 4
6/
ln y =
d ( 1,4 x ) dx
ln y =
1
y
d ( 1,4 x ) dx
· y 0 =1.4 y
y 0 =1.4 ( 4 ) x 2 2 4 +− 4 x + y =1
dy d d = ( 4 x 2 + y 2 )= 1 dx dx dx dy d d d = 4 x 2+ y 2= 1 dx dx dx dx dy =8 x + 2 yy 0 = 0 dx 2 yy 0 =−8 x
y 0 =
−8 x 2 y
2
d y d = y 0 = y 0 0 2 dx dx
( )
2
d y d −4 x = 2 d x dx y
¿ y−1
d 2 y d = −4 x y −1 2 d x dx
d d −4 x +(−4 x ) y−1 dx dx
¿ y−1 ( 4 ) + (−4 x ) (−1 ) y −2 y 0 0
¿ 4 y −1 + 4 x y −2 y 0 0
68
y 0 0 ¿
4 y 4 x y
−2
−1
y 0 0 ¿
−1
4 y 4 x y
−2
y y 0 0 ¿ −2 x y
−1
√ x 4 +− y =5
ln y = √ x ln5 1
ln y = x 2 ln 5 1
d d ln y = ( x 2 + ln5 ) dx dx 1
d · y 0 = ( x 2 + ln 5 ) y dx 1
1
d y 0 = ln5 x 2 dx
()
1
1 d 2 y 0 = ln5 x 2 dx
1
ln 5 2 5 x y 0 = 2
−1 2
6
COSTOS Cst 1argina% c = f ( - )ó f (2 )
dc =Cst 1argina% : c 0 d2 Ingres 1argina% I = f ( 2) dI = Ingres 1argina% : I 0 d2
cst /rmedic =
c n
E/8i8i,1 ". Un
#a/i8an 1i-a K6 86and, 1 /,d68 6nidad1 n d/-inad,
a/Q86, 8,1, ,a 1:
P!%$< U#$0$<(
1 3
x =¿ ( 75− x )
1 8
x =¿ x 2+ 3 x + 98 ( . ) c¿
() . (
/¿
a Haa/ 8,1, In7/1, -a/7ina1 dc x = c 0 = + 3 4 dx
In7/1,F P/8i, Unia/i, N-/, d 6nidad1 1 3
I = ( 75− x ) x
x 2
I =25 x −
3
/
. . = · ( (
. . = · ( ( 2 dI = I 0 =25 − x d2 3
"# E- 8,1, -a/7ina a/a 8a86a/ 8,1, a/a /,d68i/ a 6nidad
n,na
8 4
c 0 = + 3 =5
8 C6 1 8,1, /a d /,d68i/ a n,na 6nidad 3 c 9− 3 c 8 c=
[
1 (9 )2 + 3 ( 9 ) + 98 8
][ −
1 (8 )2 + 3 ( 8 ) + 98 8
]
c =135.13 −130 c =5.13
d Uii8 in7/1, Ma/7ina a/a 8a86a/ in7/1, ,nid, d a na d a n,na 6nidad
2 3
I 0 =25− ( 8 )
. I 0 19.67 · (
C6 1 in7/1, ,nid, d a na d n,na 6nida
3 I 9− 3 I 8
/1
[
I =
1 ( 75 −9 ) 9 3
][
1 3
− ( 75− 8 ) 9
]
[
I =
1 ( 75 −9 ) 9 3
][
1 3
− ( 75− 8 ) 9
]
I =198−178.67 I =19.33 .
'. S 1i-a K6 dn/, d a,1 a 8i/86a8i;n d 6n /i;di8, ,8a c (t )=100 t 2+ 400 t + 5000
1/
a On7a 6na /1i;n a/a a /a;n a a 86a a 8i/86a8i;n 8a-ia/a 8,n /18, a i-, dn/, a,1 dc d = ( 100 t 2+ 400 t + 5000 ) dt dt dc d d d = 100 t 2 + 400 t + 5000 dt dt dt dt dc =200 t + 400 dt
t =5 →
dc = 200 ( 5 )+ 400 =1400 dt
t =6 →
dc = 200 ( 6 ) + 400=1600 dt
/2
#UNCIÓN DEL COSTO c = f ( - ) ó f (2 )
S$ 0 H#%$&# >! %<:< >! # H0$%0#! !: ( c=
52
2
√ 22 + 3 52
c=
2
( 2 2+ 3 ) dc d = d2 d2
dc d = d2 d2
1 2
+ 5000
[ ] [ ] 2
52
( 2 +3 ) 2
1 2
+
d 5000 d2
2
52
( 2 +3 ) 2
1 2
1 2
1
( 2 +3 ) d 5 22−( 5 2 2 ) d ( 2 2+ 3 ) 2 d2 d2 2
dc = d2
+ 5000
[( 2 + ) ] 2
−1
1 2
( 2 +3 ) ( 5 ) ( 2 ) 2−( 5 2 ) ( 2 + 3 ) 2 2
dc = d2
3
1 2 2
2
2
d 2 ( 2 +3 ) d2
( 22 + 3 )
/3
1 2
−1
2 dc ( 2 + 3 ) 10 2 −( 5 2 ) ( 2 + 3 ) ( 2 2 ) = d2 ( 22 +3 ) 2
2
2
−1
1 2
2 dc ( 2 + 3 ) 10 2 −( 5 2 ) ( 2 + 3 ) ( 2 2 ) = d2 ( 22 +3 ) 2
2
2
−1
1 2
2 dc ( 2 + 3 ) 10 2 −10 2 ( 2 + 3 ) = d2 ( 22 + 3 ) 2
[
3
1 2
2
dc 10 2 ( 2 + 3 ) −2 ( 2 + 3 ) = d2 ( 22 + 3 ) 2
2
2
−1 2
]
SEUNDO HEMISEMESTRE
/'
UNIDAD III M($IMOS % MÍNIMOS
G0:<= !>$>0:= !%09<: cst→m
U$$>0>= !#0:= $#!:<: cst→1
M)TODO DE LA SEUNDA DERIVADA 3
2
f x =2 x + 3 x + 12 x −5
1. On/ a /i-/a d/iada f 0 x =6 x 2+ 6 x + 12
/5
2. On/ a 176nda d/iada f 0 0 x =12 x + 6
2. On/ a 176nda d/iada f 0 0 x =12 x + 6
3. I76aa/ a 86a8i;n a 0 f 0 x =0 6 x
2
+ 6 x + 12= 0
x 2+ x + 2= 0
x =
−# ± √ #2− 4 ac
x =
−1 ± √ 12− 4 ( 1 ) ( 2 )
x =
2a
2
−1 ± √ 1− 8 2
x 1=
x =
−1 + √ −7 2
−1 ± √ 1− 8
x 2=
2
−1− √ −7 2
'. x 1→ >0 : m x 2→ < 0 : m
/6
E/8i8i,1 ".> La -6ni8iaidad d 6na 8i6dad /nd 8,n1/6i/ 6n 1ii, d a/K6, K6 1 6n, a 6na Qa d a, /n1i,. E /a d a/K6, d 1/ d
2
5000 m
C61 1/n a1 -Qni-a1 di-n1i,n1 d 1 , E //n, 1 /8an76a/ 2 1 d n8,n/a/ a1 di-n1i,n1 -Qni-a1 a/a 8//a-in,.
DATOS •
• •
•
•
•
4 =500 m2
R!%0#0 D$?!#:$&# C!0?$!#< M#$?< a : anc!
% : %arg
!0^ 0< 0#%<
1.- 4 =%xa→4 = f ( a *% ) 5000=%a→5=
5000
a
2.- L,n7i6d d 8//a-in, F an8, G a/7, G an8, 5=a + % + a 5=2 a + %
//
5= f ( a * %) 5000
5= f ( a * %) 5=2 a +
5000
a
5= f ( a )
5=2 a + 5000 a−1
1. On/ a /i-/a d/iada −2
50 =2 + 5000 a 5000
50 0 =2 +
a2
'. On/ a 176nda d/iada 50 0 =1000 a 50 0 =
−3
1000 3
a
3. I76aa/ a 86a8i;n a 0 50 0 =2 +
0 =2 +
5000
a2
5000 2
a
a =± 50 m a =50 m
5=
500 =100 50
a =50 m
/8
5=100 500
5=100 5=2 ( 50 )+
500 50
5=200 m
4. 50 0 =
1000
a
2
>0
50 0 > 0 → 5 ∈ m
5. C,n861i;n: La1 di-n1i,n1 ;i-a1 d //n, a/a n/ 6n a/K6ad/, 5=100 1,n d: a =50 m
'.> Una i//Qa 18,a/ 6da ,n/ d di,/ d 6n i/, 6n -a/ a 6n 8,1, d 3 i/,. La i//Qa 8a86a K6 6d nd/ '00 -a/1 a 6n /8i, d "5 2 K6 ,d/Qa nd/ "0 -a/1 -1 ,/ 8ada /d688i;n d 0.5081n n /8i,. A K6< /8i, d nd/ ,1 i/,1 a i//Qa a/a a/ a -i-, 16 6iidad DATOS •
•
• •
. (
3
C<:< >! $<(
V!#0 $:&$%0( 2 L$<:K
. (
15
S!:<( 1 !!?0!: < %Q !>%%$&# >! .5` P!%$< V!#0( M$?0 U$$>0>
VARIABLE
(
x : N6 de desc(ents 0.50
. (
)
V!0: ^ N >! #$>0: P!%$< >! #$>0> 7 =( 200 + 10 x ) (15 −0.50 x )
/
7 =3000 −100 x + 150 x −5 x 2 2
7 =3000 −100 x + 150 x −5 x 2 2
7 =−5 x + 50 x + 3000
7 = f ( x ) 7 0 =−10 x + 50 7 0 0 =−10 cm7 0 0 < 0 → 7 ∈ 1
7 0 =0 x =5 8esc(ent
/) =15−0.5 x < 15−5 ( 0.5 )=12.50
. (
C,n861i;n: S /8,-inda 8,-, n6, /8i, d na a,/ d
G"%<( 2
7 =−5 x + 50 x + 3000
V 304 5 30* 0 3"0 5 3"' 0 3"' 5 3"' 0 30" 05 30* 0
$ " ' 3 4 5 ! & *
8
12.50
. ( .
3. Cada a, /,ia/i, d 6na i8,//Qa 1/a nd/ ,a1 d in, an8, 8,1, d in, 1 d 0.*5 81. P,/ ,a 8,1, d did, 1 d "0 ,/ d1a8, 2 8,1, d a-a8na-in, d 6na ,a d6/an 6n a, 1 d 0.4081 . E in, 1 8,n16- n 6na a1a 6ni=,/- d6/an 6na a, 2 8ada d1a8, 7a a na1 1 a /-inad, d 61a/ d1a8, an/i,/.
a C6na1 ,a1 d di/ 8ada d1a8, a/a /d68i/ a -Qni-, 161 8,1,1 C,n K6 =/86n8ia d di/ in, DATOS
V!#0 $:&$%0( 8 <. C<:<
( .85 %:.
C<:< !>$>< ( 1` %Q . C<:< 0?0%!#0?$!#<( .' #tt+
VARIABLE x = N6 de #te%%as -r -edid
C<:<( %<:< <!0 _C<:< A?0%!#0?$!#< _C<:< !>$>< 81
c =0.85 x + 0.40 x +
( x ) 800
10
c =0.85 x + 0.40 x +
c =1.25 x +
( x ) 800
10
( x ) 800
c = f ( x ) c 0 =1.25 −
c 0 0 =
800
x2
16000
x 3
cmc00 > 0 →c ∈ m
c 0 =0 =1.25−
800
x 2
x =80 9te%%as
C,n861i;n: S d /aia/ "0 did,1 8,n a 8anidad d *0 ,a1 C *00 ".' 5 400 '.5 0 '!& 0.4 0 '00 5.0 0 "!0 !.' 5 "34 0.* 3 ""5 ".! " "0" 0.0 0
$ " ' 3 4 5 ! & *
G"%<( c =0.85 x + 0.40 x +
( x ) 800
10
82
E?ERCICIOS ".> C,1, -a/7ina: KF"& C,1, /,-di,: c´ c´ =
c´ =
c´ =
324
√ 22 + 35
( √ +
5 19 2 18
+ +
324
2
2
5
35
324 2 2 √ 2 +35
( 2 2+ 35 ) :
c´ =
d d2
2
+
324 2
c´ =
+ +
1 2
52
2
+
+5 +
324 2
( 2 2+ 35 )
1 2
)
19 2 18
19 2 18
19 2 18
+5 +
19 2 18
83
c´ : =
d 324 2 d d 19 5 2 + + 1 d2 2 d2 d2 18 ( 2 + 35 ) 2
c´ : =
:
d 324 2 d d 19 2 + 5 + 1 d2 2 d2 d2 18 ( 2 + 35 ) 2
(
c´ =
( (
: c´ =
1 2
1
( 2 + 35 ) d 324 2 −( 324 2) d ( 2 2+ 35 ) 2 d2 d2 2
[( 1 2
( (
c´ =
:
c´ =
:
(
c´ =
c´ : =
]
1 2 2
)
19 18
−1
()
( 2 +35 ) (324 )−( 324 2 ) 1 ( 2 2+ 35 ) 2 d 22 +35 2 d2 2
2
2 + 35
324 ( 2
2
−1
1 2
+ 35 ) −162 2 ( 2 + 35 ) 2 2
c´ : =
:
2 + 35 ) 2
+0+
(
d 2 d 2 + 35 d2 d2
2
2 + 35
324 ( 2
2
324 ( 2
2
+
)+
19 18
19 18
)
−1
1 2
)
)
+ 35 ) −162 2 ( 2 + 35 ) 2 (2 2 ) 19 + 2 18 2 + 35 2
−1
1 2
)
+ 35 ) −324 2 ( 2 + 35 ) 2 19 + 2 18 2 + 35
[
324 ( 17 )
2
2
1 2
2
+ 35 ] −324 ( 17 ) [ ( 17 ) + 35 ] [ ( 17 )2 +35 ] 2
2
−1 2
)
+
19 18
630 19 + 324 18
c´ =3 :
2. Si y =2 (
2
−3 (−2 → y = f (( )
8'
2
(= x + 4 →( =f ( x )
dy
2
(= x + 4 →( =f ( x )
encntrar
dy y = f ( f ( x ) ) dx
S,68i;n =,/-a N+ " dy d = ( 2 ( 2 −3 ( − 2 ) d( d( dy d d d =2 (2−3 ( − 2 d( d( d( d(
dy =4 (−3 d( dy d = ( 2 ( 2 −3 ( − 2 ) d( d( d( 2 = x + 4 dx d( d 2 d = x + 4 dx dx dx
d( =2 x dx
dy ∗d( dy d( = dx dx
dy =( 4 (−3 )( 2 x ) dx dy =2 x ( 4 (−3 ) dx dy =2 x [ 4 ( x 2 + 4 ) −3 ] dx
85
dy =2 x ( 4 x 2+16 −3 ) dx
dy =2 x ( 4 x 2+16 −3 ) dx dy =8 x3 + 32 x dx
S,68i;n =,/-a N+ ' 2
y =2 ( x 2+ 4 ) −3 ( x 2 + 4 )−2 dy =4 ( x 4 + 4 ) · 2 x −3 ( 2 x + 0 ) −0 dx dy =8 x ( x 2+ 4 x ) −6 x dx dy =8 x3 + 32 x −6 x dx dy =8 x 3+ 32 x dx 3. y =2 ( x
2
2
+ 4 ) −3 ( x2 + 4 ) −2 2
y =2 ( x 2+ 4 ) −3 x 2−12−2 2
y =2 ( x + 4 ) −3 x −14 2
2
2 dy d = 2 ( x 2 + 4 ) −3 x 2−14 dx dx
2 dy d d d = 2 ( x 2 + 4 ) −3 x 2− 14 dx dx dx dx
[
]
2 2 d dy d = ( 2 ) ( x 2 + 4 ) + ( x 2 + 4 ) 2 −6 x −0 dx dx dx
[
]
2 dy d = 2 (2 ) ( x 2 + 4 ) ( x2 + 4 ) + ( x2 + 4 ) (0 ) −6 x dx dx
86
[
(
)
]
dy d d = 4 ( x2 + 4 ) x 2 + 4 + 0 − 6 x dx dx dx
[
(
)
]
dy d d = 4 ( x2 + 4 ) x 2 + 4 + 0 − 6 x dx dx dx dy =[ 4 ( x2 + 4 ) ( 2 X ) ] −6 X dx dy =8 x ( x 2+ 4 )−6 x dx dy =8 x3 + 32 x −6 x dx dy =8 x3 + 26 x dx 2;
e 4. f ( ; )= 2 ;
e ln f ( ; )= ln
2;
;2
ln f ( ; )= ln e
2;
− ln ;2
ln f ( ; )=2 ; ln e− 2 ln ; ln f ( ; )=2 ; −2 ln ;
1
f ( x )
f : ( ; ) =2−( 2)
1
;
( ;)
f : ( ; )=f ( x ) 2 −
2
( )
2; 1 e f ( ; )=2 2 1− ; ;
:
( )
2; 2e − ; −1 f ( ; )= 2
:
;
;
8/
:
f ( ; )=
2e
2;
( ; −1 ) ;3
:
f ( ; )=
2e
2;
( ; −1 ) ;3
E?ERCICIOS ". Un aa-/ d '08- d a/7, 1 8,/; n ' da,1 6n, d ,1 da,1 1 d,a a/a =,/-a1 6n 8i/86, 2 ,/, 6n 86ad/ad, A.> 86 1 -Qni-, /a ,a K6 6d n8//a/1 d 1a -an/a B.> /a ,a 6d n/ a1a 6n a-a, d DATOS 4 c = < r 2 4 c( =a2 4 = 4 c + 4 c(
a
VARIABLES 4 = < r 2 + a2 20=%ngit(d de% circ(% + -erimetrde% c(adrad 20=2
r=
10− 2 a
<
R!?09< !%0%$&#
88
4 =
(
10−2 a
<
)
2
+ a2
33 c m
2
4 =
(
10−2 a
<
)
2
+ a2 2
4 =
4 =
100 −40 a + 4 a
+ a2 <
< 100
<
−
40 a
<
2
+
4a
<
+
a2 < <
4 = 2.27 a2−12.73 a + 31.83
PRIMERA DERIVADA 4 0 = 4.54 a −12.73
SEUNDA DERIVADA 4 0 0 = 4.54 4 0 0 > 0 → 4 0 0 ∈ m
PRIMERA DERIVADA IUAL A CERO 4 0 = 0 4.54 a=12.73
a=
12.73 4.54
4.54 a=12.73
a =2.80
r=
10− 2 a
<
r =1.40
CONCLUSIÓN( 0: >$?!#:$<#!: &$?0: 00 !#%!0 <: ><: !>09<: >!( $ " ' 3 4 5 ! & *
r =1.4 0 % '".3& "5.45 G0"%< "4.0& "&.'3 '4.93 3&."& 53.95 &5.5&
8
a =2.80
'. Una i//Qa 18,a/ 6d ,n/ d di,/ d 6n i/, 6n -a/ a 6n 8,1, d 3 i/, a i//Qa 8a86a K6 6d nd/ '00 -a/1 a 6n /8i, d "0 2 K6 ,d/n nd/ "0 -a/1 -1 ,/ 8ada /d688i;n d 0.50 81. En /8i, . A K6< /8i, d nd/ ,1 i/,1W 6iidad DATOS . Cstde% %i#r =3 ( . 7enta!istrica =200 %i#rs " 15 ( S(-(est :10 e=em-%ares -rc /red(ccin de 0.50 . /reci de )enta= 1>x + (ti%idad
VARIABLES
(
x = N6 de desc(ents 0.50
. (
)
R!?09< !%0%$&#
(
7enta = N6 de desc(ents 0.50
)
. ∗ /( (
7 =( 200 + 10 x ) (10 −0.50 x )−( 200 + 10 x ) 3
7 3000 100
150
5
2
−( 600 30 X )
a i//Qa a/a a/ a -i-, a
7 =( 200 + 10 x ) (10 −0.50 x )−( 200 + 10 x ) 3
7 =3000 −100 x + 150 x −5 x 2−( 600 + 30 X ) 2
7 =−5 x + 50 x + 3000−600 −30 x 2
7 =−5 x + 20 x + 2400
7 = f ( x )
PRIMERA DERIVADA 7 0 =−10 x + 20
SEUNDA DERIVADA 7 0 0 =−10 7 0 0 < 0 → 4 0 0 ∈ 1
PRIMERA DERIVADA IUAL A CERO 7 0 =0 0 =−10 x + 20
x =2 /reci de )enta=10−0.5 x =10 −0.5 ( 2 )=14
. (
CONCLUSIÓN( S! !%0 %! !#0 ! 0< >!
G0"%<
$ 0 5 " 0 " 5 ' 0
% '400 '3&5 '"00 "5&5 *00
1
. (
14
3. S a did, a 6n 8a/in/, a8/ 6na 8aa 1in aa d ,6-n -i-, 2 d a1 86ad/ada 1i -a/ia d ,1 ad,1 861a 3 ,/ - ' 2 -a/ia d a a1 4 ,/ - ' a/i76a/ 1i 1 6d a8/ a 8aa 8,n 4*. DATOS 9ase c(adrada 1ateria%%ads : 3
1ateria% #ase :4
. m
2
. m2
Cst=48 .
l l
VARIABLES 7%(men= area#ase∗a%t(ra
2
2
7 =% ∗!
2
7 =% ∗!
Cstde ca=a =area %ad∗ -reci (nit + +area #ase∗ -reci(nt + 2
48 =4 !%∗3 + % ∗4
. = m+m+
. . + m2 2 2 m m
.=. 48 =12 !% + 4 % 48 + 4 % != 12 %
2
2
12 + % != 3%
2
R!?09< !%0%$&#
( )
2
12+ % 3%
7 =% ∗
2
( )
7 =% ∗
12 + % 3
2
7 = f (% )
7 =
7 = 2
(
12 % + % 3
12 %
2
3
)
− 2 %4
3% −1
12 % → 3%
4
2
4
2% − 3%
−1
7 =12 % + 3 % −2 % + 3 %
3
7 =36 % −2−6 % −4
PRIMERA DERIVADA
7 =36 % −2−6 % −4
PRIMERA DERIVADA − − 7 0 =74 % 3− 24 % 5
SEUNDA DERIVADA −4
−6
7 0 0 =222 % −120 % →cmC00 > 0 → C ∈ m
PRIMERA DERIVADA IUAL A CERO −3
0 =74 %
$ 0. 5 " ' 3 4 >" >'
−24 %−5
% 4* 30 *.!' 3.9' '.'3 30 *.!
'
UNIDAD IV INTERAL ES
5
UNIDAD IV INTERALES '.1. INTERAL INDE#INIDA L0 $#!0 !: 0 $#!:0 >! 0 >!$0>0 !: $#>!"#$>0 <! #< !:# >!"#$><: <: $?$!:. y = f ( x )
dy d = f = f 0 ( x ) dx dx ( x ) dy = f 0 ( x )
∫ f 0 ( ) dx
y =
x
P<$!>0>!: 1.
∫ ( ( + ) ) dx =∫ (dx± ∫ ) dx
2.
∫kx
n
∫
n
dx = x dx
c =(na cnstante de integración
RELA DEL E$PONENTE y =∫ x dx n
n+ 1
x +c y = n +1
E?EMPLOS: y = x 3
0
y = x
n
y 0 =3 x 2 y =3 ∫ x 2 dx 2 +1
x
y =3
2+ 1
+c
3 y = x + c
INTERALES DIRECTAS e=em-%s 1
y 0 = x dy 1 = dx x dy =
dy x
∫ dy x
y =
y =%nx
x
y 0 =e
∫
x
y = e dx x
y = e
E?ERCICIOS: ".
3 2 y 0 =3 x − 4 x + x
dy =3 x 3− 4 x 2+ x dx dy = ( 3 x − 4 x + x ) dx 3
2
dy =3 x 3 dx −4 x2 dx + xdx
dy =∫ 3 x 3 dx −∫ 4 x 2 dx +∫ xdx
∫
∫
3
∫
2
dy = 3 x dx −4 x dx + xdx 3 +1
y =3
x
−4
3+1
3 4
x
2+ 1
+
x
1+ 1
2+1 1 +1
4 3
+c
1 2
4 3 2 y = x − x + x + c
3 4
4 3
1 2
3 2 y 0 = ( 4 ) x − (3 ) x + ( 2 ) x + 0
3 2 y 0 =3 x − 4 x + x
y =√ x
'. y = x
1 2
1 y 0 = x 2
−1 2
dy 1 = x dx 2 1 dy = x 2
−1 2
−1 2
1 y = x 2
∫
dx
−1 2
dx
−1 + 1
1 x 2 +c y = 2 −1 +1 2
y =
1 2
1 x 2 1 2
+c
1 2
y = x + c 1 y 0 = x 2
−1 2
+0
1 y 0 = x 2
−1 2
4 y = x −3 √ x +
3.
1 2
4
y = x − 3 x +
1 16
1 16
∫ ( ( + ) ) dx=∫ (dx±∫ ) dx 1
3 y 0 = 4 x − x 2 2 3
∫
y =
(
1
)
3 2 4 x − x dx 2 3
1
3 2 x dx 2
∫ 4 x dx −∫ 3
y =
1
3 y = 4 x dx − x 2 dx 2
∫
y = 4
x
∫
3
−1 + 1
3 +1
3+ 1
−
3 x 2 +c 2 −1 +1 2 1 2
4
x 3 x +c y = 4 − 4
2 1 2 1 2
4
y = x −3 x + c y = x − 3 √ x + c 4
∫ ( x − 4 x + 2 x +1) dx 3
4.
∫
3
∫ 4 x dx +∫ 2 x dx +∫ 1 dx
y x dx −
y =
x 3+1 3 +1
2
−4
2
x 2+1 2 +1
+2
x 1+ 1 1 +1
+ x +c
x 4
x 3
x2
4
3
2
1 4
4 3
y = − 4 + 2
+ x + c
4 3 2 y = x − x + x + x + c
HALLAR LA INTERAL INDICADA COMPRUEBE DERIVACIÓN
∫ x
".
∫
5
dx
5
y = x dx
y =
x
5+ 1
5+1
x
+c
6
y = + c 6
1 6
6
y = x + c
1 6
5 y 0 = ( 6 ) x + 0
y 0 = x
∫ x
'.
∫ x y =
5
3 4
3 4
dx
dx
x
3 +1 4
3 +1 4
+c
7 4
x y = + c 7 4
7
4 y = x 4 + c 7
LA RESPUESTA MEDIANTE
()
7
4 7 4 −1 y 0 = x + 0 7 4
y 0 = x
3 4
∫ x1 dx
3.
2
y =∫ x dx −2
y =
x−2+1
−2+ 1
x−1
y =
−1
+c
+c −1
y =−1 x + c −1 y =− x + c
y 0 =−(−1 ) x
− 1− 1
−2
y 0 = x
1
y 0 = 2 x
∫ 5 dx
4. y =5
∫ dx
y =5 x + c y 0 =5 + 0 y 0 =5
5.
∫
∫3e
x
x
y = 3 e dx x y =3 e + c
dx
+0
y 0 =3 e x + 0 x
y 0 =3 e
∫ 21 x − x 2 + √ 3 x
!.
2
∫(
)
1 1 −2 + dx − 2 x dx + 3 x 2 x
y =
1 y = 2
∫
∫
∫
dx x x2 −2 +3 −2+ 1 −1 x 2
1 2
y = %nx −2
x
−1
+3
2
−1 + 1
−2+ 1
−1
−1
x
1 2
+c +1
+c
1 2 1
1 −1 y = %nx + 2 x + 6 x 2 + c 2
()
1 1 y 0 = 2 x
1 2 3 − 2+ 2 x x √ x
y 0 =∫
1 −2 y 0 = −2 x + 3 x 2 x
∫
&.
()
(
2
+0
)
e x 2
−1
+ x √ x dx
x
1 2
y =
e
y =∫
1 x e d x + x 2 dx 2
∫ 2 dx +∫ x + x ∫
1 y = 2
∫e
1 y = 2
∫e
x
x
1
−1 + 2 (−1 ) x + 6 1 x 2 + 0 2 −2
∫
dx
3
3 2
dx + x dx
3 2
dx +∫ x dx
dx
1 2
x y = e +
x
3 +1 2
3 +1 2
+c
5 2
1 x x y = e + + c 2
5 2
5
1 2 y = e x + x 2 + c 2 5
()
5
1 x 2 5 2 −1 y 0 = e + x + 0 2 5 2 x
e
y 0 = + x √ x 2
∫
*. 1 y = 3
∫
(
)
x 1 3 − 2 + e2 + √ dx 3 x 2 x 2
3 dx − 2
∫ x
−2
dx + e
2
∫
∫
1 +1 2
− 2+ 1
1 3
y = %nx −
3 x 1 x + e2 x + +c 2 −2 + 1 2 1 +1 2 3 2
−1
1 3
y = %nx −
1
1 dx + x 2 dx 2
3 x 1 x + e2 x + 2 −1 2 3 2
+c
3
1 3 −1 2 1 y = %nx + x + e x + x 2 + c 3 2 3
()
1 1 y 0 = 3 x
y 0 =
9.
()
1
3 1 3 2 + (−1 ) x−2 + e 2 + x + c 2 3 2
1 − 3 2 + e 2+ √ x 3 x 2 x 2
∫ (2 e + x6 + ln2 ) dx x
1
y =2∫ e dx + 6∫ dx + ln2∫ dx x x
x y =2 e + 6 %nx + ln2 x + c
1
x y 0 =2 e (1 )+ 6 + ln2 ( 1 ) + 0 x
6
x y 0 =2 e + + ln 2 x
∫ ( x −2 x ) ( x1 −5 ) dx 3
"0.
∫
y =
(
3
2
)
2
2 x − 5 x 3− x +10 x 2 dx x x
∫ ( x −5 x −2 x + 10 x ) dx 2
y =
3
2
∫ ( 11 x −5 x −2 x ) dx 2
y =
3
y =11∫ x dx −5∫ x dx −2∫ xdx 2
x
3
3
x
4
x
2
y =11 − 5 − 2 + c 3
y =
4
2
11 3 5 4 2 x − x − x + c 3 4
y 0 =
11 ( 3 ) x 2− 5 ( 4 ) x 3− 2 x +c 3 4
y 0 =11 x 2−5 x 3−2 x
PRO#LEMA DE APLICACIÓN S 1i-a K6 dn/, d -11 a ,a8i;n d 8i/, 6, 8,-ina/a a 6na /a;n d 4G5 'J5
m . Si a ,a8i;n a86a 1 d "0000 /1,na1
C6 1/ a ,a8i;n dn/, d * -11
Da,1 t =meses
(
2 3
)
2
-ersnas dra$ón = 4 + 5 t → = 4 + 5 t 3 mes dt
-#%aciónact(a% =10000 ?C(>%sera/→t =8 m@
2
d = 4 + 5 t 3 dt
(
2 3
)
d- = 4 + 5 t dt
-=∫ ( 4 + 5 t ) dt 2 3
-= f (t ) 2 3
-= 4 ∫ dt + 5∫ t dt
-= 4 t + 5
2 +1 3
t
2 +1 3
+c
5 3
-= 4 t + 5
t
5 3
+c
5 3
-= 4 t + 3 t + c * / =10000 " t = 0
10000= 4 ( 0)+ 3 ( 0 )
5 3
+c
10000= c 5 3
-= 4 t + 3 t + c 5 3
-= 4 ( 8)+ 3 ( 8 ) + 10000
-=32 + 96 + 10000 -=10128
E?ERCICIOS CON SUSTITUCIÓN
∫ ( 2 x1+1 ) dx
".
y =∫
( x + ) dx(= x + 1
2
1
2
=2 ∫ 1( + d(2 d( dx
y =
y =
1 2
∫ d(( d(2 = dx
1 2
y = %n( + c 1 2
y = ln ( 2 x + 1 )+ c 1 1 ( 2 ) +0 2 2 x + 1
y 0 = +
y 0 =
1 2 x + 1
∫ ( 2 x + 6 ) dx 5
'.
y =∫ ( ( )5
y =
1 2
d( 2
( =2 x + 1
=2 ∫ ( d( d( dx 5
5 +1
1 ( + c d( =dx y = 2 5+1 2
1
6
1( + c y = 2 6
y =
1 6 ( +c 12
y =
1 ( 2 x +6 )6+ c 12 1 ( 6 ) ( 2 x + 6 )5 ( 2)+ 0 12
y 0 =
y 0 =( 2 x + 6 )5
E?ERCICIOS
∫ ( x +1 ) ( x +2 x +5 ) 2
".
y =∫ ( ( )
y =
1 2
d(
12
2
∫(
12
12
dx
2 (= x + 2 x + 5
d(
d( = 2 x + 2=2 ( x +1 ) dx
12 + 1
1 ( + c d( =2 ( x + 1 ) dx y = 2 12+ 1 13
1( + c d( =( x + 1 )dx y = 2 13 2
y =
1 13 ( +c 26
y =
1 2 ( x +2 x + 5 )13+ c 26
y 0 =
13 1 ( 13 )( x 2 + 2 x + 5 ) (2 )( x + 1)+ 0 26
y 0 =( x + 1 ) ( x 2+ 2 x + 5 )125
INTERAL DE#INIDA #
∫ dx = x|#a =( #) − ( a ) 1
1
a
EJERCICIOS 5
∫ xdx
". 5
|
1
5 x 2 5 5 ( )2 = [ 5 −(1 )2 ]= ( 25 −1 )=6 0 5 2 1
2
2
1
∫ ( x + x ) dx 2
'.
0
1
1
¿∫ x dx +∫ xdx 2
0
¿
x
0
3
| | 1
1
2 3 0 + x 0
2
[
] [
]
( 1 )3 ( 0 )3 ( 1 )2 ( 0 )2 ¿ − + − 3
3
1 3
3
3
1 2
¿ − 0+ − 0
¿
5 6 3
∫ ( 3 x − x + 6 ) dx 2
3.
[
x
3
¿ x −
[
3
−1
2
2
¿ ( 3) −
]
+6 x 3
( 3 )2 2
1
][
3
+ 6 ( 3 ) − (−1 ) −
(− 1 )2 2
+ 6 ( −1 )
]
(
)(
9 2
1 2
¿ 27 − + 18 − −1− −6
9 2
)
1 2
¿ 45 − + 7 + = 48
E?ERCICIO CON SUSTITUCIÓN 0
∫ 1
0
x
√ 1 + x
¿∫ 1
3 4
dx ( =1 + x
4
1 d( d( =4 x 3 → x 3 dx = d( 4 √ ( 4 dx −1
0
¿∫ ( 2 d( 4
1
−1 +1
1 ( 2 ¿ 4 −1 +1 2 1
1 ¿ ( 2 2
|
1
¿ 1 ( 1 + x 4 ) 2 0 2 1
[
1
1
1 ¿ ( 1 +(1 )4 ) 2 −( 1 +( 0) 4 ) 2 2
[
1
1
¿ 1 (2 )2 − (1 ) 2 2
]
]
1 2
¿ √ 2 – 1
In7/a dnida: 1i/ a/a n8,n/a/ /a1 da, d a1 86/a1 1
∫ ( 2 x +3 ) √ 2 x −1 dx −1
(=2 x −1 → 2 x =( −1 d( =2 dx 1
1
¿ ∫ ( (−1 ) √ ( d( + 3 ∫ √ ( d( 2
−1
1
¿ ∫ ((
1 2
d( 2
−1
¿
5 2
1( 2 5 2
+
1
+∫ ( −1
3 2
1( 2 3 2
+
1 2
d( 3
+
2
3 2
3( 2 3 2
5
2
−1
1
∫(
2 −1
1 2
d(
|
1 −1
3
3
|
1 1 ¿ ( 2 x −1 ) 2 + ( 2 x −1 ) 2 + ( 2 x −1 ) 2 1 5 3 −1
¿
{
¿
23 −i 15
5
3
3
1 1 2 2 2 2 ( 1 )−1 ] + [ 2 ( 1 )−1 ] + [ 2 ( 1)− 1 ] [ 5 3
E?ERCICIO
}{ −
5
3
3
1 1 2 2 2 2 (−1)−1 ] + [ 2 (−1)− 1 ] + [ 2 (−1 )−1 ] [ 5 3
}
2
$.
y = x
%.
y = 4 x + 20
2 x =4 x + 20
2
x −4 x −20=0
x =
−# ± √ #2− 4 ac
x =
4 ± √ 16− 4 (−20 )
2a
2
x =
4 ± √ 16 + 80 2
x =
4 ± √ 96 2
x =
4 ± 9.8 2
x 1=
13.8 =6.9 2
x 2=
5 .8 =−2.9 2
y = x
2
y 1=( 6.9 )2= 47.61
y 2=( 2.9 )2= 8.41
R!%0( OA,CO 6.9
4 1=∫ ( 4 X + 20 ) dx 0
4 1= 2 x + 20 X |6.9 2
0
2 2 4 1=2 ( 6.9 ) + 20 ( 6.9 )−2 ( 0 ) + 20 ( 0)
4 1=233.22 ( m 6.9
x
2
2
4 2=∫ dx 3
0
( 6.9 )2 ( 0 )2 − 4 2= 3
3
4 2=109.50 ( m
2
4 0= 4 1− 42 4 0= 233.22−109.50
4 0=123.72 ( m
2
OAEMO −2.90
4 1=
∫ 0
( 4 X + 20 ) dx
4 1= 2 x + 20 X |−2.90 2
0
2
2
4 1=2 (−2.90 ) + 20 (− 2.90 ) −2 ( 0 ) + 20 ( 0 ) 2
4 1= 41.16 ( m −2.90
4 2=
x
∫
2
3
0
(−2.90 )2 ( 0 )2 − 4 2= 3
3
4 2=−8.13 ( m
2
4 0= 4 1− 42 4 0= 41.16 −(−8.13 ) 2
4 0= 49.29 ( m
En8,n/a/ /a n/ a1 86/a1 dnida1 ,/ : y =9 − x 2 2 y = x + 1
x =3
x =
$ % >5 >"! >4 >& >3 0 >' 5 >" * 0 y =99− x 2 y = x2 + 1 " * ' 5 3 0 4 >& 5 >"!
$ >5 >4 >3 >' >" 0 " ' 3 4 5
% '! "& "0 5 ' " ' 5 "0 "& '!
E?ERCICIOS DE APLICACIÓN Un =a/i8an d i8i8a1 1/a 8,n16-id,/1 8,-/n
K6 dn/, d -11 ,1
#icic%etas f x =500 + 60 √ x mes
a 6n /8i, d
- x = 80 + 3 √ x
. #ic
C6 1 in7/1, ,a K6 =a/i8an 6d 1/a/ d a na d a1 i8i8a1 n ,1 /;i-,1 "! -11 DATOS
#icic%etas f x =500 + 60 √ x mes - x = 80 + 3 √ x
. #ic
x =tiem- Ingres ( 16 meses ) =@ Ingres= / + Anitari∗ N6 de (nidades I =80 + 3 √ x
( )
( )
. #ic + 500 + 60 √ x #ic mes
I =400000 + 1500 √ x + 4800 √ x + 180 x I =400000 + 19800 √ x + 180 x
∫ ( 400000 +19800 √ x + 180 x ) dx
I =
∫
1 2
∫
dx +¿ 19800 x dx + 180 x dx I = 400000
∫¿
I =400000 x + 19800
x
1 +1 2
1 +1 2
+ 180
x 1+1 1 +1
+c
3 2
I =400000 x + 13200 x + 90 x 2+ c ( . ) I =f x x =0 " I =0 " c = 0 16
¿ ¿ = I x =16 400000 ( 16)+ 13200 ¿ I x =16 =7267840.00 .
CONCLUSIÓN: E $#!:< <0 >! H0$%0#! :! !>! !:$?0 !# &$?<: 16 ?!:!:.
7267840.00 .
!# <:
Haa/ a =61i;n 862a an7n in a ndin
2 x 1 −3 x
2
a/a 8ada a,/
d 2 862a 7/a8a a1a ,/ 6n, 0.5 2 x
y =∫
1−3 x
(=1 −3 x
2
dx
2
d( =−6 x dx d( =dx −6 x 2 x d( ( −6 x
y =∫
y =
y =
y =
5=
−1 d( 3
∫
−1 3
−1 3
−1 3
(
ln (
ln ( 1−3 x
2
[
)+c 2
ln 1−3 ( 0 )
]+ c
5= c
y =
−1 3
ln ( 1−3 x
2
)+5
S 1i-a K6 dn/, d a,1 a,/ d 6na 8/a1 d //n, 86ia a6-na/a a 6na /a;n d
0.4 x
3
. √ 0.2 x 4 + 800 aB
. Si 6na 8/a
a n a a86aidad 5000 C6n, ad/ dn/, d "0 a,1 /=
0.4 x
3
. √ 0.2 x 4 + 800 aB
∫
0.4 x
3
4 0.2 x + 800 √ 0.2
dx
(= 0.2 x 4 + 800 d( =0.8 x 3 dx d(
= x3 dx
0.8
∫
0.4 x
∫
0.4 x
/=
3
dx
√ (
/=
(
/=0.5
(
3
dx
1 2
1 +1 2
1 +1 2
+c
1 2
/=( + c 1 2
/=( 0.2 x + 800 ) + c 4
[
500= 0.2 ( 0 )
4
1 2
+ 800 ] + c
500−89.44 = c 410.56 =c
X =10 1 2
/=[ 0.2 ( 10 ) + 800 ] + c 4
c =100 + 410.56
c =510.56
INTERALES CON APLICACIÓN Si a ,8idad d 6n 86/, n i-, 1 /8,//ida n/ F'W F5
7 =2 t + 3 t 2 aa/ a di1an8ia
5
∫ ( 2 t +3 t ) dt 2
d=
2
2
2 t d= 3
|
3 5
3 + t 3
2
5
d = t 2+ t 3|2 5
d (t )= t + t |2 2
3
d = f (t ) 5
¿ ¿
5
¿ ¿
2
¿ ¿
2
¿ ¿ ¿ ¿ d =¿ d =25 + 15− 4− 8 d =138 m
C6 1 a di1an8ia /8,//ida /8,//ida n K6in, -in6, 2
3
d (t )=t + t d = f (t )
d (5) =(5 )2 +(5 )3=150 m 2
3
d (4 )=( 4 ) +( 4 ) =80 m
d (5 ) −d ( 4) =150 −80 =170 m
Haa/ a =6n8i;n 862a an7n an7n in ndin
2
3 x
+1
Pa/a 8ada a,/ d 2 862a 7/a8a a1a ,/ 6n, 'W ! DATOS
m =3 x 2+ 1 x =2 y =6 2
m =3 ( 2 ) + 1 m =13
2 y =∫ 3 x + 1
∫
2
∫
y =3 x dx + dx 3 y = x + x + c
3
6 =(2 )
+( 2 )+ c
6 =8 + 2 + c
−4 =c 3
y = x + x −4
y − y 1= m( x − x1 ) y −2 =13 ( x − 6 ) y =13 x −78 + 2
$ " ' 3 4 5 ! & * >" >' >3 >4 >5 >! >& >*
% 4 "3 '* 49 &! "09 "4* "93 4 "3 '* 49 &! "09 "4* "93
y =13 x −76 → Ec(aci Ec(aciónde ónde %arect %arect a
G0"%<
En a -an6=a8 -an6=a86/a 6/a d 6n /,d68, /,d68, ,1 8,1,1 8,1,1 ,1 ,/ 1-ana 1-ana 1,n d 4000 1i a =6n8i;n d 8,1, -a/7ina 1: 2 Cm= 0.000001 ( 0.002 2 −25 2 ) + 0.2 E 8,1, ,a C //1na a,/ d /,d /,d68 68i/ i/ K i/, i/,1 1 d 6n /,d /,d68 68, , ,/ ,/ 1-a 1-ana na.. En86 En86n n/ / 8,1 8,1, , d /,d68i/ "0000 i/a1 n 6na 1-ana Cm= 0.0000002 2 2−0.000025 2 + 0.2
∫ ( 0.0000002 2 − 0.000025 2 +0.2 ) d2 2
C =
C =0.0000002
C =0.0000002
∫ 2 d2− 0.000025∫ 2 d2 +0.2 ∫ d2 2
2
2 +1
2 +1
2
−0.000025
3
C =0.0000002 − 0.000025 3
−8
2
2
1+ 1
+0.2
1 +1 2
2
+ 0.2
3
2
2
C =6.6 ( 10 )( 10000 ) −0.000025 + 0.2 2
C =666.66 −1250 + 2000 + 4000 C =1416.66 + 4000
C =5416.66 .
∫ ( x x +1 )
20
(= x + 1 d( =1 dx d( = dx
∫
y = ( ( )
20
( ( ) 20+1 +c y = 20 + 1 y =
21 ( x x + 1 )
21
+c
∫ x √ x +5 dx 2
2 (= x + 5
d( =2 x dx d( =dx 2 x
∫
y = x √ (
y =∫ x (
y =
y =
1 2
d( 2 x
d( 2 x
1 +1 2
1 ( +c 2 1 +1 2 3 2
1( 2 3 2
+c
()
3
1 2 2 y = ( +c 2 3
y =
()
1 2 2 3
x 2 + 5 ) + c √ ( x 3
2 2 3
∫ t ( a−t ) 2
(= a−t
d( =−2 t dx d( = dx −2 t
∫
y = t ( (( )
y =∫ ( ( )
y =
y =
2 3
2 3
d( −2 t
d(
−2
−1 2
∫ ( ( )
2 3
d(
2 +1 3
−1 ( ( ) 2 2 +1 3
+c
5
y =
y
−1 ( ( ) 3 2
5 3
( ) ( +c
− = 1 3
y =
y =
+c
2
5
−3
5 3
10
( + c
−3 10
5 3
5 2 3
( a −t ) + c
dt
S 1i-a K6 dn/, d -11 a ,a8i;n d 8i/, 6, 8,-ina/a a 6na /a;n d
2 + 6 √ x
m
. Si a ,a8i;n a86a 1 d 5000 /1,na1
C6 1/ a ,a8i;n dn/, d 9 -11 Da,1 x =meses ra$ón =( 2 + 6 √ x )
-ersnas d→ = 2 + 6 √ x mes dt
-#%aciónact(a% =5000 ?C(>%sera/→t = 9 m@
d=( 2 + 6 √ x ) dx
(
1 2
)
d- = 2 + 6 x dx -=∫ (2 + 6 x ) dx 1 2
-= f ( x )
∫ dx +6∫ x
-=2
-=2 x + 6
-=2 x + 6
x
dx
1 +1 2
1 +1 2
x
1 2
3 2
3 2
+c
+c
()
3
2 2 -=2 x + 6 x + c 3 3 2
-=2 x + 4 x + c
3 2
-=2 x + 4 x + c * / =5000 " x = 0 3 2
5000=2 ( 0)+ 4 ( 0 )
+c
5000= c 3 2
-=2 x + 4 x + c 3 2
-=2 ( 9)+ 4 ( 9 ) + 5000 -=18 + 108 + 5000
-=5126
DEBERES
PRI MER HEM ISE MES TRE
DEBERES DE#ER N$ 1 R!:<! <: :$$!#!: !!%$%$<: 1.- E#%<#0 0 !#>$!#!= #<: >! $#!:!%%$&# @ 0"%0 x + 3 y −1
1.−
−5
−
2
=1
2 ( x + 3 )−[ −5 ( y − 1) ]
−10 2 x + 6 −(5 y + 5 ) =1 −10 2 x + 6 − 5 y − 5 =−10
2 x + 1− 5 y =−10 2 x + 1− 5 y =−10 −1
5 y =−11−2 x 5 y =−2 x −11
y =
−2 x 11 − 5
5
PENDIENTE y =mx + #
=1
m=
−2 5
INTERSECCIÓN E?E $ y =O 0 =−11−2 x
2 x =−11
x =
−11 2
E?E % x =O 5 y + 11=0
5 y =−11
y =
−11
G"%0( $ % 0 > >""J' ""J5 0
2.- R!:<! ! :$:!?0 @ !0$90 0 0"%0 2
5 x
−2 y 2= 2
5 x −2 y =4
5 x −2 y =4
5 x =4 + 2 y
x =
4 + 2 y 5
SUSTITUIR
(
2 y + 4 5 5
5
(
4 y
2
)
2
−2 y 2=2
+ 16 y +16 25
4 y
2
)− 2
+16 y +16− 10 y 2 5
4 y
2
2 y
2
=2
=2
+ 16 y + 16 −10 y 2−10= 2
−6 y 2 + 16 y + 6 =0
R!?09< !# 2 x =
4 + 2 y 5
x =
6 +4 5
x =
10 5
x 1=2
R!?09< !# 1
2
5 x
−2 y 2= 2 2
2 ( 3) + 2 x = 5 2
2 x =
18 + 2 5
√ x2= √ 4 x 2=2
y =
−# ± √ #2 −4 ac 2a
−16 ± √ ( 16 )2− 4 (−6 )( 6 ) y = 2 (−6 ) y =
−16 ± √ 256− 144 −12
y =
−16 ± 20 −12
y =
−16 + 20 −12
y 1=−0.333 y =
−16 −20 −12
$ % > &. 5 y 2=*3 > !. 4 ' > 4. 3 ! /8a: > 3 ' ". > '' " 3 0 4. " ! ' 4. 3 ! 4 !. 5 ' * &. *
$ 0 4J5
% >' 0
DE#ER N$ % 3
2
−2 t + 19 t −8 t −9 1. S ( t )= −t 2+ 8 t − 7
S F
lim s ( t )=4 t→2
G"%<( $ % " I ' & 3 "0.5 4 "5 5 '' >" > >' ".'5 >3 >3.! >4 >! >5 > *.'G > "0.5
0.9 ! S 3.8 6
0.9 * 3. 1
0.9 9 3. /
0.99 " 9 3. '
".00 " '. 2
".0 " '. 3
"." ".'
'.2 8
'.5 /
2.lim x → 1
lim x→ 1
√ x −1 x −1
√ x −1 = √ 1−1 = 0 x −1
1− 1
0
( x − 1) = 1 x −1 √ x + 1 ( x −1 )( √ x + 1) √ x + 1
√ x −1 = √ x −1 · √ x + 1 = x −1
lim x→ 1
1
1
√ x + 1 √ 1+ 1
G"%<( $ " ' 3 4 5 ! & *
=
% I 0.4" 0.3! 0.33 0.30 0.'* 0.'& 0.'!
1 2
= =0.5
DE#ER N$ 3 En86n/ ,1 Q-i1 n ,1 1i76in1 /,-a1: 2
1.− lim ( t −5 ) t →5
lim ( t −5 )=( 5) 2
2
−5=25 −5=20
t→5
P#< C$%< *5K 2+ $ " ' 3 4 5
% >4 >" 4 "" '0
2.− lim 2 x x → 3
2 x =¿ 2 ( 3 )=6 lim ¿ x → 3
P#< C$%< *3K 6+ $ " ' 3 4 5
% ' 4 ! * "0
3.− lim 3 x x→ − 2
lim 3 x
3
3
−4 x 2 + 2 x −3
− 4 x 2+2 x −3 ¿ 3 (−2 )3−4 (−2 )2 +2 (−2 )−3
x→ 3
¿ 3 (−8 )− 4 ( 4 )+ 4 −3 ¿− 24− 16− 4 −3 ¿− 47
$ " ' 3 4 >'
% >' 9 4* "3 3 >4&
P#< C$%< *3K -'/+
4. −lim x→ 9
lim x→ 9
4 r −3 11
4 r −3 4 ( 9 )−3 36 −3 33 = = = =3 11 11 11 11
P#< C$%< *K 3+ $ 3 ' 5 * 9
% 0.* 0.4 ".5 '.! 3
5.− lim √ y + 3 y → 15
lim y → 15
√ y + 3= √ 15 + 3 =√ 18= √ 32 + 2= 3 √ 2= 4.2
P#< C$%< *15K '.2+ $ " ' 3 4 5 " 5
% >'3 " ! "3 33 4.'
x 2+ 6 6.− lim x→ −6 x − 6 2
x + 6 (−6 ) + 6 36 + 6 42 = = = =−3.5 lim (−6 )− 6 −12 −12 x → − 6 x − 6 2
P#< C$%< *-6K -3.5+ $ " ' 3 4 5 >!
% >" > ".'5 > ".!! > '.'0 > '.*" >3.5
2 x − x −6 7.− lim x −3 x → 3
x − x −6 ( 3) − ( 3 ) −6 9 −3− 6 0 = = = = I lim (3 )−3 0 0 x −3 x→ 3 2
2
x − x − 6 ( x −3 )( x + 2) = = x − 2 ( x −3 ) x −3 2
lim x − 2=3− 2=1 x→ 3
P#< C$%< *3K 1+ $ " ' 3 4 5
% 3 4 I
! &
2
t − 4 8.− lim t → 2 t −2 2
t − 4 ( 2 ) −4 4 −4 0 = = = = I lim (2 )−2 0 0 x→ 2 t − 2 2
t − 4 ( t + 2 )( t −2) = =t + 2 ( t −2 ) t −2 2
lim t + 2=2 + 2 =4 x→ 3
P#< C$%< *3K 1+ $ " ' 3 4 5
% 3 4 I
! &
3
2
t + 3 t 9.− lim 3 2 t → 0 t − 4 t 3
2
t + 3 t (0 ) + 3 ( 0 ) 0 = 3 = = I lim 3 2 x→ 0 t − 4 t ( 0 ) − 4 ( 0 )2 0 3
2
2
t ( t + 3 )
(t +3 ) 2 t ( t − 4 ) ( t −4 ) =
( t + 3 ) 0 + 3 −3 = = =−0.75 0− 4 4 x→ 0 ( t − 4 ) lim
P#< C$%< *K -./5+ $ " ' 3 >" >' 0
% >".3 >'.5 >! >0.4 >0." > 0.&5
x 2−9 x + 20 10.− lim 2 x→ 4 x − 3 x − 4 lim x
2
−9 x + 20
x → 4
x2 −3 x −4
( 4 )2−9 ( 4 )+ 20 16−36 + 20 0 = 2 = = = I − − 16 12 4 0 ( 4 ) −3 ( 0 )− 4
x 2− 9 x + 20 ( x −5)( x −4 ) ( x −5 ) = = x 2−3 x −4 ( x −4 )( x −1) ( x −1) lim x −5 x → 4
x −1
=
4 −5 − 1 = =−0.2 4 −1 5
P#< C$%< *'K -.2+ $ " ' 3 >4 >! 4
% >' >" >0.5 >5 >".& >0.'
E?ERCICIOS CON IN#INITOS
11.− lim x →∞
3 x − x
3
3 x + x + 1
∞
¿ ¿ ¿3 ¿ ∞
¿ ¿ ¿ 3 + ( ∞ ) +1 3 ( ∞ )−¿ 3 3 x − x =¿ lim 3 n →∞ x + x + 1 3 x − x 3
3 x − x
3
3
3 x x
− 3
3 3
3 2
−1
x x x x = = = 3 3 3 x + x + 1 x + x + 1 x x 1 1 + 1 + 1 + 3+ 3 3 3 x 2 x3 x x x x
lim n→∞
1+
3
3
x
1
−1 2
+ 2
=
1
x x 3
$ " ' 3 4 >" >' >3 >4
( ∞ )2 −1+
−1
1
% 0.! >0." >0.5 >0.& ' >0.' >0.! >0.&
x + 8 x→ ∞ x − 3
12.− lim
( ∞ )+ 8 ∞ x + 8 =¿ = = I x − 3 ( ∞ ) −3 ∞ lim
¿
x → ∞
8 x 8 + 1− x +8 x x x = = 3 x − 3 x 3 1− − x x x
1− lim x → ∞
8
x
1−
3
x
1−
=
8
+ = 1 0 =1 3 1 −1 1− ∞ ∞
2
+ (∞ )2 ( ∞)3
=
−1 0 −1 = =−1 1 + 0 +0 1
$ " ' 0 >5 >& >*
% >3 >"0 >'.!> 0.4 >0." 0
3
13.− lim −
lim x → ∞
3
x
−
2 x
2
x→ ∞
x x 2 + 1
3
2 x
x
−
2 x
2
x 2+ 1
2
2
x + 1
=
2 (∞)
∞
( ∞ )2 + 1
= −
3 ( x
2
3
∞ ∞
= = I
3 x
2
2
2
3
3
x
$ " ' 3 4 >" >' >3 >4
% ' >0." >0.* >"." >4 >3." >'.* >'.!
3
x x
lim x→ ∞
3
3
+ 3−2
1−
1
x
2
=
+
3
∞ ( ∞ )3 1−
−2
1
( ∞ )2
=
3
−
2 x
3
3
3
x
0 + 0 −2 =−2 1+ 0
+
3
−2 x x3 +1 ) − x ( 2 x ) 3 x + 3−2 x x x x = = = 3 3 2 1 x − x x x x ( x + 1 ) 1− 2 + 3 3
2
+
3
x
x 2−1 14.− lim 3 x→ − ∞ x −4 x − 3
−∞ ¿ ¿ ¿ 3 −4 (−∞ ) −3 ¿ 2 ( − ∞ )2 − 1 x −1 = lim 3 ¿ n →−∞ x − 4 x −3 2
2
1 x −1 x 1 − 3 −1 2 3 3 x −1 x x x x = 3 = 3 = 3 4 3 x − 4 x −3 x −4 x − 3 x 4 x 3 1− 2 − 3 − − 3 3 3 3 x x x x x x
1
lim
x
n → −∞
1−
4
x
$ " ' 3 4 5 >' >5 >4
% 0 >" 0.! 0.3 0.' >" >0.' >0.3
2
1
−1 =
− 33 −1− x
( −∞ ) 4
( −∞ )
2
−1 −
3
(− ∞)3
=
0− 0 =0 1− 0 − 0
4 − 3 x
15.− lim
3
3
x −1
x→ ∞
∞
¿ ¿ ¿ 3 −1 ¿ lim n →∞
4 − 3 x
3
=
x3 −1
4 −3 (−∞ )
¿
4 −3 x 3
4 −3 x
3
4
3
3
−
3 x 3
3
4 3
−3
x x x = 3 = 3 = 3 1 x −1 x −1 x 1 1− 3 − 3 3 3 x x x x lim n→ ∞
x
3
4
4
x
( ∞ )3
1−
$ 4 ' 3 >4 >5 >! 5
% 3 >'.* >'.9 >3 >3." >3 >'.9
−3 3 1
x
3
=
−1−
−3 1
( ∞)3
=
0 −3 −3 = =−3 1 −0 1
3
x 16.− lim 2 x→ ∞ x + 1 lim x
3
( ∞ ) + 8 ∞ = = I 2 ∞ − ( ) 3 ∞ x + 1
x →∞
¿
3
x 3 x
3
x = 2 2 x + 1 x x
+ 3
1
lim x → ∞
1
−
1
x x 3
$ " ' 3 4 5 ! >" >' >3 >4 >5 >! 0
% 0.5 ".! '.& 3.& 4.* 5.* >0.5 >".! >'.& >3.' >4.* >5.* 0
1
x
=
=
3
1 1
− 13
x x 1
3
∞
−
3
∞
3
=
1 0+0
= I
+ x 3 17.− lim 2 x→ ∞ 4 + 5 x −7 x 6 −4 x
2
∞
¿ ¿ ¿3 ¿ ∞
¿ ¿ ¿2 2 6 −4 ( ∞ ) −¿ 2 3 6− 4 x + x =¿ lim 2 n →∞ 4 + 5 x − 7 x 2
6− 4 x + x
3
lim n→∞
4
x
$ " ' 3 4 5 >" >' >3
3
6
x
3
4
− +1 3 x
5
7
x
x
+ 2−
% ".5 0." 0." > 0.0! >0.' > 0."' 0.5' 0.&&
3
=
(∞ ) 4 3
( ∞)
+
−
3
(∞ )
5 2
( ∞)
+1
−
2
(∞)
=
0 −0 + 1 1 = = I 0 + 0 −0 0
4 x
2
3
x 6 4 − + − +1 2 3 3 3 3 3 3 x x x x x x 6 −4 x + x = = = 2 2 2 4 5 7 4 + 5 x −7 x 4 + 5 x −7 x 4 5 x 7 x + 2− + − 3 3 3 3 3 x x x x x x x 6
18.− lim x→ ∞
5 x 7
3 x
− x 3+ 4
∞
¿ ¿ ¿3+ 4 7 3 ( ∞ ) −¿ 5 (∞) ¿ lim 5 x x → ∞
3 x
7
− x 3 + 4
=¿ 5 x
5 x
7
5 6
x x = 7 3 = 7 3 1 4 3 x − x + 4 3 x x 4 − 7 + 7 3− 4 + 7 7 x x x x x
$ " ' 3 4 > " > ' > 3 > 4
% 0.* 0.0' 0.00 ' 0.00 04" >'.5 0.0' ! 0.00 ' 0.00 04
lim n→∞
3−
1
x
4
5
5
x
(∞ )6
6
+
4
x
7
= 3−
1
(∞ )
4
+
4
( ∞ )7
=
0 0 = =0 3− 0 + 0 3
2
−3 x + 4 x→ ∞ 5 x + 7 x − 1 2 x
19.− lim
2
∞
¿ ¿ ¿ 2 + 7 ( ∞ )−1 5¿ 2 2 2 x −3 x + 4 2 ( ∞ ) −3 ( ∞ ) + 4 = lim 2 ¿ x → ∞ 5 x + 7 x −1 2 x
2
−3 x + 4
2 x
2
− 3 x2 + 12
2
2
x
lim 2−
3 1
5+
7
−
1
x x
$ " ' 3 4 >" >' >3 >4
% 0.' 0." 0.' 0.'' >'.5 0.0' 0.003 0.000 4
2−
+
x x 2
x →−∞
2
x
= 5+
3
+
−
x
x x
1
( ∞ ) ( ∞ )2 7
x
2−
3 1
+ x x 2 −3 x + 4 x x x x = 2 = 2 = 2 5 x + 7 x −1 5 x + 7 x −1 5 x 7 x 1 5 + 7 − 1 + 2 − 2 2 2 2
2 x
2
1
( ∞ ) ( ∞ )2
=
2− 0 + 0 2 = =0.4 5 + 0 −0 5
20.− lim x→ ∞
3
=
3
√ x
3
√ x x 12 lim x → ∞
$ " ' 3 4 5 ! & 9
3
3
x 2
( ∞ )2
= 1
% 3 '." ".& ".5 ".3 ".' "." "
3 0
= =0 1
DE#ER N$ &
DERIVADAS
R!0$90 5 !!%$%$<: %<# 0: !0: >! >!$0%$&# E$PONENTE
dy d 6 = x =6 x6 −1 =6 x5 dx dx
80
2.− y = x
6
1.− y = x
dy d 80 = x =80 x 80−1=80 x 79 dx dx
53
3.− y = x
dy d 53 = x =53 x 53−1 =53 x 52 dx dx
3
4. − y = 4 x
dy d = 4 x 3=( 3 )( 4 ) x 3−1=12 x2 dx d x
21
5.− y = x
dy d 21 = x =21 x 21−1=21 x 20 dx dx
CONSTANTE 6.− y =6
dy d = 6 =0 dx dx
7.− y =8
dy d = 8 =0 dx dx
8.− y =−5
dy d = −5 =0 dx dx
9.− y =−9
dy d = −9 =0 dx dx
10.− y = 75
dy d = 75 =0 dx dx
COCIENTE 1
11.− y =
x
1
−1
y = = x x
−1 dy d −1 = x =(−1 ) x −1−1=− x−2= 2 dx dx x
12.− y =
8
x 5
x 8 (¿¿ −5) 8
y = 5 =¿ x x 8 (¿¿−5 )=( 8 ) (−5 x
−5 −1
dy d = ¿ dx dx
) =−40 x−6
13.− y =
1
x
1
y = 2 = x x
2
−2
dy d −2 = x =(−2 ) ( x−2−1 ) =−2 x−3 dx dx
2 4 14.− y = x 3
()
8 dy d 2 4 2 = x = ( 4 ) ( x 4−1 )= x3 3 3 dx dx 3
15.− y =
y =
t 7 25
=
t 7 25 1 7 + t 25
7 dy 1 d 7 1 = t = + ( 7 ) ( t 7−1 )= t 6 25 25 dt 25 dt
DE LA CADENA
6
16. y =(3 x + 2 )
dy d = ( 3 x + 2)6 dx dx dy d =6 ( 3 x + 2)6−1 · ( 3 x + 2 ) dx dx
(
dy d d =6 ( 3 x + 2)5 3 x + 2 dx dx dx
)
dy =6 ( 3 x + 2)5 .3 dx dy =18 ( 3 x + 2 )5 dx
17. y =(5 − x
2 3
)
dy d = ( 5− x 2 )3 dx dx
dy d =3 ( 5− x 2 )3−1 · ( 5− x 2 ) dx dx
(
dy =3 ( 5− x 2 )2 d 5− 2 d x 2 dx dx dx
)
dy =3 ( 5− x 2 )2( 2) dx
dy =6 ( 5 − x2 )2 dx
18. y =( x
2
−2 )3
dy d = ( x 2−2 )3 dx dx dy d =3 ( x 2−2 )3 −1 · ( x 2−2) dx dx
(
dy =3 ( x 2−2 )2 2 d x 2− d 2 dx dx dx dy =3 ( x 2−2 )2( 2 x ) dx
)
dy =6 x ( x 2−2 )2 dx 19. y =(7 − x
4 5
)
dy d = ( 7 − x 4)5 dx dx dy d =5 ( 7 − x 4)5−1 · ( 7− x 4 ) dx dx
(
dy d d =5 ( 7 − x 4) 4 7− 4 x 4 dx dx dx
)
dy =5 ( 7 − x 4) 4 (−4 x 3) dx dy =−20 x 3 ( 7 − x 4) 4 dx
20. y =(5 x
2
−9 )8
dy d = ( 5 x2 −9 )8 dx dx dy d =8 ( 5 x 2−9 )8−1 · ( 5 x 2−9 ) dx dx
(
dy d d =8 ( 5 x 2−9 )7 5 ( 2 ) x 2− 9 dx dx dx
)
dy =8 ( 5 x 2−9 )7 (10 x ) dx dy =80 x ( 5 x 2−9 )7 dx
DE#ER N$ ' R!:<! <: :$$!#!: !!%$%$<: %<# 0 !0 >! <>%<
1. f x =( 4 x + 1)( 6 x + 3 )
dy d d =( 4 x + 1 ) ( 6 x + 3 ) +( 6 x +3 ) ( 4 x + 1) dx dx dx
[
]
[
]
[
]
[
]
dy =( 4 x + 1 ) d 6 x + d 3 +( 6 x +3 ) d 4 x + d 1 dx dx dx dx dx dy =( 4 x + 1 ) 6 d x + d 3 +( 6 x +3 ) 4 d x + d 1 dx dx dx dx dx dy =( 4 x + 1 ) ( 6 ) +(6 x +3 )( 4 ) dx dy =24 x + 6 + 24 x + 12 dx dy =6 ( 8 x + 3 ) dx 2. f x =(3 x −1 )( 7 x + 2)
dy d d =( 3 x −1 ) ( 7 x + 2 )+( 7 x + 2) ( 3 x + 1) dx dx dx
[
]
[
]
[
]
[
]
dy =( 3 x −1 ) d 7 x + d 2 +(7 x +2 ) d 3 x + d 1 dx dx dx dx dx
dy d d d d =( 3 x −1 ) 7 x + 3 +( 7 x + 2 ) 3 x + 1 dx dx dx dx dx dy =( 3 x −1 ) ( 7 ) +( 7 x + 2)( 3 ) dx
dy =21 x −7 + 21 x + 6 dx dy =42 x −1 dx 3
2
3. f t =(5 −3 t )( t −2 t )
dy d d = ( 5−3 t ) ( t 3−2 t 2 ) +( t 3− 2 t 2 ) ( 5−3 t ) dt dx dx
[
]
[
]
dy = ( 5−3 t ) d t 3+ d 2 t 2 + ( t 3 −2 t 2 ) d 5 + d 3 t dt dt dt dt dt
[
]
[
]
dy = ( 5−3 t ) 3 d t 3+ 2 d t 2 + ( t 3− 2 t 2 ) d 5 + 3 d t dt dt dt dt dt dy =( 5−3 t ) ( 3 t 2+ 4 ) + ( t 3−2 t 2 ) (3 ) dx
dy =15 t 2−9 t 3−20 t + 12 t 2−3 t 3+ 6 t 2 dx dy =12 t 3−33 t 2+ 20 t dx
4. f x =( 3 + x )( 5 x
2
−2 )
dy d d =(3 + x ) ( 5 x 2−2 )+( 5 x 2−2 ) ( 3 + x ) dx dx dx
[
]
[
dy =(3 + x ) d 5 x 2− d 2 +( 5 x 2−2 ) d 3 + d x dx dx dx dx dx
[
]
[
]
dy =(3 + x ) 5 ( 2 ) d x2 − d 2 +( 5 x 2−2 ) d 3 + d x dx dx dx dx dx dy =(3 + x )( 10 x )+( 5 x 2−2 )( 3) dx dy =30 x + x + 15 x2 −6 dx dy =15 x 2 + 31 x −6 dx
5. f x =(3 x − x
2
)( 3 − x − x 2)
]
dy d d =(3 x − x 2) (3 − x − x 2)+( 3− x − x 2 )+ ( 3 x − x 2) dx dx dx
[
]
[
dy =(3 x − x 2) d 3− d x − d x 2 +( 3− x − x 2 ) d 3 x − d x 2 dx dx dx dx dx dx
[
]
[
dy =( 3 x − x2 ) d 3 − d x −2 d x 2 +( 3− x − x 2 ) 3 d x −2 x dx dx dx dx dx dy =(3 x − x 2 ) (−1−2 x )+( 3 − x − x 2)( 3 −2 x ) dx dy =−3 x −6 x 2 + x 2+ 2 x 3+ 9− 6 x −3 x + 2 x 2−3 x2 + 2 x 3 dx dy =4 x 3 + 4 x 2−12 x + 9 dx
]
]
DE#ER N$ (
R!:<! <: :$$!#!: !!%$%$<:( 1.−
x " x =1 2 x − 6
dy = dx
dy = dx
2 x −6
d d x − x 2 x −6 dx dx
( 2 x − 6 )2 2 x −6
(
d d d x − x 2 x − 6 dx dx dx 4 x
2
−24 x + 36
dy 2 x −6 ( 1 )− x ( 2 ) = dx 4 x2 −24 x + 36 2 x −6 −2 x dy = 2 dx 4 x −24 x + 36
−6 dy = 2 dx 4 x −24 x + 36 −6 dy = dx → 1 4 ( 1 )2−24 ( 1)+ 36 −6 dy = dx → 1 16 −3 dy = dx → 1 8
)
2.−
1− x " x =5 1 + x
dy = dx
dy = dx
(1+ x )
d d ( 1− x )−( 1 − x ) (1 + x ) dx dx
( 1 + x )2
[
( 1 + x ) ⌈ d 1− d x ⌉ −(1 − x ) d 1− d x dx
dx
( 1+ x )2
dy 1 + x (1 )−( 1 − x )( 1 ) = dx ( 1+ x )2 dy 1 + x −1 + x = dx ( 1 + x )2 2 x dy = dx ( 1 + x )2 2 (5 ) dy = dx → 5 ( 1 + 5 )2 10 dy = 2 dx → 5 6
dy 5 = dx → 5 18
3. y =( x
2
+ 9 )3 " x = 4
dy d 2 3 = ( x + 9 ) dx dx dy d =3 ( x 2+ 9 )3 −1 · ( x 2+ 9) dx dx
(
dy =3 ( x 2+ 9 )2 2 d x 2− d 9 dx dx dx
dy =3 ( x 2−9 )2 ( 2 x ) dx
)
dx
dx
]
dy =6 x ( x 2−2 )2 dx 2 dy =6 ( 4 ) [ ( 4 )2−2 ] dx → 4
dy =24 ( 14 )2 dx → 4 dy =4704 dx → 4
4. y =
y =
1 2
3
( x −1 ) 1 2
3
( x −1 )
" x =2
=( x 2−1 )−3
dy d 2 = ( x −1)−3 dx dx dy d =−3 ( x2 −1)−3−1 · ( x 2−1 ) dx dx
(
dy =−3 ( x2 −1)−4 2 d x 2− d 1 dx dx dx dy =−3 ( x2 −1)−4 ( 2 x ) dx
dy =−6 x ( x 2−1 )−4 dx −4 dy =−6 ( 2) [( 2)2−1 ] dx → 2
dy =−12 ( 3 )−4 dx → 2
( )
1 dy =−12 dx → 2 81
−4 dy = dx → 2 27
)
5. y = x
2
+ 3 X −7 " x =2
dy d 2 = ( x + 3 X −7 ) dx dx
dy d 2 d d = x + 3 X − 7 dx dx dx dx dy =2 x + 3 dx dy =2 ( 2 ) + 3 dx → 2
dy =7 dx → 2
4
x + 3 x " x= 8 6. y = 3 x 4
x 3 x y = 3 + 3 x x 3
y = x + 2 x −2
y = x + 3 x
dy d = ( x +3 x−2 ) dx dx dy d d = x + 3 x−2 dx dx dx dy =−6 x−3 dx dy =−6 ( 8 )−3 dx → 8
−3 dy = dx → 8 256
2
7. y =3 x
+7 " x=2
dy d = ( 3 x 2 +7 ) dx dx dy d d = 3 x 2 + 7 dx dx dx dy =6 x dx
dy =6 ( 2 ) dx → 2 dy =3 dx → 2 8. y = x
2
+ 3 X − 4 " x =−1
dy d 2 = ( x + 3 X −4 ) dx dx dy d 2 d = x + 3 X − 4 dx dx dx
dy =2 x + 3 dx dy =2 (−1 ) + 3 dx → 2
dy =1 dx → 2 3
9. y =8 − x " x =1
dy d = ( 8− x 3 ) dx dx dy d d = 8 + x3 dx dx dx dy =3 x 2 dx
dy = 3 ( 1 )2 dx → 1 dy =3 dx → 1
10. y =
5 x
y =
x
−3 x 3
x
2
5 x
2
−
3 x
" x =1
3
x
y =5 x −3 x
2
dy d = ( 5 x − 3 x 2 ) dx dx dy d d = 5 x + 3 x2 dx dx dx dy =5−6 x dx dy = 5 − 6 ( 1) dx → 1 dy =−1 dx → 1
DE#ER N$ ) D/-ina/ a 86a8i;n d a /8a an7n
1.− y =2 x
/8,: $ % " >3 ' 3 3 "3 4 '& 5 45 >" >3 >' 3 >3 "3 >4 '& >5 45
2
−5 " (−2,3 )
RECTA DE LA
$ 0
% >" 0
y =2 x 2−5 " x =1 dy d = ( 2 x 2−5 ) dx dx
dy d d = 2 x 2− 5 dx dx dx dy =4 x dx dy = 4 (−2 ) dx →−2 dy =−8 dx → 1 m =−8 y − y 0 =m ( x − x 0)
y −( 3 )=−8 [ x −(−2) ]
y −3 =−8 ( x + 2 ) y −3 =−8 x −16 y =−8 x −13
2
2. y =3 x
/8,: $ " ' 3 4 5 ! >" >' >3 >4 >5 >!
% ' "4 3' 5! *! "'' >4 ' "4 3' 5! *!
+ 3 x − 4 " (−1,4 )
RECTA DE LA
$ % 0 " >"J3 0
y =3 x + 3 x −4 " (−1,4 ) 2
dy d = ( 3 x 2 +3 x −4 ) dx dx dy d d d = 3 x 2 + 3 x − 4 dx dx dx dx dy =6 x + 3 dx
dy = 6 ( −1 ) + 3 dx →−1 dy =−3 dx →−1
m =−3 y − y 0 =m ( x − x 0) y −(−4 )=−3 x −(−1 )
y + 4 =−3 ( x + 1 ) y + 4 =−3 x −3
y =−3 x + 1
3. y = x
2
+ 4 " ( 3,7 )
/8,: $ " ' 3 4 5 ! >" >' >3 >4 >5
% 5 * "3 '0 '9 40 5 * "3 '0 '9
RECTA DE LA
$ % 0 >' >'J3 0
2 y = x + 4
dy d 2 = ( x + 4 ) dx dx dy d 2 d = x + 4 dx dx dx dy = x dx dy =3 dx → 3
m =−3 y − y 0 =m ( x − x 0) y −7 =3 ( x −3 ) y −7 = 3 x −9
y =3 x −2
2
4. y =( 2 x + 3 ) " (−2,1 )
y =( 2 x + 3 )
2
y =( 2 x )2+ ( 2 x ) 3 + (3 )2 2
y = 4 x + 12 x + 9
2
y = 4 x + 12 x + 9
/8,: $ " ' 3 4 5 >" >' >3 >4 >5 >! >& >*
% '5 49 *" "'" "!9 " " 9 '5 49 *" "'" "!9
$ % 0 >& >".& 0
2 y = 4 x + 12 x + 9 " (−2,1 )
dy d = ( 4 x 2 +12 x +9 ) dx dx dy d d d = 4 x 2+ 12 x + 9 dx dx dx dx dy =8 x + 12 dx dy = 8 (−2 ) + 12 dx →−2 dy =− 4 dx →−2
m =−4 y − y 0 =m ( x − x 0)
y −1 =−4 [ x −(−2 ) ] y −1 =−4 ( x + 2 ) y −1 =−4 x −8 y =−4 x −7
2
5. y =( x −1 ) " ( 0,1 )
y =( x −1 )2 2
y =( x ) −2 ( x ) 1 + ( 1 )
2
2 y = x −2 x + 1
2 y = x −2 x + 1
/8,: $ " 0 ' 3 4 5 & >" >' >3 >4 >5
% 0 " " 4 9 "! 3! 4 9 "! '5 3!
$ 0 0.5
% " 0
y = x 2−2 x + 1 dy d 2 = x −2 x + 1 dx dx dy d 2 d d = x − 2 x + 1 dx dx dx dx
dy =2 x −2 dx dy = 2 ( 0 )− 2 dx → 0 dy =−2 dx →− 0
m =−2
y − y 0 =m ( x − x 0) y −1 =−2 [ x −0 ] y −1 =−2 x
y =−2 x + 1
DE#ER N$ * D/ia8i;n i-Q8ia R1,/ ,1 1i76in1 /8i8i,1
1.− x
2
+ y = x 3 + y 2
d 2 ( x + y ) = d ( x 3 + y 2) dx dx d 2 d d 3 d 2 x + 2 y =3 x + 2 y dx dx dx dx 2 x + 2 y : =3 x
2
+ 2 yy 0
2 : y ( 1 + 2 y )=3 x −2 x
2
3 x −2 x y 0 = 1 + 2 y 2
2.5 x + x y
3
= 2 y
d (5 x + x 2 y 3) = d (2 y ) dx dx d d 2 3 d 5 x + 2 x y =2 y dx dx dx
5
d 2 d 3 3 d 2 x + x y + y x = 2 y 0 dx dx dx
5−( 3 x y y 2
2
2
+ 2 y 3 x )=2 y 0
:
:
2
3
5−3 x y y −2 y x =2 y 0
−3 x 2 y2 y : −2 y 0 =2 y 3 x −5 2 2 3 : y (−3 x y −2 ) =2 y x −5
3
y 0 =
3. x
3
2 y x −5
−3 x 2 y 2−2
+ y 3= xy
d 3 3 d ( x + y )= ( xy ) dx dx d 3 d 3 d d x + y = x y + y x dx dx dx dx 2
3 x
+ 3 y 2 y : = xy0 + y 2
:
xy0 + 3 y y = y − 3 x
2
y 0 ( 3 y 2 − x ) = y −3 x 2 2
y −3 x y 0 = 2 3 y − x
4. −
1
1
+ =1
x y
d −1 −1 d ( x + y ) = 1 dx dx d −1 d −1 x + y =0 dt dt
−1 x−1−1 + (−1 y−1−1 y : )= 0 x−2+ y −2 y : =0 −
−2
:
y 2 y =− x −2
x y = −2 y :
− y 2
:
y =
x
2
5.−√ x + √ y =1 1 2
1 2
x + y =1
(
1
1
)
d 2 d x + y 2 = 1 dx dx 1
1
d 2 d 2 x + y =0 dx dx 1
1
1 2 −1 1 2 −1 x + y y : = 0 2 2
1 x 2
−1 2
1 2 x
1 2
−1
+ 1 y 2 y : =0 2 1
+
1 2 √ y
2 y
:
y =
:
1 2
y = 0
1 2 √ x
1 :
y =
2 √ x 1 2 √ y
y : =
2 √ y 2 √ x
y y : = √ √ x
DE#ER N$ + R1,/ ,1 1i76in1 /8i8i,1 1. g =
5 x
2
−8 x
2 x 2
5 x 8 x g= − 2 x 2 x
g=
5 x −4 2
(
dg d 5 x = −4 dx dx 2
)
dg d 5 x d = − 4 dx dx 2 dx dg 5 d d = x − 4 dx 2 dx dx
dg 5 = dx 2 2. f ( x )=5 x
3
√ 3 +2 x 4
f (x ) =5 x ( 3 + 2 x 3
1 4 2
)
1
1
dy =5 x 3 d ( 3 + 2 x 4 ) 2 + ( 3 + 2 x 4 ) 2 d 5 x 3 dx dx dx
[( )
]
1
−1 1 dy 3 4 2 ( 3 + 2 x ) + d 3 + 2 x 4 + ( 3 + 2 x 4 ) =5 x 2 dx dx
[( )
dy 1 ( 3 + 2 x 4 ) =5 x 3 2 dx
−1 2
]
+ 8 x + ( 3 + 2 x 3
1 4 2
) [ 15 x 2 ]
()
−1
1
()
−1
1
1 dy =40 x 6 ( 3 + 2 x 4 ) 2 −15 x 2 ( 3 + 2 x 4 ) 2 dx 2
dy =20 x 2 1 ( 3 + 2 x 4 ) 2 −15 x 2 ( 3 + 2 x 4 ) 2 dx 2
3. y =
√ √ x
2
+
y =
( ) +(
y =
(
1 x 2
1 2
2 2
2 x
1 −1 x + 2 x 2
−1
)
1 2
)
1 2
−1 2
[
d 3 5 x dx
]
(
)
dy 1 1 − = x + 2 x 1 dx 2 2
−1
dy d 1 = x + 2 x−1 dx dx 2
1 2
(
)
(
) (
dy 1 1 = x + 2 x−1 dx 2 2
(
dy 1 1 −1 = x + 2 x dx 2 2
(
dy 1 1 = x + 2 x−1 dx 2 2
4. y = √ ( 7 −3 x 3
y =( 7 −3 x
2
−1 2
·
(
d 1 − x + 2 x 1 dx 2
) −1
d 1 d · x + 2 x dx 2 dx
−1
)
) (− ) 2
·
1 2
−1
) ( 2
·
−2
2 x
1 − x −2 4
)
2 2
)
2 2 3
)
[( )
2
−1 dy = 2 d ( 7 −3 x2 ) 3 + d 7− 3 x 2 3 dx dx dx
[( )
2
]
−1 2 d dy 2 3 ( 7 −3 x ) + d 7 − d 3 x 2 = dx dx dx 3 dx
[( )
2
−1 2 dy ( 7 −3 x 2) 3 +( 0 −6 x ) = 3 dx
−1
dy =−4 x ( 7 −3 x 2 ) 3 dx
5. g( x ) =
g( x )=
−3 4 3 4 ( $ + 2 $ −5 )
−3 4
−4
3 · ( $ + 2 $ −5 )
]
]
dy −3 d 3 = ( $ + 2 $ −5 )−4 −1 +( $ 3 + 2 $ −5 )−4 d − 3 d$ 4 d$ d$ 4 −5 d dy −3 = (−4 ) ( $3 + 2 $ −5 ) ( $ 3 + 2 $ −5 ) + ( $3 + 2 $ −5 )−4 d$ 4 d$
(
−5 d 3 d dy −3 d = (−4 ) ( $3 + 2 $ −5 ) $ + 2 $ − 5 d$ 4 d$ d$ d$
)
−5 dy =3 ( $3 + 2 $ −5 ) ( 3 $ 2+2 ) d$
−5 dy =( g ) ( $ 3 + 2 $ −5 ) d$
x + 4 x + 3
6. y =( ( + 3 ) √ ( + 6 "
)√
y =
(
y =
(
)√
y =
x + 4 + 3 x + 9 x + 3
(
)√
(
)(
x + 4 x + 4 +3 +6 x + 3 x + 3 x + 4 + 3 ( x + 3 ) x + 3
x + 4 + 6 ( x + 3 ) x + 3
x + 4 + 6 x + 18 x + 3
x + 4 + 3 x + 9 x + 4 + 6 x + 18 y = x + 3 x + 3
y =
(
4 x + 13 x + 3
)(
7 x + 22 x + 3
)
1 2
y 0 =
(
4 x + 13 d 7 x + 22 x + 3 dx x + 3
y 0 =
(
4 x + 13 1 7 x + 22 x + 3 2 x + 3
d dx
(
) (
( x + 3 )
)(
)
1 2
)
) ( 1 2
+
1 −1 2
7 x + 22 x + 3
(
) (
)
)(
)
1 2
d 4 x + 13 dx x + 3
+ d 7 x + 22 + 7 x 22 dx x + 3 x + 3
d d ( 4 x + 13 )−( 4 x + 13 ) ( x + 3 ) dx dx
( x + 3 )2
)
1 2
y 0 =
(
+
(
(
(
(
(
d dx
)( 1 2
dx
( x + 3 )2
dx
dx
( x + 3 )2
)(
4 x + 13 1 7 x + 22 x + 3 2 x + 3
)
1 −1 2
[
( x + 3 ) ( 7 ) −( 7 x + 22 ) ( 1 ) ( x + 3 )2
)
1 −1 2
[
)
]
)
7 x + 21 −7 x −22
( x + 3 )2
]
) ( x +( x−+ x) − ) 1 2
4
1 4
3
13
2
)(
4 x + 13 1 7 x + 22 x + 3 2 x + 3
(
)(
7 x + 22 x + 3
− x x e +e
3
1 3
− x x y = ( e + e )
dy 1 d x − x = ( e + e ) dx 3 dx
(
[
dx
( x + 3 ) ( 4 )−( 4 x + 13 ) (1 ) ( x + 3 )2
1 4 x + 13 y 0 = 2 x + 3
7. y =
(
)
( x + 3 ) d ( 7 x + 22 )−( 7 x + 22 ) d ( x + 3 )
( x +3 ) d ( 4 x + 13 )− ( 4 x + 13 ) d ( x +3 )
)(
7 x + 22 x + 3
y 0 =
)
1 2
4 x + 13 1 7 x + 22 x + 3 2 x + 3
7 x + 22 x + 3
y 0 =
+
)(
7 x + 22 x + 3
y 0 =
+
(
4 x + 13 1 7 x + 22 x + 3 2 x + 3
1 −1 2
x
dy 1 d d − x = e + e dx 3 dx dx
)
dy 1 x = [ e ( 1)+ e− x (−1 ) ] dx 3
−1
)
2
−1
)
2
[ ](
1 2
)(
−1 ( x + 3 )2
)
[ ](
)(
−1 ( x + 3 )2
)
−1 7 x + 22 + x + 3 ( x + 3 )2 −1 7 x + 22 + x + 3 ( x + 3 )2
1 2
]
dy 1 x − x = ( e −e ) dx 3
3
2 x
8. y =5
%ny= ln 52 x
3
3 %ny= 2 x ln5
dy d 3 ln y = 2 x + ln 5 dx dx 1
y
2
y 0 ¿ 6 x + ln 5
2
y 0 ¿ 6 x + ln5. y x
y 0 ¿ 6 x 2 + ln5.52
3
2;
e 9. f ( ; )= 2 ; 2;
f ( ; ) = e + ;
−2
df ( ; ) d 2 ; −2 = ( e + ; ) d; d; df ( ; ) d; df ( ; ) d;
= e2 ;
d − 2 −2 d 2 ; ; + ; e d; d;
(
= e2 ; (−2 ;−3 )+ ;−2 e2 ;
d 2 ; d;
df ( ; ) 2 ; = e (−2 ;−3 )+ ;−2 ( e2 ; )( 2) d; df ( ; ) d;
= e2 ; (−2 ;−3 )+ 2 ;−2 ( e 2 ; )
df ( ; ) 2 ; = e (−2 ;−3 + 2 ;−2 ) d;
)
x
e −1 10. y = x e +1
dy = dx
dy = dx
dy = dx
x e +1
d x ( e −1 ) −( e x −1 ) d ( e x +1 ) dx dx
( e x + 1 )2 ( e x + 1 )
(
)
(
d x d d d e − 1 −( e x − 1 ) e x + 1 dx dx dx dx
( e x + 1 )2 ( e x + 1 )
(
x
e
)
(
d x x d x −( e − 1 ) e x dx dx 2 x
x
e + 2 e +1
dy ( e + 1 ) e −( e −1 ) e = 2 x x dx e + 2 e +1 x
x
x
x x 2 x 2 x dy e + e − e + e = 2 x x dx e +2 e + 1
x
dy 2e = 2 x dx e + 2 e x + 1 x
x
dy 2 e 2 e = 2 x + x + 2 e x dx e 2e dy 2 = x +1 + 2 e x dx e
x
)
)
SEGUNDO EMISEMESTRE
DEBERES SEUNDO HEMISEMESTRE DE#ER N$ 1 PROBLEMAS ". CERCADO 6na -/1a di1,n d 9000 a/a 8/8a/ 6na ,/8i;n /8an76a/ d //n, ad2a8n a 6n /Q, 86a 1 61a 8,-, 6n, d ,1 ad,1 d /a 8/8ada. E 8,1, d a 8/8a a/aa a /i, 1 d "5 ,/ i in1aad, 2 d a 8/8a a/a ,1 d,1 ad,1 /1an1 1 d 9 ,/ i in1aad,. En86n/ a1 di-n1i,n1 d /a -i-a 8/8ada.
DATOS
U#0 <%$&# ` C<:< >! %!%0 00!0 `15 < $!
ab
C<:< >! <: 2 0><: !:0#!: ` < $!
a@b
!0 ^ 0< c 0#%< 4 =% x a 15 x + 9 ( 2 y )= 9000
y =
9000 −15 x 18
VARIABLE
!0 ^ %<:< >! 0 %!%0 00!0 c C<:< >! <: ><: 0><: !:0#!: 4 = x + y
(
9 − 15 x 4 = x 18
4 =
2
)
1 ( 9 x −15 x 2) 18
4 = f ( xy)
PRIMERA DERIVADA
4 =
1 ( 9 x −15 x 2) 18
4 0 =
1 ( 9 −30 x ) 18
SEUNDA DERIVADA 4 0 0 =
1 (−30 ) 4 00 <0 → 4ϵ 1 18
PRIMERA DERIVADA IUALAR A 0 4 0 = =
1 ( 9 −30 x ) 18
30 x 9000 = 18 18 30 x =18 18
x =
9000 18
x =300
REEMPLA@AR y =
9000 −15 x 18
y =
9000 −15 ( 300 ) 18
y =
9000 −4500 18
y =250
CONCLUSION: 0: >$?!#:$<#! >! !0 ?$?0 %!%0>0 !: >! 3? ! 0< @ 25? ! 0#%< >! 0 %!%0.
G0"%< $ " ' 3 4 5
% 499." & 99!.! & "49'. 5 "9*!. !& '4&9. "&
'. C,1, /,-di, 6n =a/i8an d/-ina K6 8,1, ,a 8 d /,d68i/ 6n a/Q86, 1 dad, ,/ a =6n8i;n d 8,1, 2
c =0.05 2 + 5 2 + 500
dPa/a K6< ni d /,d688i;n 1/ -Qni-, 8,1, /,-di, ,/ 6nidad
DATOS 2
c =0.05 2 + 5 2 + 500
2 ´c = 0.05 2 +5 2 + 500
2
c´ =0.05 2 + 5 +
500
2
PRIMERA DERIVADA c´ 0 = 0.05 +
500
2
2
SEUNDA DERIVADA c´ 0 0 =
500
2
3
> 0→ c m
PRIMERA DERIVADA IUALAR A 0 c´ 0 = 0.05 +
0 =0.05 +
−0.05=
2 2=
500
2
2
500 2
2
500
22
500 0.05
2
2 =10000
√ 22= √ 10000 2 =± 100 2 =100
CONCLUSION: E #$! >! <>%%$&# :! ?#$?< %0#>< ! %<:< $< :!0 1 < #$>0>
G0"%< $ " ' 3 4 5 ! &
% 505.0 5 '55." 0 "&".* ' "30.' 0 "50.' 5 **.!3 &!.&*
3. UTILIDAD Pa/a /,d68, d 6n -,n,,i1a a =6n8i;n d d-anda 1 -=85 −0,05 2 2 a =6n8i;n d 8,1, 1 c =600 + 35 2 . A K6< ni d /,d688i;n 1 -ai-ia a 6iidad A K6< /8i, ,86// 1, 2 86 1 a 6iidad -i-a DATOS
P!%$<^ -=85 −0,05 2 C<:< ^ c =600 + 35 2 VARIABLE
;^ %0#$>0> >! <>%< >!?0#>0><. U^ $$>0>. (ti%dad =tta% ingress− csts
(ti%idad = -recis Dcantidad − csts
(= - D 2− c (= ( 85−0,05 2 ) 2−( 600 + 35 2 ) 2 (= 85 2− 0,05 2 −600 −35 2
(=−0,05 2 2+ 50 2 −600
PRIMERA DERIVADA 2
( : =−0,05 2 + 50 2−600 ( : =−0,1 2 + 50
SEUNDA DERIVADA : : : : : : ( =−0,1 cm( < 0 → c ϵ 1
PRIMERA DERIVADA IUALAR A CERO ( : = 0 0 =−0,1 2 + 50 0,1 2 =50
2 =500 -=85 −0,05 2
-=85 −0,05 ( 500 ) . -=20 ( (=−0,05 2 2+ 50 2 −600 (=−0,05 ( 500 )2 + 50 ( 500 )−600 (= . 11900,00
CONCLUSION:
L0 $$>0> >! <>%< :! 0 ?$?0 %0#>< :! <>9%0# 5 #$>0>!: 0 !%$< >! 2 ` %0>0Q #$>0> <!#$!#>< #0 $$>0> >! ` 11=.
G0"%< $ " ' 3 4 5 ! &
% >550.05 >500.'0 >450.45 >400.*0 >35".'5 >30".*0 >'5'.45
4. UTILIDAD Pa/a /,d68, d 6n -,n,,i1a a =6n8i;n d d-anda 1 40 1 200 -= ´ = + c 2 a =6n8i;n d 8,1, /,-di, 1
√ 2
3
2
En86n/ /8i, 2 a /,d688i;n K6 -ai-ian a 6iidad. A 1 ni d-61/ K6 in7/1, -a/7ina 1 i76a a 8,1, -a/7ina. DATOS
-=
P!%$<
40
√ 2 1 200 3 2
c´ = +
C<:< $< VARIABLES
C^ %<:< <0 U^ $$>0> ;^ %0#$>0> >!?0#>0>0 c c´ = =´c 2 =c 2 cst tta%=cst -rmediD cantidad
c=
(
)
1 200 + 2 3 2
1 3
c= 2 +
200
2
2
2 c = + 200 3
(ti%idad =ingress −csts (ti%idad = -reci Dcntidad demandada− cst
(=
40
√ 2
2 D2 − + 200 3
2 (= 40 2−1 /2 2 − + 200 3
2 1 /2 (= 40 2 − + 200 3
PRIMERA DERIVADA 2 1 /2 ( : = 40 2 − + 200 3
:
( =20 2
−1/ 2
1 3
− =
20
√ 2
−
1 3
S76nda d/iada 1 3
: :
−1 / 2
−
: :
−3 /2
cm( < 0 →(
( =20 2
( =10 2
::
::
1
ϵ
PRIMERA DERIVADA IUALAR A CERO (: =0 O=
20
√ 2
−
1 3
1 20 = 3 √ 2 1 20 = 3 √ 2
√ 2 =20 D 3 √ 22= 602 2 =3600
-=
-=
40
√ 2 40
√ 3600
=
40 =0,67 . / (nidad 60
INRESO MARINAL I =
40
√ 2
D2
− 1/ 2
I =40 2
2
I =40 2 1/ 2 1
40 − 1 I : = 2 2 2 −1
I : =20 2
I : =
I : =
2
20
√ 2 20
√ 3600
I : =
20 60
I : =
1 3
COSTO MARINAL 2 c = + 200 3
:
c=
3 ( 1 ) −2 ( 0 ) 9
3 9
c : = =
+0
1 3
CONCLUSION:
S! !%0 <>%$ 36 #$>0>!: >! <>%< @ !#>! 0 # !%$< >! =6/ `Q#$>0> 00 ?0$?$90 0 $$>0>. G0"%< $ " ' 3 4 5 ! &
% '39.!& '55.90 '!*.'* '&*.!& '*&.&* '95.9* 303.50
5. In7/1,. Una -/1a d in1 /aQ81 ,1 "00 da/a-n,1. Cada 6n, 6d /na/1 a 40000 ,/ -1. Sin -a/7, ,/ 8ada "000 -n16a1 d in8/-n, a/ d,1 da/a-n,1 a8Q,1 1in ,1iiidad d /na/,1. 6< /na ,/ da/a-n, -ai-ia/a in7/1, -n16a DATOS
1 >!00?!#<: `'= ` 1= 2 >!00?!#<: PLANTEAMIENTO DE LA ECUACION r =( 100 −2 x ) ( 400 + 10 x )
PRIMERA DERIVADA r =( 40000 + 1000 x −800 x −20 x 2 ) :
r=
d (−20 x2+ 200 x +40000 ) dr
:
r =−40 x + 200
SEUNDA DERIVADA : : r =
d (−40 x + 200 ) dr
r : : =−40 < 0 →maxim
PRIMERA DERIVADA IUALAR A 0
( 100−2 x ) ( 10 ) + ( 400 + 10 x ) (−2 )=0 ( 1000−20 x −800 −20 x )=0 200− 40 x =0 40 x = 200
x =
200 40
x =5
. d
REEMPLA@O EN LA ECUACIÓN. 400 + 10 ( 5 ) 400 + 50 = . 450,00
CONCLUSIÓN
L0 !#0 < >!00?!#< ! ?0$?$900 ! $#!:< ?!#:0 !: >! . 450,00 G0"%< $ " ' 3 4 5 ! & * 9 " 0
% 40"* 0 403' 0 404' 0 404* 0 4050 0 404* 0 404' 0 403' 0 40"* 0 4000 0
!. In7/1,: Una -/1a d i1i;n ,/ 8a in 4*00 1618/i,/1 K6 a7an "*.00 -n16a1 8ada 6n, 2 6d 8,n176i/ "50 1618/i,/1 -1 ,/ 8ada /d688i;n K6 0.50 n a /na -n16a C6 1/ a /na K6 -ai-i8 in7/1, 2 86a 1/ 1 in7/1, DATOS:
S:%$<!: '8 P0<
18
. m
S<:$%$&# 15 ::%$<!: < %0>0 !>%%$&# >!
0.50
. m
VARIABLES:
)^ N >! >!:%!#<: 00 ?0$?$90 ! $#!:<. R!#0( *!#0 ?!#:0 !>%%$&# >! !#0+ *::%$<!: _ :<:$%$<#!:+ r =( 18− 0.50 x )( 4800 + 150 x )
r = 86400+ 2700 x + 2400 x −75 x
2
r =−75 x 2+ 300 x + 86400
PRIMERA DERIVADA r =−150 x + 300
SEUNDA DERIVADA r =−150 r < 0 → r 1
PRIMERA DERIVADA IUALAR A 0
r =0 0 =−150 x + 300
150 x =300
X =
300 150
x =2
r =18 −0.50 ( 2 )=17
. m
I =4800 + 150 ( 2)∗17 I =( 4800 + 300 )∗17 I =(5100 )∗17 I =. 86700
CONCLUSIÓN
L0 !#0 ! ?0$?$%! ! $#!:< !: >! 1/` @ !:! $#!:< !: . 86700 G0"%< $ " ' 3 4 5 ! & * 9 " 0
% 40"* 0 403' 0 404' 0 404* 0 4050 0 404* 0 404' 0 403' 0 40"* 0 4000 0
&. Di1, Di1, d 6n /8ii /8iin n.> .> 6n =a/i8 =a/i8an an d /8ii /8iin n1 1 di1a di1a 6na 8aa 8aa /8an76a/ 1in aa 2 8,n a1 86ad/ada K6 d n/ 6n ,6-n d 3' i13. 6< di-n1i,n1 d n/ a 8aa 1i 1 /K6i/ K6 1 6ii8 a -n,/ 8anidad d -a/ia
DATOS:
V<?!# >! 0 %00( 32 $!: 3 C0#$>0> >! ?0!$0( ?#$?0 )( 0#%< < $>0>. Y( Y( 00 VARIABLES 2
x y =32
y =
32
x
2
A^ !0 >! ?0!$0 PRIMERA DERIVADA 2
4 = x + 4 xy
2 4 = x + 4 x
( ) 32
x 2
2 4 0 = x +
128
x
SEUNDA DERIVADA
(
d 2 128 x x + dx x
4 0 0 =
4 0 0 =2 x −
)
128
x
2
PRIMERA DERIVADA IUALADA A CERO x 2+
128
128
= 2 x
x2
x
=0
128=2 x 64 = x
3
3
3 3 64 64 =√ x √ 3
x =4
REEMPLA@O 2 x y =32 ( 4 )2 y =32 y =
32 =2 16
L0 %00 :$# 00= %<# !0 >! ?0!$0 ?#$?0 ?$>!( ,0:!^ ' $!: ' $!: =A0^ 2 $!:
G0"%<
$ " ' 3 4 5 !
% "'9 !* 5".! 4* 50.! 5&.33
'4.> DISEXO DE UN CARTEL: 6n 8a/ /8an76a/ d 8a/;n d n/ "50 2
-(%g
a/a -a/ia i-/1, -/7n1 d 3 67 a//ia 2 da, d ' 67 a
8ada ad,. En86n/ a1 di-n1i,n1 d Ca// d -an/a K6 a 8anidad d 8a/;n K6 1 61 1a a -i1-a. 3ee
2ee 2ee
DATOS:
C0! !%0#0 !0 15 0>0:
3ee
M0!# :!$< ! $#H!$< 3 0>0: M!#!: >! <: 0><: 2 0>0: VARIABLES:
)^ 0< Y^ 0#%< !0 ^ 0:! f 00 150= x ∗ y 150
x
= y
area =( x + 4 )( y + 6 )
4 =( x + 4 )
4 = ( x x + 4 )
4 =
150 x
x
( x + ) 150
(
6
150 + 6 x
x 2
+
6 x
x
4 =174 + 6 x +
)
+ 600 + 24 x x
x
600
x
PRIMERA DERIVADA :
4 =174 + 6 x +
600
x
x ( 0 )−600 ( 1 ) : 4 =0 + 6 (1 )+ x2 :
4 =6 −
600
x
2
SEUNDA DERIVADA : : 4 =6 −
600
x
2
0− x ( 0 )−600 ( 2 ) x x 2
: :
4 =
: : 4 =
: :
4 =
2 ( x x2 )
1200 x
x
4
1200 3
x
PRIMERA DERIVADA IUALAR A CERO :
4 =0 0 =6 −
600
x
2
x 2=
600
x
2
=6
600 6
2 x = √ 100 100 √ x
x =± 10
REMPLA@O y =
150 10
L0: >$?!#:$<#!: >! %0! 00 ! 0 %0#$>0> >! %0&# :!0 ?#$?0: :<#( L0< ^1 0>0: A#%< ^ 15 0>0:.
G0"%< $ " ' 3 4 5 ! & * 9 " 0
% &*0 4*! 39' 34* 3'4 3"0 30" '9& '94.! '94
DE#ER N$ % D/-in a in7/a indi8ada 8,-/6 a1 /161a1 d/iand, 1.
∫ √ 2t dt −1
y =∫ 2 t 2 dt −1
y =2∫ t 2 dt −1 +1
y =2
t 2
−1 2
+c +1
y =2
1 2
t
1 2
+c
1 2
y = 4 t + c
()
1
1 2 −1 y = 4 t + c 2 −1
y = 2 t 2
y =
2.
2
√ t
∫ x1
2
y =∫ x dx −2
−2+ 1
y =
x
−2+ 1
x−1
y =
−1
+c
+c −1
y =−1 x + c
y =− x−1+ c y 0 =−(−1 ) x −2
y 0 = x
1
y 0 = 2 x
3.
∫3 e
y =3
x
dx
∫e
x
dx
− 1− 1
+0
y =3 e x + c y 0 =3 e x
∫
− 0.3
4. x
dx
−0.3− 1
y =
x
x
y =
y =
+c
−0.3−1 0.7
+c
0.7
1 0.7 x + c 0.7
y 0 =
1 ( 0.7 ) x 0.7 −1+ 0 0.7 −0.3
y 0 = x
5.
(
)
∫ x 1 − x1 2
∫
3
−2
y = x − x
dx
−3
dx
y =∫ x dx −∫ x dx −2
−3
−2+ 1
y =
y =
x
x
−3 + 1
− +c −2+1 −3 + 1 −1 x
−
1 2
2 x
+c
− 1− 1
y 0 =−(−1 ) x
−2
−3
y 0 = x − x 1
1
y 0 = 2 − 3 x x
1 2
− ( 2 ) x−2−1
6.
∫ ( 3 t − √ 5 t + 2 ) dt 2
y =3
∫ t dt −∫ ( 5 t ) 2
2+ 1
t
y =3
2+ 1
−
( 5 t )
1 2
1 +1 2
1 +1 2
+2
3 2
t ( 5 t ) + 2 t + c y =3 − 3
3
3 2
3
2 y =t − ( 5 t ) 2 + 2 t + c 3 3
1 2
y =3 t −( 5 t ) + 2 :
2
y =3 t 2 −√ 5 t + 2 :
7.
∫(( )
−2 5
−2 + 1
( () 5 +c y = −2 + 1
5
y =
( () 3 5
3 5
+c
3
5 y = ( ( )5 + c 3
()
3 −1 5 3 ( ) = y 0 ( 5 +c
3 5
∫
0
dt + 2 x dt
t 0+1 0 +1
+c
−2
y 0 =( ( )
5
8.∫ ( x − 3 x 1 3
)
−2
+ 6 dx
3
−2
1 3
y =∫ x dx −3∫ x 3 dx + 6∫ x0 dx
y =
x
1 +1 3
1 +1 3
x
−3
−2 + 1 3
−2 3
4 3
+6
x 0 +1
+1
0 +1
+c
1 3
x x y = −3 + 6 x + c 4 3
1 3
4
1
3 y = x 2 − 9 x 3 + 6 x + c 4
()
()
4
1
3 4 3 − 1 1 3 −1 y 0 = x − 9 x + 6 x + 0 4 3 3
:
1 2
2 3
y = x −3 x + 6
9.
∫ ( 3 √ x − 2 x− ) dt 3
(
)
1 2
y =∫ 3 x −2 x−3 dt
y =
y =
( 3 x )
1 +1 2
1 +1 2
( 3 x )
3 2
+2
3 2
−3 + 1
−2
x
+2
3 2
x−3 +1
−2
−2
+c
y =2 ( x ) + x + c
+c
()
3
−1 3 ( x )2 − (−2 ) x−2−1 y =2 2 :
1 2
−3
y : = 3 x −2 x
y 0 =3 √ x −2 x
∫
10.
1 y = 2
(
−3
)
1 2 3 − 2+ dt 2 t t √ t
∫
dt −2 t
−1
∫ t
−2
dx + 3
1 2
t
−2 + 1
+3
t 2
−1 2
1
1 −1 y = ln t + 2 t + 6 t 2 + c 2 −1
1 −2 y = −2 t + 3 t 2 2 t :
:
y =
1 2 3 − 2+ 2 t t √ t
11.
(
1 y = 2
∫e
∫
1 2
)
x
e
2
+ x √ x d x
x
x y = e +
3 2
dx +∫ x dx
x
3 +1 2
3 +1 2
+c
5 2
1 x x y = e + + c 2
5 2
5
1 x 2 y = e + x 2 + x 2 5
2
−1 + 1
− 2+ 1
y = ln t −2
∫ t
+c +1
e x
y = + √ x 2 x :
2
x
e
:
y = + x √ x 2
(
)
3 x + 2 x + 1 dx 2 x
∫
12.
−1 −2 y =∫ ( x + 2 x + x ) dx
+∫ x− dx ∫ dx x
∫
2
y = xdx + 2
y =
x
1+ 1
1 +1
x
−2 + 1
+ 2 ln x +
x
−2 + 1
+c
2
− y = + 2 ln x− x 1+ c
2
1 2
2−1
y : = ( 2 ) x
1
+ 2 − x−1−1 x
−1 −2 : y = x + 2 x + x
∫ (2 e + 6( +ln 2 )d( (
13.
y =2
∫ e + 6∫ (− + ln 2∫ d( (
1
1
( y =2 e + 6 + ln 2 ( + c (
−1 : ( y =2 e + 6 ( + ln 2
6
: ( y =2 e + + ln 2 (
∫
3
( )
14. x 2 x
1
x
dx
y =∫
(
3
)
x 2 x + dx x 4
4 2 y =∫ ( 2 x + x ) dx
∫
4
x
4+ 1
∫
2
y =2 x dx + x dx
y =2
+
4+1
x
y =2
5
x
x
2 +1
2+ 1
+c
3
+ + c
5
3
2 5
1 3
5 3 y = x + x + c
2 5
1 3
y : = ( 5 ) x5−1+ ( 3 ) x 3− 1 + 0
:
4
2
y =2 x + x
(
∫ √ x − 2 1√ x
15.
5
5 2
1 x dx − 2
∫
y =
x
)
dx
∫ x
−1 2
dx
−1 +1
5 +1 2
1 x 2 − +c y = 5 2 −1 +1 +1 2 2
x y =
−1 + 1
5 +1 2
−
7 2
1 x 2 2 1 2
7
+c
1
2 y = x 2 + x 2 + c 7
()
7
1
2 7 2 1 2 y : = x + x + 0 7 2 2
5 2
1 2
x y : = x + + 0 2
1
y =√ x + :
5
2 √ x
∫
2
16. x ( 2 x + 1 ) dx
y =∫ x ( 4 x 2 + 4 x + 1 ) dx
∫ ( 4 x + 4 x + x ) dx 3
y =
2
y = 4 ∫ x 3 dx + 4∫ x 2 dx +∫ xdx y = 4
+1
x 3
+
x 2
+1
+
x 1
+1
3+ 1 2 +1 1 +1
x
4
x
3
2
y = 4 + + 4
+c
x
3
1 3
2
+c
1 2
y = x 4 + x 3 + x 2+ c
y : = x
:
4 −1
1 3
1 2
+ (3 ) x 3−1+ (2 ) x 2−1+ 0
3
2
y =4 x + 4 x + x
∫ x1 ( x +1 ) dx 2
17.
y =∫
1
( x 2+ 2 x +1 ) dx
x 1
2
1
1
y =∫ + x + .2 x + .1 dx x x x
∫
−1
y = x + 2 + x dx
y =∫ x dx + 2∫ dx +∫ x dx −1
y =
x 1+1 1 +1
1
+ 2 x + + c x
x 2
1
y = + 2 x + + c 2 x 1
1
y = x 2+ 2 x + + c 2 x 1 2
y : = ( 2) x 2−1 + 2 x + 1+ 0
y = x + 2+ x−1 :
∫−5 x +12 x −2 xdx 3
18.
2
∫ (−5 x +12 x −2 x ) dx 3
y =
2
y =−5∫ x3 dx + 12∫ x 2 dx −2∫ x dx y =−5
y =
x 3+ 1 3 +1
−5 4
+ 12
4
x 2+1 2+1
3
−2
2
x + 4 x − x + c
:
3
19.
∫ √ t ( t −1 )
2
y =−5 x + 12 x − 2 x 2
y =∫ √ t ( t 2−1 ) dt
∫
y = t √ t − √ t dt 2
1 2
1 2
y =∫ t + t −t dt 2
5 2
1 2
y =∫ t −t dt
y =
5 +1 2
t
5 +1 2
−
1 +1 2
t
1 +1 2
+c
x 1+1 1+1
+c
y =
7 2
t
7 2
3 2
t
− +c 3 2
7
3
2 2 y = t 2 − t 2 + c 7 3
()
()
7
3
2 7 2 −1 2 3 2 −1 y : = t − t + 0 7 2 3 2 5 2
1 2
y : =t −t
∫ ( 2 x +5 ) dx 8
20.
(=2 x + 5
d( =2 dx = d( 2 dx
∫
y = ( ( )
y =
1 2
8
d( 2
∫ ( d( 8
8 +1
1 ( +c y = 2 8+1 9
1( + c y = 2 9
y =
1 9 ( +c 18
y =
y =
1 ( 2 x + 5 )9 + c 18
1 9− 1 2 ( 2 x + 3 ) ( 2)+ 0 18
y = (2 x + 5 )8
DE#ER N$ 3 RESOLVER LOS SIUIENTES E?ERCICIOS % COMPROBARLOS
∫
5
1. X dx
y =
x
5+ 1
5+1
x
+c
6
y = + c 6
1 6
6
y = x + c
1 6
6− 1 y = ( 6 ) x + c
:
y = x 5
2
∫ x1 dx 2
y =∫ x−2 dx −2+ 1
y =
x
+c
−2+ 1
− y =− x 1+ c
−2
:
y = x :
1
y = 2 x 3.
∫ ( 3 x − √ 5 x + 2 ) dx 2
∫
1 2
∫
∫
0 y =3 x dx − ( 5 x ) dx + 2 x dx
y =3
x
2
2 +1
2+ 1
−
( 5 x )
1 +1 2
0 +1
+2
1 +1 2
x
0 +1
+c
3 2
x ( 5 x ) + 2 x + c y =3 − 3
3
3 2
3
2 y = x − ( 5 x ) 2 + 2 x + c 3 3
1 2
y =3 x −( 5 x ) + 2 :
2
2 : y =3 x −√ 5 x + 2
4.
(
∫ 3 √ x − x2 + x1
y =3
y =3
3
∫ ( x ) x
1 2
1 +1 2
1 +1 2
)
dx
∫
−3
∫ dx x
dx − 2 x dx +
−2
x
−3 + 1
−3 + 1
+ ln x + c
y =3
x
3 2
3 2
−2
3 2
x −2
−2
+ ln x+ c
−2
y =2 x + x + ln x + c 1 2
:
1
−3
y =3 x −2 x + x 2
:
1
y =3 √ x − 3 + x x
5.
∫ ( x
1 2
)
2 3
−3 x + 6 dx 1 2
2 3
y =∫ x dx −3∫ x dx + 6∫ x0 dx
y =
x
1 +1 2
1 +1 2 3 2
−3
x
2 +1 3
2 +1 3
+6
x 0+ 1 0+1
+c
5 3
x x y = −3 + 6 x + c 3 2
5 3
3
5
2 9 y = x 2 − x 3 + 6 x + c 3 5 1 2
:
2 3
y = x −3 x + 6
6.
∫
(
1 y = 2
)
1 2 − 2 + 3 dx 2 x x √ x
−2∫ x−2 dx + 3 ∫ x ∫ dx x − 2+ 1
−1 + 1
−1 2
x x2 1 +3 +c y = ln x − 2 −2 + 1 −1 2 +1 2
1
1 y = ln x + 2 x −1 + 6 x 2 + c 2 1 −2 y = − 2 x + 3 x 2 x :
y : =
7.
∫
−1 2
1 2 3 − 2+ 2 x x √ x
(
1 y = 2
)
x
e
2
+ x √ x dx
∫e
x
1 2
x y = e +
3 2
dx +∫ x dx
x
3 +1 2
3 +1 2
+c
5 2
1 x x y = e + + c 2
5 2
5
1 x 2 y = e + x 2 + x 2 5 x
e
y = + √ x 2 x :
2
:
e x
y = + x √ x 2
8.
∫
(
1 y = 3
)
1 3 x − 2 + e 2+ √ dx 3 x 2 x 2
∫
1
dx 3 1 − x−2 dx + e 2 dx + x 2 dx x 2 2
∫
∫
∫
−2+ 1
1 3
y = ln x −
1 +1 2
2+ 1
3 x 1 x e + + +c 2 −2 + 1 2 + 1 2 1 +1 2 3
1 3 1 1 y = ln x + x−1+ e3 + x 2 + c 3 2 3 3 1
1 3 −2 2 1 y = − x + e + x 2 3 x 2 2 :
y : =
9.
∫
1 3 x − 2 + e2 + √ 3 x 2 x 2
(
)
x 2+ 2 x + 1 dx x 2
∫ ( 1 +2 x− + x − ) dx 1
y =
∫
2
∫ xdx +∫ x− dx
0
2
y = x dx + 2
y =
x
0+ 1
0 +1
−2+ 1
+ 2 ln x +
x
−2 + 1
+c
−1
y = x + 2 ln x− x + c 2
y : =1 + + x−2 x
∫ ( x +2 x ) ( x1 −5 ) dx 3
10.
2
y =∫ (−5 x 3+ 12 x 2−2 x ) dx
∫
∫
3
∫
2
y =−5 x dx + 12 x dx −2 x dx
y =−5
y =
x
−5 4
3+ 1
3 +1
4
+ 12
3
x
2+ 1
2+1
2
−2
x + 4 x − x + c
x
1+ 1
1+1
+c
:
3
2
y =−5 x + 12 x − 2 x
∫ √ x ( x −1 ) dx 2
11.
(
5 2
1 2
)
y =∫ x − x dx 5 2
1 2
y =∫ x dx −∫ x dx
y =
5 +1 2
x
5 +1 2
−
x
1 +1 2
1 +1 2
7
+c
3
2 2 y = x 2 − x 2 + c 7 3 5 2
:
y = x + x
1 2
1 2
y = x ( x 2−1 ) :
2 : y =√ x ( x −1 )
12.
∫ (√ x − 2 √ 1 x +√ 2) dx 3
3 2
1 x dx − 2
∫
y =
x
∫ x
−1 2
dx + ( 2 )
−1 +1
3 +1 2
1
1 2
∫ x dx
0+ 1
0
1 x 2 x − +( 2 ) 2 +c y = 3 2 −1 0 +1 +1 +1 2 2 5
1
1
2 y = x 2 + x 2 + ( 2 ) 2 x + c 5 3 2
1 y = x + x 2 :
: 3 y =√ x +
−1 2
+ (2)
1 2 √ x
1 2
+ √ 2
∫
13.
(
3
(
5 2
)
x + √ x −2 x dx √ x 1 2
)
y =∫ x + 1−2 x dx 5 2
∫
∫
∫
0
1 2
y = x dx + x dx −2 x dx
y =
5 +1 2
x
+
x
0 +1
5 0+ 1 +1 2 7
−2
3
x
1 +1 2
2 4 y = x 2 + x − x 2 + c 7 3 5 2
:
y = x + 1 −2 x
1 2
∫ 2 X 1+1 dx
14.
(=2 x + 1
d( d( =2 dx = 2 dx y =∫
y =
1 2
1 d( ( 2
∫ d((
1 2
y = ln ( ( ) + c
1 2
y = ln ( 2 x + 1 )+ c
y : =
1 1 ( 2) +0 2 2 x + 1
1 +1 2
+c
y : =
1 2 X + 1
∫ 3 x2+2 dx
15.
(=3 x + 2 d( =3 dx dx =
d( 3
y =∫
y =
2 3
2 d( ( 3
∫ d((
2 3
y = ln ( ( )+ c
2 3
y = ln ( 3 x + 2 ) + c
y : =
y : =
16.
2 1 ( 3 ) + 0 3 3 x + 2 2 3 X + 2
∫ ( 2 x + 3 ) dx 5
(=2 x + 3 d( =2 dx dx =
d( 2
y =∫ ( ( )
y =
1 2
5
d( 2
∫ ( d( 5
5 +1
1 ( +c y = 2 5 +1 6
1( + c y = 2 6
y =
1 6 ( +c 12
y =
1 ( 2 x + 3 )6+ c 12
y : =
1 6 −1 6 ( 2 x + 3 ) 2 + 0 12
y : =
12 ( 2 x + 3 )5 12
y : =( 2 x + 3 )
5
∫ 9 ( x +3 x + 5 ) (2 x +3 ) dx 8
2
17.
2
(= x + 3 x + 5
d( =2 x + 3 dx d( =( 2 x + 3 ) dx
∫
8
y = 9 ( ( ) d(
y =9 ∫ ( d( 8
y =9
(
8+ 1
8+1
+c
9
(
y =9 + c 9
9 y =( + c
2
9
y =( x + 3 x + 5 ) + c
y =( x 2+ 3 x + 5 )9+ c 2 y : = 9 ( x + 3 x 5 )
9− 1
( 2 x + 3 )+ 0
8
y : = 9 ( x 2 + 3 x 5 ) ( 2 x + 3 ) 8
2 y : = 9 ( 2 x + 3 ) ( x + 3 x 5 )
∫
18.
3 x + 6
√ 2 x
2
+ 8 x + 3
dx
2 (=2 x + 8 x + 3
d( =4 x + 8 =4 dx dx =
d( 4 ( x + 2)
∫ 3 ( x√ +( 2 ) 4 ( xd(+2 )
y =
3 y = 4
∫(
−1 2
d(
−1 + 1
y =
3 (2 +c 4 −1 +1 2
y =
1 2
3 ( 4 1 2
+c
1
3 y = ( 2 + c 2 1
3 2 y = ( 2 x + 8 x + 3 )2 + c 2 1
3 y = ( 2 x 2 + 8 x + 3 )2 + c 2 1
−1 31 ( 2 x 2+ 8 x + 3 ) 2 ( 4 x + 8 )+ 0 y : = 22
1
−1 31 ( 2 x 2+ 8 x + 3 ) 2 ( 4 x + 8 )+ 0 y : = 22
3 ( 4 x + 8
y : =¿
y : =
y : =
19.
¿ 2 4 √ 2 x + 8 x + 3 12 x + 24
4 √ 2 x
2
+ 8 x + 3
3 x + 6
√ 2 x 2 + 8 x + 3
∫ √ 4 x −1 dx
(= 4 x −1 d( =4 dx dx =
d( 4
∫
y = (
1 2
d( 4
y =
1 +1 2
1 ( +c 4 1 +1 2 3 2
y =
1 ( 4 3 2
+c
3
1 y = ( 2 + c 6 3
1 y = ( 4 x −1) 2 + c 6 3
1 y = ( 4 x −1) 2 + c 6 3
−1 13 ( 4 x −1) 2 4 + c y : = 62
1
12 y : = ( 4 x −1 ) 2 + 0 12
y : = √ 4 x −1
∫
20. x ( x
2
+1) dx
2 (= x + 1
d( =2 x dx dx =
d( 2 x
y =∫ x (( )5
d( 2 x
y =
1 2
∫ ( ( ) d( 5
5 +1
1 ( +c y = 2 5 +1 6
1( + c y = 2 6
y =
1 6 ( +c 12
y =
1 2 ( x + 1 )6+ c 12
y =
1 2 6− 1 6 ( x + 1 ) 2 x + 0 12
y =
12 x 2 ( x + 1)5 12 2
5
y = x ( x + 1)
∫
21. x ( x 2
3
3/ 4
+ 1 ) dx
3
(= x + 1 d( =3 x2 dx
dx =
d( 2 3 x
y =∫ x 2 ( ( )3 4 /
y =
y =
1 3
∫ (() /
3 4
3 +1 4
d( 2 3 x d(
1 ( +c 3 3 +1 4
7 /4
1( y = 3 7 4
+c
7
4 y = ( x 3 + 1 ) 4 + c 21 7
4 y = ( x 3 + 1 ) 4 + c 21 7
−1 4 7 3 y : = ( x +1 ) 4 ( 3 x 2)+ 0 21 4
3
2
21 x y : = ( x 3 + 1 ) 4 21
y : = x ( x + 1 ) 2
3
3 4
∫ ( x 3+ x2− x 3+6 ) dx
22.
2
2 (= x + 2 x + 6
d( =2 x −2 dx
d( =2 ( x −1 ) dx dx =
d( 2 ( x −1 )
y =∫ 3 ( x −1)
y =
3 2
d( ( 2 ( x −1) 1
∫ d((
3 2
y = ln ( ( )+ c
3 2
y = ln ( x 2 + 2 x + 6 ) + c
3 2
2 y = ln ( x + 2 x + 6 ) + c
y : =
y : =
y : =
y : =
3 1 ( 2 x −2 ) +0 2 2 x + 2 x + 6 3 ( 2 x − 2 ) 2 ( x
2
+ 2 x + 6 )
6 x −6 2 ( x
2
+ 2 x + 6 )
3 x −3 2
( x + 2 x + 6 )
∫ x −x 1 dx
23.
(= x −1 x =( + 1
d( =1 dx dx = d(
∫ ( +( 1 d(
y =
∫ (( d( +∫ d(( + c
y =
y =( + ln ( ( ) + c y = x −1 + ln ( x − 1 )+ c
y = x −1 + ln ( x −1 )+ c 1
(1 ) y : = 1 − 0 + x −1
1
y : =1 + x −1 x −1 + 1 y : = ( x −1 ) x y : = x −1
24
∫ ( x −x 5 ) dx 6
x =( + 5 (= x −5
d( =1 dx y =∫
(( +5 )
y =∫
( d( 6 (
y =∫
5 ( d(+∫ 6 d( 6 ( (
(
∫
6
d(
−5
y = ( d( + 5 −5+ 1
y =
y =
(
−5 + 1 (
−4
−4
x −5
+5
¿ ¿ x −5 ¿ ¿ −1 ¿ y = 4
+5 (
∫ (− d(
(
6
− 6+ 1
−6 + 1
−5
−5
+c
+c
y =
−1 4
−4 −5 ( −( + c
x − 5
¿ ¿ x − 5 ¿ ¿ : −1 (−4 )¿ y = 4
x −5
¿ ¿ ¿
:
y =
1
+5 ( x −5 ) ¿ 5
: x − 5 + 5 y = ( x −5 )6
x
:
y =
( x −5 )6
∫ ( xx−+43) dx
25.
2
(=2 x −1 2 x =(−1
d( =2 dx y =∫
( +4 +3 d( 2 (
∫ ((+7 d(
y =
S:$%$<#( (= x −4 x =( + 4
2
y =∫
7 ( + d( d( ∫ 2 2 ( (
y =∫
d( + 7∫ (−2 d( (
d( =4 dx
y =%n( + 7
(−1
−1 −1
y =%n(−7 (
−1
y = ln ( x − 4 )−¿−7 ( x − 4 ) −1
y = ln ( x − 4 ) −7 ( x − 4 ) x − 4
¿ ¿
1
:
y =
( x − 4 ) 1
:
y =
+7 ¿ 7
+
( x − 4 ) ( x −4 )−2
: x − 4 + 7 y = ( x −4 )2
:
y =
26.
x +3 ( x − 4 )2
∫ ( 2 x +3 ) √ 2 x −1 dx
y =∫ 2 x √ 2 x −1 dx +∫ 3 √ 2 x −1 dx d( d( y =∫ 2 x √ ( +∫ 3 √ ( 2
∫
y = ( ( + 1) √ (
y =∫ ( √ (
1 y = 2
1 y = 2
∫ ∫
d( 2
2
d( 2
+∫ √ (
1 2
1 ( ( d( + 2 3 2
+∫ 3 √ (
1 ( d( + 2
∫
∫
d( 2
d( 2
+∫ 3 √ (
1 2
3 ( d( + 2 1 2
3 ( d( + 2
d( 2
∫(
∫(
1 2
1 2
d(
d(
y =
y =
3 +1 2
1 +1 2
1 +1 2
1 ( 1 ( 3 ( + + +c 2 3 2 1 2 1 +1 +1 +1 2 2 2 5 2
1( 2 5 2
+
3 2
1( 2 3 2
5
+
3 2
3( 2 3 2
3
+c
3
1 1 y = ( 2 + ( 2 + ( 2 + c 5 3 5
3
3
1 1 y = ( 2 x −1 ) 2 + ( 2 x −1 ) 2 + ( 2 x −1 ) 2 + c 5 3
()
()
3
1
1
1 5 ( 2 x −1 ) 2 .2 + 1 3 ( 2 x −1 ) 2 .2+ 3 ( 2 x −1 ) 2 .2+ 0 y 0 = 5 2 3 2 2
3 2
1 2
y 0 =( 2 x − 1 ) + ( 2 x − 1 ) + ( 2 x −1 )
y 0 =( 2 x − 1 )
1 2
1 2
[ ( 2 x −1 ) + 1 +3 ]
y 0 =( 2 x + 3 ) √ 2 x −1
DE#ER N$ & T/aa, d /86/a8i;n R1,/ ,1 1i76in1 /,-a1 DEBER DE RECUPERACION
21.- S 1i-a K6 dn/, d -11 a ,a8i;n d 8i/, 6, 8a-ia/a a 6na /a;n d
2 3
4 + 5 t
/1,na1 ,/ -1. Si a ,a8i;n a86a 1 "0000
/1,na1 C6 1/ a ,a8i;n dn/, d * -11
DATOS t =meses 2 3
ra$n =4 + 5 t
-ersnas mes
&'o"lación dentro de ( meses) 2
d = 4 + 5 t 3 dt 2 3
-=∫ 4 + 5 t dt
-= 4 t +
2 +1 3
5 t +c 2 +1 3 5 3
-= 4 t +
5 t 5 3
+c
5 3
-= 4 t + 3 t + c 5 3
-= 4 t + 3 t + 10000
P<0%$&# >!#< >! 8 ?!:!:. 5 3
/( 8) =4 ( 8 )+ 3 ( 8 ) + 10000
/( 8 )=32 + 96 + 10000
/( 8 )=10128 -ersnas
CONCLUSION( 0 <0%$&# >!#< >! 8 ?!:!: :! >! 1128 !:<#0:.
''.> Un ,, 1 -6 d -an/a K6 16 ,8idad d16<1 d -in6,1 1 ) ( t )=1 + 4 t + 3 t 2 -/,1 ,/ -in6, 6 di1an8ia /8,// ,, d6/an /8/ -in6,
DATOS t =min(ts ) ( t )=1 + 4 t + 3 t
2
dD$:0#%$0 ! !%<! ! <!< >0#! ! !%! ?$#<] d =∫ ( 1 + 4 t + 3 t 2 ) dt 2
4 t d =1 t + 2
3
+ 3 t 3
+c
d =t + 2 t 2 + t 3+ c t = 0 " d = 0 c = 0
2 3 d (3) =3 + 2 ( 3 ) +( 3 )
d (3) =3 + 18 + 27 =48 m
2 3 d (2) =3 + 2 ( 2 ) +( 2 )
d (2) =2 + 8 + 8 =18 m
La di1an8ia /8,//ida d6/an /8/ -in6,
d (3 ) =48 −18=30 metrs
CONCLUSION( D0#! ! !%! ?$#< 0 >$:0#%$0 !%<$>0 !: >! 3 ?!<: . '3.> Un ,, 1 -6 d -an/a K6 16 ,8idad d16<1 d -in6,1 1 2 ) ( t ) =3 + 2 t + 6 t -/,1 ,/ -in6,. 6< di1an8ia /8,// ,, d6/an 176nd, -in6, DATOS
t =min(ts ) ( t ) =3 + 2 t + 6 t
2
dD$:0#%$0 ! !%<! ! <!< >0#! ! :!#>< ?$#<] d =∫ ( 3 + 2 t + 6 t 2) dt 2
2 t d =3 t + 2 2
3
6 t + +c 3 3
d =3 t + t + 2 t + c t = 0 " d = 0 c = 0
2 3 d (2) =3 ( 3 )+ ( 2 ) + 2 (2)
d (3) =6 + 4 + 16 =26 m
2 3 d (1) =3 ( 1 )+ (1 ) + 2 ( 1 )
d (1) =2 + 8 + 8 =6 m
L0 >$:0#%$0 !%<$>0 >0#! ! :!#>< ?$#<.
d (3 m ) =26 −6=20 metrs
CONCLUSION( D0#! ! :!#>< ?$#< 0 >$:0#%$0 !%<$>0 !: >! 2 ?!<: . '4.> S 1i-a K6 dn/, d a,1 a,/ d 8i/a a/8a 1 in8/-na/a a 6na /a;n d r ( t ) d;a/1 ,/ a,. Ha 6na /1i;n a/a a 8anidad n K6 a6-na/a a,/ d a i//a d6/an ,1 /;i-,1 5 a,1. DATOS t =aBs
ra$ón =r ( t )
d%ares aB
dr = r ( t ) dt 5
r =∫ r ( t ) 0
'5.> L,1 /,-,,/1 d 6na =/ia d di1/i, 1i-an K6 ,/a1 d16<1 d a/i/ a1 6/a1 a a1 9:00 a- ,1 i1ian1 n/a/n a a =/ia a 6na /a;n d r ( t ) /1,na1 ,/ ,/a. Ha 6na /1i;n a/a d/-ina/ n-/, d /1,na1 K6 n/a/ a a =/ia n/ a1 "":00 a- 2 a ":00 - .
DATOS t = !ras
a#re =9:00 am ra$ón =r ( t )
-ersnas !ra
11( 0? 1( ?
dr = r ( t ) dt 13
r =∫ r ( t ) 11
'!.> Un =a/i8an d i8i8a1 1/a K6 dn/, d -11 ,1 8,n16-id,/1 8,-/n 5000 i8i8a1 ,/ -1 a 6n /8i, d / ( x )= 80 + 3 √ x d;a/1 ,/ i8i8a. C6 1 in7/1, ,a K6 =a/i8an 6d 1/a/ d a1 na1 d a1 i8i8a1 n ,1 /;i-,1 "! -11 DATOS x =meses
)enta estimada=5000
/ ( x ) = 80 + 3 √ x
#icic%etas mes
d%ares #icic%eta
dI#!:< !# <: &$?<: 16 ?!:!:] :
I = cantidad ∗ -reci I =5000 ( 80 + 3 √ x ) :
. mes
I = 400000 + 15000 √ x :
∫
I = 400000 + 15000 √ x dx 1 2
I =∫ 400000 + 15000 x dx
I =400000 x +
I =400000 x +
15000 x 1 +1 2
15000 x 3 2
1 +1 2
+c
3 2
+c
3 2
I =400000 x + 10000 x + c I =0 " x =0 " c = 0
I ( 16)= 400000 ( 16 ) x + 10000 ( 16 )
3 2
I ( 16)=6400000 + 640000 :
I ( 16)=7 040000 dó%ares
CONCLUSION: E $#!:< <0 !# <: &$?<: 16 ?!:!: /e' >&0!: < 0 !#0 >! 0: $%$%!0: '&.> Un =a/i8an d i8i8a1 1/a K6 dn/, d -11 ,1 8,n16-id,/1 - ( x )= 80 + 3 √ x 8,-/n f ( x )=5000 + 60 √ x Bi8i8a1 ,/ -1 a 6n /8i, d
d;a/1 ,/ i8i8a C6 1 in7/1, ,a K6 =a/i8an 6d 1/a/ d a na d a1 i8i8a1 n ,1 /;i-,1 "! -11
DATOS x =meses )eta estimada= f ( x ) =5000 + 60 √ x
#icic%etas mes
. = 80 + 3 √ x ( ) x P!%$<^ #icic%eta
:
I = cantidad ∗ -reci
I : =( 5000 + 60 √ x )( 80 + 3 √ x ) :
I = 400000 + 15000 √ x + 4800 √ x + 180 x
I = 400000 + 19800 √ x + 180 x :
. mes
. mes
∫
I = 400000 + 19800 √ x + 180 x dx 1 2
I =∫ 400000 + 19800 x + 180 x dx
19800 x I =400000 x + 1 +1 2
19800 x I =400000 x + 3 2
1 +1 2
3 2
2
180 x + 2
180 x + 2
+c
2
+c
3 2
I =400000 x + 13200 x + 90 x 2+ c I =0 " x =0 " c = 0 3 2
I ( 16)= 400000 ( 16 )+ 13200 ( 16 ) + 90 (16 )
2
I ( 16)=6 : 400000 + 844800 + 23040 :
I ( 16)=7 267840 dó%ares
CONCLUSION: D!#< >! <: 16 ?!:!: :! !:!00 /e26/8' >&0!: >! 0 !#0 >! 0: $%$%!0:.
'*.> Un -in,/i1a /8i 6n 8a/7a-n, d "'00 i/a d 1-ia1 d 1,2a K6 1 8,n16-i/n a 6na /a;n 8,n1an d 300 i/a1 ,/ 1-ana. Si 8,1, d a-a8na-in, d a1 1-ia1 d 1,2a 1 0'0 8na,1 ,/ i/a a a 1-ana C6n, nd/ K6 a7a/ -in,/i1a n 8,1,1 d a-a8na-in, n a1 /;i-a1 40 1-ana1
DATOS cm-ra =1200 %#
cns(m= 300
%# semana
ct)s cstde a%macena=e =0,20 %#
VARIA,LE^ ^ #?!< >! :!?0#0: CT : = c + a%macena= e∗%i#rasa%macenadas :
C T =0,20 ( 1200−300 t )
C T : =( 2400−60 t )
%# semanas
∫
CT = ( 2400 −60 t ) dt 2
60 t +c CT = 2400 t − 2 2
CT = 2400 t −30 t + c CT = 0 " t = 0 " c =0
%# semanas
2
CT ( 40)= 2400 ( 40 )−30 ( 40 ) CT ( 40)= 96000 −48000 CT ( 40)= 48000 dó%ares
CONCLUSION: P< ! %<:< >! 0?0%!#0?$!#< >! 0: &$?0: ' :!?0#0: ! ?$#<$:0 !#> ! 00 '8 >&0!:. '9.>S 1i-a K6 dn/, d a,1 a ,a8i;n d 8i/a 8,-6nidad a a ,/ia 2 d 6n a7, 8a-ia/ a 6na /a;n d 0,6 t + 0,2 t + 0,5 -i1 d /1,na1 ,/ a,. L,1 18iai1a1 n -di, a-in an n8,n/ad, K6 ni d 8,na-ina8i;n n a7, a6-na a 6na /a;n a/,i-ada d 5 6nidad1 ,/ 8ada "000 /1,na1 En 86n, 1 in8/-na/ a 8,na-ina8i;n n a7, d6/an ,1 /;i-,1 d,1 a,1
DATOS t =aBs 2
ra$ón =0,6 t + 0,2 t + 0,5
mi%es de -ersnas aB
a(ent cntm+ =5 (nidades -r cada 1000 -
&En cánto incrementa la contaminación en % a*os) 2
/=0,6 t + 0,2 t + 0,5 2
0,6 t + 0,2 t + 0,5
(¿) dt /=∫ ¿ 3
0,6 t /= 3
2
+ 0,2 t 2
+ 0,5 t
/=0,2 t 3+ 0,1 t 2 + 0,5 t + c /=0 " t =0 c =0 3 2 /( 2) =0,2 ( 2 ) + 0,1 ( 2) + 0,5 ( 2 )
/( 2) =1,6 + 0,4 + 1 /( 2 )=3
30. 6n 16di, a-ina /aiad, n 8i/a 8,-6nidad /a K6 dn/, d a,1 ni d -,n;id, d 8a/,n, n ai/ 8a-ia/a a 6na /a;n an6a d 0.1 t + 0.1 a/1 ,/ -i;n. Si ni d -,n;id, d 8a/,n, n ai/ 1 d 3.4 a/1 ,/ -i;n C6 1/ ni dn/, d 3 a,1 DATOS
R09&# >! %0?$<(
0.1 t + 0.1
-ersnas mes
T$!?<( 3 07<: >!:4:
∫
f ( t ) = ( 0.1 t + 0.1 ) dt 2
0.1 t f ( t ) = 2
+0.1 t + c 2
f ( t ) =0.05 t + 0.1 t + 3.4 t = 3 2
f ( t ) =0.05 ( 3 ) + 0.1 ( 3 )+ 3.4 f ( t ) =0.45 + 0.3 + 3.7
f ( t ) =4.15
CONCLUSIÓN: E #$! >! %<#0?$#0%$&# :! >! '.15 0!: < ?$&# !# 3 07<: 3". Un =a/i8an a 8,n/aa K6 8,1, -a/7ina 1 6 2 +1 1,a/1 ,/ 6nidad 86nad, 1 an /,d68id, K 6nidad1. E 8,1, ,a in86id,1 indi/8, d /,d688i;n d a /i-/a 6nidad 1 d "30Y.00 C6 1 8,1, ,a d /,d688i;n d ,1 "0 /i-/a 6nidad1 DATOS :
c =6 2 + 1
^ #$>0>!: C >! 0 $?!0 #$>0> 13. C >! 1 U ^] dc =∫ ( 6 2 + 1 ) d2
dc = 6
∫ 2 d2 +∫ d2
2
2
dc =6 + 2 + c 2
2
dc = 3 2 + 2 + c c =130 2 =1 2
130=3 ( 1)
+1 +c
126= c 2
dc = 3 2 + 2 + 126
2 =10 2
130=3 ( 10)
+ 10 + c
c = 436
CONCLUSIÓN: E %<:< >! <>%$ 1 #$>0>!: !: >! `'36. >&0!: −1
3'. Un =a/i8an 1i-a K6 in7/1, -a/7ina 1
100 2
2
d;a/1 ,/ 6nidad
8and, ni d /,d68n 1 K 6nidad1 S a 1a8id, K6 8,1, -a/7ina 8,//1,ndin 1 d 0.4 K d;a/1 ,/ 6nidad . S6,n7a K6 a 6iidad d =a/i8an 1 d US5'0 86and, ni d /,d688i;n 1,n"! 6nidad1 C6 1 a 6iidad =a/i8an 86and, ni d /,d688i;n 1,n '5 6nidad1 DATOS −1 :
I =100 2
2
:
c =0.4 :
:
A = I −C : −1
:
A =100 2 2 − 0.4
∫
−1
A = 100 2 2 −0.4 d2
2
A =100
1 2
1 2
22
+ 0.4 + c 2
1 2
2
A =200 2 + 0.2 2 + c
2 =16 A =520 1 2
520=200 ( 16 )
+0.2 (16 )2 +c
520=800 −51.20 + c
−228.80 =c
1 2
2
A =200 2 + 0.2 2 + c 2 =25 1 2
2
A =200 ( 25 ) + 0.2 ( 25) − 228.80 A =1000 + 352.80 A = 646.20
CONCLUSIÓN: 0 U$$>0> >! H0$%0#! 0 <>%$ 25 #$>0>!: !: >! `6'6.2 >&0!:
33. LA 6iidad -a/7ina a d/iada d a 6iidad d 8i/a 8,-aQa 1 d 100−2 2 d;a/1 ,/ 6nidad1 86and, 1 /,d68n K 6nidad1 . Si a 6iidad d a 8,-aQa 1 d &00 86and, 1 /,d68 "0 6nidad1 C6 1 a -i-a 6iidad ,1i d a 8,-aQa DATOS :
A = 100 −2 2
2 =(nidades A =700 2 =10
A =∫ 100 −2 2 d2 2
2
A =100 2 + 2 + c 2
A =100 2 + 2 2+ c A =700 2 =10 700=100 ( 10 ) −(10 )
2
+c
700=1000 − 10000 + c
−200 =c A =100 2 + 2 2+ c 2 A =100 ( 50 ) + ( 50 ) − 200
A =5000 −2500 + 200
A =2300
CONCLUSIÓN: 0 U$$>0> ?$?0 >! H0$%0#! :! >0 0 <>%$ 5 #$>0>!: >0#>< # 0< `23 >&0!:
3' Haa/ a =6n8i;n 862a an7n in ndin 2 862a 7/a8a a1a ,/ 6n, "W ' DATOS m = 4 x + 1 x =1
y =2 m = 4 ( 1 )+ 1 m= 4 + 1 m =5
4 x + 1
Pa/a 8ada a,/ d
y =∫ 4 x + 1
∫
∫ dx
y = 4 xdx +
x
y = 4
1+ 1
+ x + c
2 2
y =2 x + x + c
y =2 x =1 2
2= 2 ( 1 )
+(1 )+ c
2=2 + 1 + c
−1 =c 2 y =2 x + x −1
y − y 1= m( x − x1 ) y −2 =5 ( x − 1 )
y =5 x −5 + 2 y =5 x −3 → Ec(ación de%a recta
$ " >" 0.5 0 >".5 >'
% ' 0 0 >" ' 5
35.
Haa a/ a =6n =6n8i;n 8i;n 862 862a an7 an7n n in in nd ndin in
a,/ d 2 862a 7/a8a a1a ,/ 6n, 0W ! DATOS m =3 x 2+ 6 x −2 x =0 y =6 2
m =3 ( 0) + 6 ( 0 )−2 m =−2
y =∫ ( 3 x 2 + 6 x −2 ) dx
∫
∫
2
∫ dx
y =3 x dx + 6 x dx − 2
y =3
x
3
3
x
2
+ 6 − 2 x + c 2
y = x 3 + 3 x2 −2 x + c x =0 y =6 3
6 =(0 )
+ 3 (0 )2− 2 (0 )+ c
2
3 x
+ 6 x −2 Pa/a 8ada
6 =c 3 2 y = x + 3 x −2 x + 6
y − y 1= m( x − x1 ) y −6 =−2 ( x −0 ) y =−2 x − 6 → Ec(ac Ec(ación ión de%are de %arecta cta
$ " >" 0.5 0 >".5 >'
% ' 0 0 >" ' 5
3!. Haa/ Haa/ a =6n8i;n =6n8i;n 862a 862a an7n an7n in in ndin ndin d 2 862a 7/a8a a1a ,/ 6n, "W 3 DATOS 2
m = x 3− 2 + 2 x
2
3 x − 2 + 2 Pa/a 8ada a,/ x
x =1 y =3 2
m =( 1)3−
( 1)2
+2
m =1
(
)
2
y =∫ x 3− 2 + 2 dx x y =∫ x 3 dx −2∫ x −2 dx −2∫ dx x
−1
4
y = − 2
x
4
−1
−2 x +c
1 4
− y = x 4 + 2 x 1−2 x + c
x =1 y =3
3=
1 ( 1) 4+ 2 ( 1)−1− 2 ( 1)+ c 4
3= 0.25 + 2− 2 + c 2.75= c
1 4
4
−1
y = x + 2 x − 2 x +
11 4
y − y 1= m( x − x1 ) y −3 =1 ( x − 1 ) y = x + 2 → Ec(aci Ec(aciónde ónde %arecta %arecta
$ " >" 0.5 >".5 >'
% ".5 >'.5 3.'! >0.*" '.'5
3!. 3!. Ha Haa/ a/ a =6n8 =6n8i; i;n n 862a 862a 7/a 7/a8a 8a in in 6n -Qni -Qni-, -, /a /ai i, , 86an 86and, d, F" F" 2 6n -i-, /ai, 86and, F4 Da,1
{
re%ati) x = 1 minim re%ati) maxim re%ati) re%ati) 4 maxim
dy =( x −1 )( x − 4 )
∫
y = ( x x −1 ) ( x x −4 ) dx
∫ ( x x − 4 x + x + 4 ) dx 2
y =
∫
∫
2
y = x dx − 5 xdx + 4
x
y =
3
− 5 x 3 2
2
+ 4 x + c
∫ dx
CORRECCIONES DE LAS PRUEBAS
CORRECCIÓNES DE LAS PRUEBAS PRIMER HEMISEMESTRE CORRECCIÓN DE LA PRUE#A N$ 1 C<!%%$&# >! 0 !0 !0 FILA A C0%! @ 0"! 1. lim lim x →∞
(
(
3
−2 √ x −3
4 x
)
)
3 −2 4 ( ∞ ) + 2 ∞ = = = I lim x → ∞ √ x x −3 √ ∞ +3 ∞
4 x
3
3 x + 3 −2 √ x + 3 2 ( 2 x −1 ) √ x = · x − 9 x −3 √ x + 3 √ x
4 x
3
lim 2 ( 2 x x →∞
3
−1 ) √ x + 3
x −9
G"%<( $ " ' 3 4 5
% >" >"*.9 >*3.5 >'54 >!".*
=
(
−1 ) √ ( ∞ )+ 3 ∞ = = I ∞ ( ∞ ) −9
2 2 (∞)
3
C0%! @ 0"! 2
− x −10 x → 2 x − 15 x − 14
2.lim
3 x 2
29−10
¿
2
¿ ¿ ¿ 2 −15 ( 2 )−14 ¿ 2 4 ( 2 ) −¿ 2 3 lim 3 x + x x →2 =¿ 2 x −15 x −14
$ " ' 3 4 5 ! *
% 0.' 0 >0.' >0.* >0.9 >".3 >'.4
P#< C$%< *2K +
3. Ra8i;n 6<1d a/a1i, : a/a 6na /a8i;n a/i86a/ 6<1d Z a/a1i, 1 d/-ina K6 86and, a dn1idad d 6<1d N-/, d 6<1d1 ,/ 6nidad d /a 1 $ n,n81 n-/, d a/1i,1 a , a/7, d /i,d, 1
(
y =23 1−
1 1+ 2 x
)
1i a dn1idad d
6<1d a6-na/a indnida-n A K6< a,/ 1 a/,i-a/Qa %
(
y =23 1−
1 1+ 2 x
(
lim ¿ 23 1 − x → ∞
(
lim ¿ 23 1 − x → ∞
)
1 1 +2 ( ∞)
1 1+∞
)
)
( )
lim ¿ 23 1 − x → ∞
1
∞
lim ¿ 23 ( 1−0 ) x → ∞
lim ¿ 23 x → ∞
C,//88i;n d a /6a #ILA B
C0%! @ 0"!
2 x + 1 √ 1. lim x →∞ x + 1
2 2 x + 1 √ ( ∞ ) + 1 ∞ √ = = = I lim x → ∞ x + 1 ( ∞ ) +1 ∞
1 x+1 √ x 2 + 1 · √ x 2−1 = = 2 x + 1 √ x −1 x + 1 ( √ x2 −1 ) √ x 2−1 lim 1 x→ ∞
=
1
= 1 =0
√ x −1 √ ( ∞ ) −1 ∞ 2
2
G"%<( $ " ' 3 4 5 !
2.-
% 0.&0 0.&4 0.&9 0.*' 0.*4 0.*!
lim x →− 3
2
+ 14 x −3 2 x + 3 x
5 x
x + 14 x −3 5 (−3) + 14 (− 3 )− 3 45− 42−3 0 = = = = I lim 2 2 − 9 9 0 x→ 1 (−3 ) + ( 3 ) (−3) x + 3 x 2
2
x =
x =
( 5 x + 15)( 5 x −1 ) 5
5 ( x + 3)( 5 x − 1 ) 5
x =( x + 3 )( 5 x −1 )
+ 14 x −3 ( x + 3 )( 5 x −1 ) ( 5 x −1 ) = = 2 x ( x + 3) x x + 3 x
5 x
2
( 5 x −1) 58−39 −1 16 = = −3 3 x x → − 3 lim
P#< C$%< P *-3K 5.33+
G"%<(
$ " ' 3 4 5 >" >' >3 >4 >5 > 0.4 > 0.'
% 4 4.5 4.! 4.& 4.* ! 5.5 5.3 3 5.' 5.' & "0
3.> Ra8i;n /1a Z d/dad,/: a/a 6na /a8i;n a/i86a/ /1a Z d/dad,/ 1 d/-ina K6 n-/, % d /1a1 8,n16-ida ,/ 6n d/dad,/ a , a/7, d 6n /i,d, =6 6na =6n8i;n d a dn1idad d /1a1 n-/, d /1a1 ,/ 6nidad d /a 16,n7a K6
(
y = f x =
10 x 1 + 0.1 x
)
1i a
dn1idad d /1a1 a6-na/a indnida-n A K6< a,/ 1 a/,i-a/Qa %
(
10 x 1 + 0.1 x
¿
(+
)
¿
∞ ∞
y = f ( x )=
lim x → ∞
lim x → ∞
10 ( ∞ )
1 0.1 ( ∞ )
)
lim ¿ I x → ∞
C,//88i;n d a /6a #ILA C
C0%! @ 0"!
3
".>
x −3 x + 2 lim 4 x → ∞ x − 4 x + 3 3
x −3 x + 2 ∞ = = I lim 4 ∞ x → ∞ x − 4 x + 3 3
x 3 x 2 1 3 2 − 4 + 4 − 3+ 4 3 4 x x −3 x + 2 x x x = 4 = x x 4 x −4 x + 3 x 4 x 3 1 − 4 + 3 − + 3 4 x x x 4 x 4 x 4
1 lim n →∞
¿
−
x x
3
+
x 3
x
x
% 0.3 0.' 0." 0.! 0.! 0.5 0 "
+ 3
1
2
4
1−
G"%<( $ ' 3 5 0.5 > 0.5 >" >' >3
3
4
4
=
∞
−
1−
3 3
(∞) 4
+
2
(∞ )4 3
+ ( ∞)3 ( ∞ )4
=
0 −0 + 0 0 = =0 1 −0 + 0 1
2
− x −10 x → 2 x − 5 x − 14
2.lim
3 x 2
29−10
¿
2
¿ ¿ ¿ 2 −5 ( 2 )−14 ¿ 2 4 ( 2 ) −¿ 2 3 lim 3 x + x x → 2 =¿ 2 x −5 x −14
$ " ' 3 4 5 ! >" >' >3 >4 >5 >5
% 0.4 0 0.' 4 0." 5 0.0 0.0 & 0.& 5 0.3 0 0.' 4 0." ! 0." " 0.0 *
G"%<(
3.> C,-, ,1 aan81 d a 8n,,7Qa /16an n a /,d688i;n d 8a86ad,/a1 8,-a8a1 8ada -1 ,d/,1a1 a86a-n 16 /8i, n -/8ad, 1 d18ndind,. S6,n7a K6 $ -11 a a/i/ d a,/a /8i, d 8i/, -,d, 1/ P d;a/1 ,/ 6nidad
30
/ ( x )= 40 + x + 1
a C6 1/ /8i, dn/, d 5 -11 a a/i/ d a,/a 30
/ ( x )= 40 + x + 1 / ( x )= 40 +
30 5 +1
/ ( x )= 40 +
30 6
/ ( x ) = 45
C6n, di1-in6i/ /8i, d6/an K6in, -1 / ( x )= 40 + / ( x )= 45
30 5 +1
/ ( x ) = 40 +
30 6+1
/ ( x ) = 44.29
8 C6an, /8i, 1/ 43.00
/ ( x ) = 40 +
30 9+1
/ ( x )= 40 +
30 10
/ ( x )= 40 + 3 / ( x ) = 43
S<%$&#( E# ! #<!#< ?!: ! !%$< :! >! `'3= d 6 168d 8,n /8i, n a/7, a, 86and, $ lim ¿ 40 + x → ∞
¿ 40 +
30
∞ +1
30
∞
¿ 40 + 0 ¿0
CORRECCIÓN DE LA PRUE#A N$ % C<!%%$&# >! 0 "0 A
1.− y =
2 x + 3 4 x −1
[
(
dy d 2 x + 3 = dx dx 4 x −1
dy = dx
y =
)
( 4 x + 1 ) d ( 2 x + 3 )−( 2 x + 3 ) d ( 4 x − 1 ) dx
dx
( 4 x −1 )2
2 x + 3 4 x + 1
dy = dx
(
( 4 x + 1 ) d 2 x + d dx
dx
3
)
−( 2 x + 3 )
(
d d 4 x − 1 dx dx
)
( 4 x −1 )2
dy ( 4 x + 1 ) ( 2 )− (2 x + 3 ) ( 4 ) = dx ( 4 x −1 )2 dy 8 x −2 −8 x −12 = dx ( 4 x −1 )2
−14 dy = dx ( 4 x −1 )2
[
2.− y =( 2 x + 3 ) 2 ( x
4
−5 x 2+ 4 ) ] " ( 0.24 )
y =( 2 x + 3 ) ( 2 x 4 −10 x 2 + 8 ) dy d = [ ( 2 x + 3 ) ( 2 x 4−10 x 2+ 8 ) ] dx x dy d d =( 2 x + 3 ) ( 2 x 4 −10 x 2+ 8 ) + ( 2 x 4−10 x 2+ 8 ) ( 2 x + 3 ) dx x dx dy =( 2 x + 3 ) ( 8 x 3−20 x )+ ( 2 x 4 −10 x 2+ 8 ) ( 2 ) dx dy =16 x 4 + 24 x 3−40 x 2−60 x +4 x 4 −20 x 2+ 16 dx dy =20 x 4 + 24 x 3−60 x2 −60 x + 16 dx
dy =20 ( 0 )4 + 24 ( 0)3 −60 (0 )2−60 ( 0 )+16 dx → 0
dy =16 dx → 0
m =1 6
y − y 1= m ( x − x 1 ) y −24 =16 ( x −0 )
y −24 =16 x y =16 x + 2 4
G"%<( $ 0 " ' 3 4 5 >" >' >3 >4 >5
$ % 0 '4 > 0 ".5
% >'9 0 0 &'0 39!0 "3"0 4 0 0 >'40 > "*00 > &05!
3. y =ln
√
3
x −1 3
x + 1
( ) 3
x − 1 y = ln 3 x + 1
1 3
dy 1 = ln ( x 3−1 )− ln x3 + 1 dx 3
[
]
[
dy 1 d 3 d 3 1 1 = − − ( ) ( x +1 ) x 1 3 dx 3 x 3−1 dx dx x + 1
[ (
)
(
]
3 1 d dy 1 1 d 3 d d = x − 1 − 3 x + 1 3 dx 3 x −1 dx dx dx x + 1 dx
[
dy 1 1 = ( 3 x2 ) − 31 ( 3 x 2 ) 3 dx 3 x −1 x + 1
[
dy 1 3 x ( x + 1 ) − 3 x ( x −1 ) = dx 3 ( x 3− 1 )( x 3 + 1 ) 2
3
2
3
[
dy 1 3 x 5 + 3 x2 −3 x 5 + 3 x 2 = dx 3 ( x 3−1 ) ( x 3 + 1 )
[
2
dy 1 6 x = dx 3 ( x3 − 1 )( x3 + 1 )
[
2
dy = 3 2 x 3 dx ( x −1 )( x + 1 ) 4
]
]
− x
4. y =3 x e
dy d = ( 3 x 4 e− x ) dx dx − x dy d 4 d =3 x e + e− x 3 x 4 dx dx dx
12 x (¿¿ 3 )
dy d =3 x 4 e− x ( − x ) + e− x ¿ dx dx
]
]
]
)]
12 x ( ¿¿ 3 )
dy =3 x 4 e− x (−1 )+ e− x ¿ dx
dy =−3 x 4 e− x + 12 x 3 e− x dx dy =3 x 3 e− x (− x + 4 ) dx
C,//88i;n d a a B
1.− y =
a − x a + x
( )
dy d a− x = dx dx a + x
2
a + x ¿
dy = dx
¿ ( a + x ) d ( a − x ) −( a− x ) d ( a + x ) dx
dx
¿ 2
a + x ¿ dy = dx
¿
( dx
)
( a + x ) d a− d x −( a − x ) d dx
¿ 2
a + x ¿
¿ ( ) + (− a x 1 )−( a − x ) ( 1) dy = ¿ dx 2
a + x ¿
¿
dy −a − x −a + x = ¿ dx 2
a + x ¿
¿
dy −2 a = ¿ dx
(
d d a + x dx dx dx
)
x +1 "( 2, −1.5 ) 2 ( − ) x x 4
2.− y =
x +1 y = 3 x −4 x2
(
dy d x +1 = 3 dx dx x −4 x 2
)
x x 3 x − 4 x 2 ¿2
¿
d d ( ¿ ¿ 3 −4 x 2 ) ( ¿ ¿ 3− 4 x ) ( x x + 1 )−( x x + 1 ) ¿ dx dx 2
¿
dy =¿ dx x x 3 x − 4 x 2 ¿2
¿
d d ( ¿ ¿ 3 −4 x 2 ) ( ¿ ¿ 3− 4 x ) ( x x + 1 )−( x x + 1 ) ¿ dx dx 2
¿ dy =¿ dx
x d x dx x 3− 4 x 2 ¿2
¿
2
( ¿ ¿ 3− 4 x 2)
(
)
d d x + 1 −( x x + 1 ) dx dx
¿ dy =¿ dx
d ( ¿¿ 3 −4 x ) dx
¿
x 3 x 3 2 2 x − 4 x ¿
¿
( ¿ ¿ 3− 4 x 2)( 1 )−( x x + 1 ) ¿ dy =¿
( ¿¿ 2−8 x ) ¿
dx
x 3 x 3 2 2 x − 4 x ¿
¿
(¿ ¿ 3− 8 x 2 +3 x 2−8 x ) (¿ ¿ 3− 4 x )− ¿ ¿ dy =¿ 2
dx
3
2
2
x −4 x ¿ 3
¿
2
3
2
2
dy x − 4 x − 3 x + 8 x −3 x + 8 x = ¿ dx 3 2 2 x −4 x ¿
¿
dy −2 x 3 + x 2+ 8 x = ¿ dx 2
¿ ¿ ¿ ¿
3
2
−2 (2 ) +( 2 ) + 8 (2 ) dy = ¿ dx → 2 dy 1 = dx → 2 16
m=
1 16
y − y =m ( x − x )
y −1.5 =
1 ( x − 2) 16
16 ( y −1.5)= x −2
16 y + 24 = x −2
16 y = x −26
y =
x 16
−
13 8
G"%<( $ " ' 3 5 >" >' >3 >5
$ 0
% >0.!& >0.3* >0.44 >0.'4 0 0.04 0.03 >0.4
3.− y =ln
√ 4
1 + x
2
1− x
2
% >"
( ) 2
1 1+ x y = ln 2 4 1 − x
( 1 + x 2 )−¿ ln ( 1− x 2) ln ¿ dy 1 = ¿ dx 4
[
dy 1 d d 1 1 2 ( ) ( 1− x2 ) x = + − 1 2 2 dx 4 ( 1 + x ) dx (1 − x ) dx
[
1 1 dy 1 ( ) (−2 x ) = − 2 x dx 4 ( 1 + x 2 ) ( 1− x 2 )
[
dy 1 2 x 2 x = + dx 4 ( 1 + x 2 ) ( 1− x 2)
]
[
2 2 dy 1 2 x ( 1 − x ) + 2 x ( 1+ x ) = dx 4 ( 1 + x 2 ) ( 1− x 2 )
(
3
3
dy 1 2 x −2 x + 2 x + 2 x = 2 2 4 dx 4 1 − x + x − x
( )
dy 1 4 x = dx 4 1− x 4 dy x = dx 1− x 4
x
4. − y =2 x
2
ln y = x ln 2 + 2 ln x
2
d 2 ln x dx d d ln y = x ¿ dx dx ln ¿ +
)
]
]
]
2 ln ¿ + 0 1 y : =1 ¿
y
y : = ln 2 y x
y : = ln22
CORRECCIÓN DE LA MICRO PRUE#A
c´ =
c´ =
c´ =
324
√ 2 + 35 2
( √ +
324
2
2
5
35
324 2
√ 22 + 35
( 2 2+ 35 ) :
c´ =
c´ =
+ + 2
+
324 2
c´ =
:
5 19 2 18
+ +
1 2
52
2
)
19 2 18
+
19 2 18
+ 5 + 19 2 18
19 d 324 2 + + 5 2 1 18 d2 2 ( 2 + 35 ) 2
d d2
324 2
( 2 2+ 35 )
1 2
+
d d 19 2 5 + d2 d2 18
:
(
c´ =
:
( (
c´ =
1 2
1
( 2 +35 ) d 324 2 −( 324 2 ) d ( 2 2+ 35 ) 2 d2 d2 2
[( 2 + ) ] 2
1 2
:
( (
c´ =
:
c´ =
:
(
c´ =
:
c´ =
19 18
−1
()
2
2 + 35
324 ( 2
2
−1
1 2
(
+ 35 ) −162 2 ( 2 + 35 ) 2 d 2 2 + d 35 2
d2
d2
2
2 + 35
324 ( 2
2
324 ( 2
2
)
−1
1 2
+ 35 ) −162 2 ( 2 +35 ) 2 ( 2 2 ) 19 + 18 2 2+ 35 2
−1
1 2
)
+ 35 ) −324 2 ( 2 + 35 ) 2 19 + 18 22 + 35
[
324 ( 17 )
630 19 + 324 18
c´ =3 :
)
+0 +
( 2 +35 ) (324 )−( 324 2 ) 1 ( 2 2+ 35 ) 2 d 22 +35 2 d2 2
c´ = :
35
1 2 2
2
2
1 2
2
+ 35 ] −324 ( 17 ) [ ( 17 ) + 35 ] [ ( 17 )2 +35 ] 2
2
−1 2
)
+
19 18
)
)
+
)+
19 18
19 18
CORRECCIÓNES DE LAS PRUEBAS SEUNDO HEMISEMESTRE CORRECCIÓN DE LA PRUE#A N$ 1 ".> En8,n/a/ -i-,1 , -Qni-,1 d a =6n8i;n 3 2 y = x − x −6
2 : y =3 x −2 x
: :
y = 6 x − 2 :
y =0 0 =3 x
2
−2 x
0 = x ( 3 x −2 )
x 1=0 x 2=
2 3
y = x 0 =6 ( 0 )−2 =−2 : :
:: C
2 3
y : : = x =6
( )− = − = 2 3
2 4 2 2
:: C 0 ∈ m
G"%<( $ " ' 3 4 5 >" >' >3 >4 >5
% >! >' "' 4' 94 >* >"* >4' >*! >"5!
'.> U1d 1 a 8a/7, d 6n 1/i8i, d /8,//id,1 6/Q1i8,1 K6 ,=/8 a1 1i76in1 a/i=a1: '00 ,/ /1,na 1i an 50 /1,na1 a /8,//id, n-/, -Qni-, a/a 8,n/aa/ 161 1/i8i,1. P,/ 8ada /1,na adi8i,na a1a 6n -i-, d *0 /1,na1 n ,a ,/ /1,na 1 /d68 '. E 8,1, d /8,//id, 1 d !00 8,1, , -1 3' ,/ 8ada /1,na. C6na1 /1,na1 dn 8,n/aa/ 1/i8i, a/a -ai-ia/ a 6iidad d a -/1a DATOS . Cst =200 /
. /r cada -ersna adicina% se red(ce 2 -
. Cst Fi= =600 .mas 32 Ati%idad =@
VARIABLE x = N 6de -ersnas adicina%es
Ati%idad = I −C cst= cst fi=+ cstadicina% A = N de -ernas ∗ -reci− csts A =200∗50 −c A = (200 −2 x ) ( 50 + x )−( 600 + 32 x ) 2 A =10000 + 200 x −100 x −2 x −600 −32 x
2
A =−2 x + 68 x − 9400
PRIMERA DERIVADA A =−4 x + 68
SEUNDA DERIVADA A =−4 cm( < 0 ( ∈ 1
PRIMERA DERIVADA IUALADA A CERO ( = 0 0 =−4 x + 68
x =
68 4
x =17 50 + 17 =67
CONCLUSIÓN( 6/ !:<#0: >!!# %<#00 ! :!$%$< 00 ?0$?$90 0 $$>0> >! 0 !?!:0
3. S /K6i/ 8,n1/6i/ 6na 8aa d -ad/a 8,n aa d a1 86ad/ada 8,n 300 d a -an/a K6 ,6-n 1a -i-,. Si 8,1, d -a/ia d ,1 ad,1 1 d " ,/ - ' 2 8,1, d -a/ia ,/ - ' d a aa 2 d a a1 1 d, d 8,1, d -a/ia d ,1 ad,1. C61 1/n a1 di-n1i,n1 ;i-a1 d a 8aa
DATOS( +ase y tapa cadradas. Materiales lados
1
. m2 , "ase y tapa
2
. m
2
-osto de la caa / 011
2olmen máximo 2olmen 3 área de la "ase x altra 2
¿
% +! % ( % *! )∧¿ 7 =¿ Escri#a a2(G %a ec(ación+
-osto de la caa 3 costo lados 4 costo "ase y tapa -osto 3 5rea de los lados 6 p. nit. 4 5rea de "ase y tapa 6 p. nit . 2
300= 4 ! + % .1 + 2 % .2
. = mm
. m
2
+ m2
300= 4 !% + 4 % 300 − 4 % != 4%
!=
75 −%
. m
2
2
2
2
% 2
!=
75 −(5 ) 5
! =10 m
VARIABLE 7 =% 2 + !
7 =%
2
(
75−%
%
2
)
7 =75 %−%3
PRIMERA DERIVADA :
7 =75− 3 %
2
SEUNDA DERIVADA : : :: 7 =−6 %cm7 > 0 →7 ϵ 1
PRIMERA DERIVADA IUALAR A CERO :
7 =0 =75−3 % 3%
2
=75
√ %2 =
√
% =5 m
75 3
2
C<#%:$&#( L0: >$?!#:$<#!: &$?0: 00 0 %00 :<#( L0><:( 1 ? >! 00. ,0:! @ 00( 5 ? • •
'.> Ra8i,n a1 7/8a1 8,n a 8,//1,ndin 86a8i;n.
y =
( e x −e− x )
y =3
2 3 2 y = x + x − x
1
y = x + x
a ¿ y =
( e x −e− x ) 2 1
# ¿ y = x + x c ¿ y =3 3 2 d ¿ y = x + x − x
CORRECCIÓN #ILA B ". En8,n/a/ -i-,1 2 -Qni-,1 d a =6n8i;n 3 2 y = x −3 x −8
3
2
y = x −3 x −8
y 0 =3 x 2−6 x y 0 0 =6 x −6 y 0 =0 0 =3 x
2
−6 x
0 = x ( 3 x −6 )
x 1=0 x 2=
6 3
x 2=2
y 0 0 =6 ( 0 )−6 =−6 cm y 0 0 < 0 ∈ 1
y 0 0 =6 ( 2 )− 6=6 cm y 0 0 < 0 ∈ m
G0"%< $ " ' 3 4 >" >' >3 >'.5
% >"0 >"' >* * >"' >'* >!' >4'.3&
'. Un //n, d 1i-/a /8an76a/ n 6na 7/ana 1a/ i-iad, n 6n, d 161 ad,1 ,/ 6n /i, 2 ,/ ,1 ,/,1 3 ad,1 ,/ 6na 8/8a 8/i8ada 8,n 6n 1,, aa-/ 2 8,n 6na dii1i;n d a -i1-a 8/8a a/aa a ad, 8,/, d //n, d a -an/a K6 K6dan ' ,1 i76a1 . Si 1 86na 8,n *00- d aa-/ C6 1 -a2,/ /a K6 6d ,86a/ 8ada a/8a 2 861 1,n a1 di-n1i,n1
DATOS: 7erreno rectanglar % lotes igales (11m de alam"re 5rea3 largo 6 anc8o 8^ 0 800 %= a
LONGITUD DEL CERRAMIENTO= ANCHOLARGOANCHOANCHO 5=3 a + %
5=3 a +
800
a
PRIMERA DERIVADA −1 5: =3 a + 800 a
5: =3− 800 a−2
SEUNDA DERIVADA − : : 5 =−800 a 3 : : 5 =
−800 a
3
PRIMERA DERIVADA IUALAR A CERO Le^ −2 0 =3−800 a 0 =3 −
800 2
a 2
a=
800 2
a
=3 800 3
√ a2=
√
800 3
a =± 16.32 %=
800
%=
800 16.32
a
% = 49.01
R!0%$<#! 0 "%0 %<# 0 %<!:<#>$!#! !%0%$&#
•
3 2 y =3 x − x + 1
3.
1
•
•
y= x
'.
y = x
5.
− x
•
2.
y =−e
6. /.
a ¿ y =−e−
8.
# ¿ y = x
x
3
2
.
c ¿ y =3 x − x + 1
1.
d ¿ y = x
1
11. 3. S K6i/ 8,n1/6i/ 6na 8aa d -ad/a 1in aa d a1 /8an76a/ K6 d n/ 6n ad, d, d a di-n1i;n d ,/, d a -an/a K6 ,6-n 1a -i-,. Si 8,1, d -a/ia d ,1 ad,1 1 d 3 d;a/1 ,/ -' 2 8,1, d -a/ia ,/ -a' d a a1 1 d 4 d;a/1 ,/ -'. C61 1/n a1 di-n1i,n1 ;i-a1 d a 8aa a/a 8,n1/6i/a 8,n n, -1 d &5 USD "'.
Da,1 +ase rectanglar 9ado do"le del otro -osto
1!" 1#" 1$" 1%"
9ado3 0
. m2 . m2
1&"
+ase3 :
1'"
-osto3 / ;<,11
$=. 2olmen3 máximo 0"
21.
VARIABLE
{
! : %ad a : #ase
22. 7%(men= >reade #ase x a%t(ra 23. 7 = a .2 a + ! 2 2'. 7 =2 a + ! ⟹ 7 =f ( % * ! )
%<. cst ca=a=cst %ads + cst #ase 2 %>. 75=( 2 a! + 2.2 a! )3 + 2 a .4 2
%;.
75=18 a! + 8 a
%(.
75 −8 a != 18 a
2. 3. 31. 32. 33. 34. 35.
2
2
7 =2 a + ! 75 − 8 a 7 =2 a + 18 a 2
2
25
8
3
36. 7 = 3 a− 9 a 3/. 3*.
PRIMERA DERIVADA 25 8 2 = − a 7 0 3. 3 3 40. '1. '2.
SEUNDA DERIVADA −8
'3.V : : = 3 a 44. PRIMERA DERIVADA IUALAR A CERO '5. 7 0 =0
'6.
0=
25 8 2 − a 3 3 25
2 a '/. = 8 cm
'8. a =1,77 : : :: '. Cm7 < 0 → 7 ∈ 1 5. 51. 52. ". '. % 53. 5'. $ 55. 3. 11.&. 56. " 44 5/. 4. 12.9. 58. ' 55 5. 5. 13." 6. 3 1'.> 61. 62. !. '3 63. 4 15.> &. &. >" 44 *. 16.> >' 9. 9. 55 >3 1/.>" "0. 18.0 0
6'.
('.
CORRECCIÓN DE LA PRUE#A N$ %
!!. !&.
#ILA A
∫
1.
!*.
3 $ + 6
√ 2 $ + 8 $ + 3 2
2 6. (=2 $ + 8 $ + 3
/.
d( = 4 $ + 8 d$
/1.
d( =d$ 4 ( $ + 2 )
/2.
y=
∫ 3 ( $√ +( 2 ) 4 ( $d(+2 ) ∫(
/3.
y =3
/'.
3 y= 4
/5.
−1
∫( 1 2
3 ( 4 1 2
y = +
2
4 −1 2
d(
+c
( ) +c 1
/6.
d(
3 y = .2 ( 2 4
d$
[
1
]+
//.
3 2 2 y = ( 2 $ + 8 $ + 3 ) 2
/8.
3 1 y 0 = 2 2
/.
3 y 0 = ( 2 $ 2+ 8 $ + 3 ) 4
c
( )[( $ + $ + ) 2
2
8
3
[
1 −1 2
−1 2
]
( 4 $ + 8 ) + 0
4 ( $ + 2 )
]
−1
8. y 0 =3 ( 2 $ + 8 $ + 3 ) 2 ( $ + 2 ) 2
81.
82.
3 ( $ + 2 )
y 0 =
( 2 $ + 8 $ + 3 ) 2
y 0 =∫
1 2
3 $ + 6
√ 2 $ 2+ 8 $ + 3
83.
∫ ( $ + 2) d(
*4.
85.
2
2.
∫ ( $ +4 $+ 4 ) d( 2
y=
2 86. y =∫ $ d$ + 4 ∫ $d$ + 4∫ d$
$
2+1
$
1+ 1
8/. y = 2 + 1 + 4 1+ 1 + 4 $ + c $
3
$
3
$
2
88. y = 3 + 4 2 + 4 $ +c 8.
y = + 2 $ 2+ 4 $ + c
.
y 0 = ( 3 ) $
1.
y 0 = $ + 4 $ + 4
9'.
3.
3
1 3
3− 1
+ 2 ( 2 ) $2−1 + 4 $ + 0
2
2
y 0 =( $ + 2 )
2
'.
∫ √ 10 −3 - d-
3+
1 3
5. (=10 −3 6.
d( =−3 d-
/. d( =(−3 ) d8.
−d( 3
=d -
2
.
∫ √ (− d(3
y=
1 3
2
1.
y =∫
−1
1 3
3
√ ( d(
|
3 2 2
11.
1 ( 3 3 2
y ¿− +
1 3
|
3 2
12.
13.
1'.
1 2 y ¿− + ( 2 3 3
y=
y=
15.
y=
16.
y=
"0&.
y=
−2 ( 9
−2 9
−2 9
−2 9
10 −3 - )
|
3 2 2 1 3
( 10−3 (2))
3 2
( 4) −
( 8) + 2 3
[ [
( )] ]
2 1 − − 10− 3 9 3
( − )(
3 2
−10 9
1 3
2 27 ) 9
3 2
18. 1. ""0.
4. Si y 0 =( 3 x + 1 ) √ 3 x
2
+ 2 x + 5 " y( 1)=@
111. 112.
(=3 x 2 + 2 x + 5
113.
d( =6 x + 2 dx
11'.
d( =2 ( 3 x + 1 ) dx
115.
d( =dx 2 ( 3 x + 1 )
""!.
y = ( 3 x + 1 ) √ 3 x + 2 x + 5 dx
11/.
y = ( 3 x + 1 ) √ (
""*.
y =∫
""9.
1 y= 2
12.
121.
∫
2
∫
y=
y=
d( 2 ( 3 x + 1 )
1 √ ( d( 2
∫(
1 2
d(
1 +1 2
1 ( +c 2 1 +1 2 3 2
1( 2 3 2
+c
3
122.
1 y= (2 + c 3
"'3.
1 y = ( 3 x 2 + 2 x + 5 ) 2 + c 3
"'4.
1 y = ( 3 ( 1)2 + 2 ( 1 )+ 5 ) 2 + c 3
3
3
3
125.
1 y = ( 10 ) 2 + c 3
"'!.
y =10.54 + c
"'&. "'*. "'9. "30. "3". "3'. "33. "34. "35. "3!. "3&. "3*. "39. "40. "4".
#ia B
∫ 9 ( $ +3 $ +5 ) ( 2 $ + 3 ) d$ 8
2
1'2.
1+
1'3.
(= $ 2 + 3 $ + 5
1''.
d( =2 $ + 3 d$
1'5. 1'6.
d( =d$ 2 $ + 3
1'/.
∫
8
1'8.
y = 9 ( ( ) ( 2 $ + 3 )
1'. 15. 151.
y =∫ 9 ( ( ) d(
152.
( ( )8+1 + c y =9 8 +1
153.
8
d( 2 $ + 3
15'. 155. 156. 15/. 158. 15. 16. 161. "!'.
y =9
( ( )9 9
+c
9 y =( ( ) + c
9
y =( $ + 3 $ + 5 ) + c 2
2 : y =( $ + 3 $ + 5 )
9− 1
( 2 $ +3 )
8
y : =( $ 2+ 3 $ + 5 ) ( 2 $ + 3 )
"!3.
∫ ( 3 ( +2 ) d( 3
"!4.
2.
165.
x =3 ( + 2
166.
dx =3 d(
16/.
dx 3
= d(
∫
168.
y = ( x )
16.
y=
1 3
3
dx 3
∫ ( x ) dx 3
3 +1
1/.
1 ( x ) + c y= + 3 3+ 1
1/1.
y=
1 ( x )4 +c 12
1/2.
y=
1 ( 3 ( + 2 )4 + c 12
1/3.
y=
1/'.
y = ( 3 ( + 2 )3
1/5.
:
:
1 ( 4 ) ( 3 ( + 2 )4 −1 .3 +0 12
1
∫ 2 √ 2 +3 d2 2
"&!.
3.
1//.
(= 2 + 3
1/8.
d( =2 2 d2
1/.
d( =d2 22
−1
2
18. 1
d( 22
181.
y =∫ 2 √ (
182.
1 y= √ ( d( 2 −1
183.
1 y= ( 2 d( 2 −1
−1
1
∫ 1
1
∫
|
1 1 +1 2
18'.
185.
1 ( 2 1 +1 2 −1
y= +
1 2
( 2 + 3 ) 2
y= +
3 2
|
3 1 2
−1
|
3 1
186.
1 2 2 y = + ( 2 + 3 ) −1 3
[
18/.
y=
188.
y= −
18.
y =0
1. 11.
3
1 (1 )2+ 3 ) 2 ( 3
8 3
8 3
][ −
3
1 (−1 )2+ 3 ) 2 ( 3
]
1
12.
− x x e −e
∫
4.
0
1
13.
dx
2 x
e
∫ 2 −
y=
− x
e
2
0
1
1 2
dx
1 2
1'.
− x x y =∫ e − e dx
15.
1 y= 2
16.
1 1 y = e x + e− x 2 2 0
0
1
∫ 0
1 e dx − 2 x
1
∫ e− dx x
0
|
1
[
1 2
[
1 1 ( 2.72)+ ( 0.37 ) 2 2
1 2
][
1 0 1 −0 e + e 2 2
1/.
y = e1 + e−1 −
18.
y=
1.
y =[ 1.36 + 0.19 ] −[ 1 ]
2.
y =0.55
21. 22. 23. 2'. 25. 26. 2/. 28. 2. 21. 211. 212. 213.
][ −
]
1 1 ( 1 )+ ( 1) 2 2
]
21'. 215. 216.
E$P OSI CIÓ NES
'"&. '"*. 21.
''0. ''". 222. 223. 22'.
''5. 226. 22/. 228. 22.
'30.
E$POSICIÓN N+ "
'3".
TEMA: MATEM,TICAS EN LA VIDA COTIDIANA
232.
C6n, =aa a/a
•
•
•
•
•
;! !?$!%! : :!$! >! >$<: H0<$<:. ;! :!0 0 <0 >! 00 0 0. ;! :!0 ! %?!07<: >! 0$!# $?<0#! *H0?$$0= 0?$<= ! :@<+. M!:!: 00 N0$>0> < ! %?!07<:. ;! !#0 ! 000 0 $990 >! <#<...
233. •
•
O/7ania/ 6na 1a:
d%#<: $#$0><: !#!?<:] d%#0: %<:0: #!%!:$0?<: < $#$0><]
•
O0#$90 %0#$>0>!: >! %0: < >! !<: *# >! #$7<: < !<+.
23'. '35.
236. •
En a 8,8ina:
0%! # 0:! < #<: ?0%0<#!:= :<#! ?!>$ %00%$>0>= !:<= $!?<...$#%:< !:$?0 ! !%$< >!0 00 0!#>! ?0!?$%0: !# 0 %<%$#0.
23/. '3*.
MATEM(TICAS EN OOLE MAIL
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2'6. 2'/. 2'8.
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APLICACIÓN DE LAS MATEM(TICAS EN LA ECONOMÍA.
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253.
INVESTIACIÓN DE ASESINATOS
25'.
dE# 4 ?
255.
E ?!0<$:?< ?0#< 0:!0
! ?0#!#$?$!#< >! 0 !?!00 >!
%!< >! #0 !:<#0 $0 %!%0 >! <: 36.5 ℃ = !< 0 ?<$ !:! %0< >!0 >! <>%$:!= < < 0#< ! %!< !?$!90 0 !#H$0:!. L<: >!!%$!: !>!# >!!?$#0 ! ?! 0:!:$#0< %<# 0 :$$!#! H<?0
256. 25/. '5*.
T =T 4ire +
[
( T c (er- −T aire ) t
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CONCLUSIÓN
25. L0: ?0!?$%0: :$!?! :! 0# $:< %0 %<$>$0#0 @ < !:< #< %!!?<: ! :!0# #!%!:0$0: 00 #!:< >!:!#<$?$!#< !# 0 :<%$!>0>. P!<= 0: ?0!?$%0: %0 >!?<:0>< $!#!# ?$!: 0$%0%$<#!: >!#< >! ?%:$?<: :!%<!: @0 ! !: #0 %$!#%$0 >! 0# 0$%0%$&# !# ?%<: ?$<: >! 0 $>0. 26.
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E$POSICIÓN N+ '
262.
TEMA: SIMETRÍA.
263. E: # 0:< %00%!:$%< >! H<?0: !0>!: 0:0%0:= !0%$<#0>0 %<# : $#0$0#%$0 0< %$!0: 0#:H<?0%$<#!:= ?<$?$!#<: < $#!%0?$<:. 26'. L0 :$?!0 ! !:%$!< !# ! ?#>< #00= $?!:0 < 0 <$0 #00!90 $>$?!#:$<#0 >! !:0%$< !# ! ! $!. 265. L0 000 :$?!0 <%!>! >! $!< s*+ ! :$#$"%0 0 0 !9 @ >! 0 000 ,etr-+ ! :$#$"%0 ?!>$>0 266. S$?!0 #0 "0 $!#! :$?!0 :$ :! !>! <0 :<! # #< %!#0 @ %<#:!0 0 ?$:?0 00$!#%$0 !# < < ?!#<: ><: <:$%$<#!:. P< < 0#<= 0 <0 0 "0= !:0 ?0#$!#! : H<?0 < !: %<#!#! %<# 0 "0 $#$%$0. S! >$%!= !#<#%!: ! 0 "0 $!#! :$?!0 <0%$<#0. '!&.
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268. '!9.
SIMETRÍA A$IAL
2/. L0 :$?!0 0$0 :! >0 %0#>< <: #<: >! "0 %<$#%$>!# %<# <: #<: >! <0= 0 0 ! ?$:?< H!#&?!#< ! !# #0 $?0!# !!0>0 !# ! 2/1.
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'&'.
E?E DE SIMETRIA
2/3. U# !! >! :$?!0 !: #0 #!0 $?0$#0$0 ! 0 >$$>$ #0 H<?0 %0$!0 < 0%! !# ><: 0!: %@<: #<: <!:<: !>0# :$?4$%<:. 2/'. C0>0 !9 ! #0 "0 !>! >$$>$:! H<?0#>< %<# !:< ><: 0!: $0!: >!%$?<: ! !: #0 "0 :$?4$%0 2/5. 2/6. 2//. 2/8. 2/.
S! !>! >0 0?$!# !# # <!< %<# !:!%< 0 #< ?0: !!: >! :$?!0
28. 281. 282. 283. 28'. 285. 286. 28/. S$ :! ><00 0 "0 :<! ! !! >! :$?!$0 090><= :! <>$0 <:!0 %<# <>0 %0$>0> ! <: #<: >! 0: 0!: <!:0: %<$#%$>!#= !: >!%$= 0?0: 0!: :<# %<#!#!:. 288. '*9.
CARACTERÍSTICAS:
P0!%! ! %0>0 #< :! ?$0 !# ! !:!< L0 >$:0#%$0 >! # #< 0 !! !: 0 ?$:?0 ! 0 >$:0#%$0 >! !! 0 #<
%<!:<#>$!#! !# 0 :$?!0 L0 !%0 ! %<#$!#! <: #<: :$?4$%<: !: !!#>$%0 0 !! 2. 21. 22. 23. 2'. 25. '9!.
E?EMPLOS
2/. 28. 2. 3. 31. 30'.
SIMETRÍA EN LA NATURALE@A
33. E# 0 #00!90 !:0 !:!#! 0 :$?!$0 0$0 !# <><: <: :!!: $<:= %! 0?<#$0 @ !!90= ?@ 00>0! 0 0 $:0 @ ! :! 0 %<#!$>< !# # %<#%!< ?@ !0%$<#0>< %<# 0 !!90. 3'. 35. 36. 3/. 38. 3. 31. 311. 312. 313. 3"4. 3"5.
316. 31/. 318. 31. 32. 321. 322. 323. 32'. 325.
ARUITECTURA % DISEXO R(#ICO
326. SIMETRÍA DE TRASLACIÓ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
33.
331. 332. SIMETRÍA DE ROTACION 333. A#0: !%!: 0?0>0 :$?!0 >! $<. U#0 H<?0 !!>< >! # #< O :$ :! !>! 0%! ! %<$#%$>0 !0%0?!#! :<! ! <$$#0 %0#>< :! <0 0!>!>< >! O # %$!< #< <:$$< ?!#< 0 # %$%< %0>< #< %0?$0 :$ :! <0 0!>!>< >! : %!#< @ # $#< !$!< !?0#!%! :$# %0?$< 0 $0 12 :<! : %!#<. 335.
336. 33&. SIMETRÍ A A$IAL
3 3 8 . U # 0 : $ ? ! 0 0 $0 > ! ! ! ! ! : # 0 0 # : H< ?0 % $ &# = < 0 # < 0 < > < #< P >! 0#< ! %<!:<#>! << #< Pg 0?$4# >! 0#<= >! ?0#!0 ! ! !! ! :!0 0 ?!>$0$9 >! :! ?!#< AAg. 3 3 . L 0: 1 i- /Q a1 a i a 1 1 ,n i 1, - / Qa 1 < ! >$:0#%$0: !#! <: #<: @ ::
% <#: ! 0 #
0:
3'. 3'1.
3'2. 3'3. 3''. 3'5. 34!.
EOMETRÍA % ARTE
3'/. 3'8. 0# :$>< >!:>! 0%! $#"#$>0> >! 07<: ><: >$:%$$#0: !0%$<#0>0: >! #0 H<?0 ?@ !:$%0. L0: ?0!?$%0: !:# !:!#!: !# ! >$:!7< >! ?<#?!#<: 0 04: >! 0 !$! 00 0#>!: !:%<!: @ $#<!: !# ! >!:0<< >! :: <0:. 3'. E <>!# >! #$!:< 0 04: >! <: #?!<: :! <#! >! ?0#$"!:< %<# P$<0: @ 0 E:%!0
35. P$0&$%0= !: 00 !<: <: #?!<: @ 0: "0: :<# 0 !:!#%$0 >! 0: %<:0:. P$<0: >!:%$& 0 !0%$&# ! !$:! !#! 0 <#$> >! 0: %!>0: >! #0 $0 @ <: 0%<>!: >! 0 ?:$%0. P00 4 <>< !:00 !0%$<#0>< @ !$>< < <: #?!<:( ! 0!= 0 0$!%0 @= !# !:?!#= ! #$!:<. 351. 352. E 0! $#<%0 0#< 0 0: !:<#0: ! < 0%$%0# %! ?$:?< !>! :! >! $< $#!!%0= !?<%$<#0= !:4$%< < $!# #0 ?!9%0 >! <><: !<:. 353.
35'. 355. 35!.
CUADRADO M(ICO
35/. 358. P$?! %0>0>< ?$%< #0%$& !# ! :$< ))III 0. C. @ H! !#%<#0>< < ! !?!0>< %$#< >! 0 4<%0 !# ! %0009&# >! #0 <0 !. L< ! !#0 !0# <: #?!<: #00!: >! 1 0 = >$:!:<: !# H<?0 >! %0>0><= >! 0 H<?0 ! 0 :?0 <$9<#0= !$%0 @ >$0<#0 >! <: <: >! !: %$H0: H<?0><: >00 :$!?! 0 ?$:?0 %0#$>0> 0 0 ! :! ! 0?0 %<#:0#! ?$%0 359.
TESELADOS
36. S<# <: >$:!7<: >! "0: !!0 !%<: #$ :!<#!:!= < :!0= ! %$?$!#< >! 0#< %<# "0: 361. 3!'.
#RACTAL
363. 36'. E: # <!< !0 < $!0= :! !$! 0 >$H!!#!: !:%00:. E 4?$#< H! <!:< < ! ?0!?$%< ,!#!< !# 1/5 @ >!$0 >! 0# tractos= ! :$#$"%0 !0>< < H0%0><. M%0: !:%0: #00!: :<# >! $< H0%0. L0 <$!>0>
?0!?$%0 %0! >! # <!< !#$#0?!#! H0%0 !: ! : >$?!#:$&# ?4$%0 H0%0 !: # #?!< #< !#!<. 3!5.
ALORITMO DE COLOR
366. C0>0 :$:!?0 >$#?$%< <>%! #0 :!%!#%$0 >! 0<!: ? 1, ? $, ? %, ? 0,... ? n. L0: $?!#!: H0%0!: :! %!0# !#!0#>< #0 >! !:0: :!%!#%$0: 00 %0>0 ! *#< >! 0 0#00+ !# 0 $?0!#. P<:!$<?!#!= ! 0<$?< >! %<< !: ! !#%00>< >! $#!!0 0 :!%!#%$0 #?4$%0 00 <>%$ # %<< "#0 ! 0 !!:!#! 36/. 368. 3!9.
DESCUBRIENDO EL MUNDO DE LOS MOVIMIENTOS
3&0.
TESELACIÓN
3/1. E: !?0><:0 #0 :!"%$! %<# "0: !0!:. A !:!0 # 0#<= !#! 0: "0: #< !>0 !:0%$<: @ 0?<%< :! :!<#!#. 3/2. E$:!# !: !:!0%$<#!: < ?<:0$%<: !0!:( 0 ?00 >! $#<: !$!<:= ! !$%0>< %0>0>< %! 0!< >! 0!>!9 @ 0 %<#"0%$&# !0<#0= %! <: 0#!!:. 3/6. 3/5.
3/'. 3/3.
T1a8i;n d T/in76,1
3//. 3/. 3/8.
T1a8i;n d C6ad/ad,1
381. 38.
382. 38'. 383.
T1a8i;n d H7,n,1
386. 385.
38/. 388.
E -ad,1ad, 8,n T/an1=,/-a8i,n1 I1,-</i8a1
38. L0 :$?! <:!0%$&# @ 0#$:$: >! !?0><:0><:= #<: !?$! %! 0#:H<?0%$<#!: $:
3.
33. E?0> <:0>< < 0:0%$&#
32. 31.
3'. E?0> <:0>< < <0%$&#
35. E?0>< :0>< < !!$&#
36. 39&.
TESELACIONES DE MAURITS CORNELIS ESCHER
38.
C!0%$&# >! 0 "0 >! # 0< 0 0$ >! <<#< A,CD.
3. '.
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