Limites, derivadas
-
Upload
christian-medina -
Category
Documents
-
view
268 -
download
0
Transcript of Limites, derivadas
-
7/26/2019 Limites, derivadas
1/214
UNIVERSIDAD CENTRAL DEL ECUADORFACULTAD DE CIENCIAS ADMINISTRATIVAS
PORTAFOLIO MATEMTICASII CURSO: CA2-2
MEDINA AYMACAA CHRISTIAN STALINSEGUNDO SEMESTRE
-
7/26/2019 Limites, derivadas
2/214
NOMBRE:Christian Medina
CEDULA DE IDENTIDAD:
17252!"#$
FECHA DE NACIMIENTO:
$"%11%2$15
ESTADO CIVIL:
SOLTERO
DOMICILIO:
A&' A()ert* S+en,er Oe-. S2#!7
CIUDAD:
/UITO
E-MAIL:
0ristianesta(inh*tai(',*
-
7/26/2019 Limites, derivadas
3/214
FILOSOFA CO RPORATIVA DE LA FACULTAD
Visin de la UCE
La Uni&ersidad Centra( de( E,3ad*r ,*ntin3ar4 en e( (idera.* de (a ed3,a,i6n s3+eri*r de (a +r*d3,,i6n de,ien,ia te,n*(*.8a ,3(t3ra 9 arte 9 en (a :*ra,i6n de +r*:esi*na(es ,*n +r*:3nda res+*nsa)i(idad s*,ia('
Misin de la UCE
La Uni&ersidad Centra( de( E,3ad*r :*ra +r*:esi*na(es ,r8ti,*s de ni&e( s3+eri*r ,*+r*etid*s ,*n (a &erdad;3sti,ia e9 ,rea es+a,i*s +ara e( an4(isis 9 s*(3,i6n de +r*)(eas na,i*na(es'
Visin de la FCA
Mantener a (a ?a,3(tad de Cien,ias Adinistrati&as ,** (a +riera de( +a8s 9 3na de (as e;*res de A=ri,ai+artiend* 3na :*ra,i6n e@,e(ente
-
7/26/2019 Limites, derivadas
4/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
UNIVERSIDAD CENTRAL DEL ECUADOR
FACULTAD DE CIENCIAS ADMINISTRATIVAS
CARRERA DE CONTABILIDAD Y AUDITORIA
SLABO
EJE BSICO
MATEMTICA II
SEMESTRE: SEPTIEMBRE 2015 FEBRERO 2016
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina1P!"#$# 2%1& - 2%1'
-
7/26/2019 Limites, derivadas
5/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
$' DATOS INFORMATIVOS
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina2P!"#$# 2%1& - 2%1'
1(1( FACULTAD: CIENCIAS ADMINISTRATIVAS
1(2( CARRERA: CONTA)ILIDAD Y AUDITORIA
1(*( ASIGNATURA: MATEMATICAS II
1(+( CDIGO DE ASIGNATURA:&*%+(,, O!a. /Ma0ia II&(CA2(&(&
1(&( CRDITOS: '
1('( SEMESTRE: 2
1(3( UNIDAD DE ORGANI4ACINCURRICULAR:
).ia
1(5( TIPO DE ASIGNATURA: O67iga#!ia
1(,( PROFESOR COORDINADOR DE
ASIGNATURA: F!ani.# )a8a0#n$1(1%( PROFESORES DE LA ASIGNATURA: F!ani.# )a8a0#n$9 Ma!8a R#a.9
1(11( PER;ODO ACADMICO:SORAS DE TUTORIAS:Presenciales
1'Virtuales:
1(1+( PRERRE?UISITOSAsi!naturas:
MATEMATICA I&(CA1(&((&
1(13( CORRE?UISITOSAsi!natur
as:
ADMINISTRACION IIINTRODUCCION ALDEREC>OCONTA)ILIDAD GENERAL II
ECONOM;A
&(CA(2(1(&&(CA(2(+(2
&(CA(2(2(+
&(CA(2(*(2
-
7/26/2019 Limites, derivadas
6/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
1( DESCRIPCI%N DE LA ASI&NATURA
E7
!a#na0in# 7gi#9 a." #0# a06iHn
A
D.a!!#77a! 0#$7#. $ #
&( RESULTADOS DE APRENDI*A'E DE LA ASI&NATURA:
Ana7ia
-
7/26/2019 Limites, derivadas
7/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
R.7
'( PRO&RAMACI%N DE UNIDADES CURRICULARES
DATOS INFORMATIVOS DE LA UNIDAD CURRICULAR N"# $NOMBRE DE LAUNIDAD:
LIMITES Y CONTINUIDAD DE FUNCIONES
OB%ETIVO DE LAUNIDAD:
Aplicar el conocimiento de lmites a la solucin de problemas administrativos yfinancieros de empresas pblicas y privadas
RESULTADOS DEAPRENDI&A%E DE LAUNIDAD: Analiza comportamientos de problemas empresariales plasmados en
grficas, en relacin a funciones del mbito empresarial. Resuelve problemas con aplicacin de lmites y continuidad aplicados al
sector empresarial, para estimar lmites de beneficio y prdida.
C'LCULO DE (ORASDE LA UNIDAD
ESCENARIOS DEAPRENDI&A%E
N)# ("ras a*ren+i,a-e Te.ricas1%
N)# ("ras Prcticas/la0"rat"ri"
1+
TUTOR1ASN)# ("ras Presenciales
+
N)# ("ras A*ren+i,a-e AulaVirtual
2
TRABA%OAUT2NOMO
("ras +e Tra0a-" Aut.n"3"2+
PRO4RAMACI2N CURRICULAR
CONTENIDOS
ACTIVIDADES DETRABA%O AUT2NOMO5
ACTIVIDADES DEINVESTI4ACI2N Y DEVINCULACI2N CON LA
SOCIEDAD
MECANISMOS DEEVALUACI2N
1.1 mites! "efinicin y #ropiedades
1.$ mites infinitos, lmites al infinito ylmites laterales.
%&aminar el concepto de lmite ysus propiedades.
'alcular limites empleando unavariedad de tcnicas y
procedimientos
#articipacin en clase%&posicin(areas individuales! clase y e&traclase.(raba)os grupales
#articipacin en el aula virtual.#ruebas
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina+P!"#$# 2%1& - 2%1'
-
7/26/2019 Limites, derivadas
8/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
1.* 'ontinuidad. +alor intermedio
1.* Aplicaciones.
E.$ia! 7a #nini$a$ $a!ia. @ni#n.(
"esarrollar aplicaciones yresolver problemas relacionadoscon los negocios y la economa, la
vida y las ciencias sociales.
METODOLO41AS DEAPRENDI&A%E:
MH#$# 8!".i#9 A
OBRAS F1SICAS
DISPONIBILID
AD ENBIBLIOTECA VIRTUALNOMBRE
BIBLIOTECAVIRTUAL
SI NO
B'SICA http://www.epulibre.o
rg/
COMPLEMENTARIA
DATOS INFORMATIVOS DE LA UNIDAD CURRICULAR N"# 6NOMBRE DE LAUNIDAD:
DIFERENCIACION UNO
OB%ETIVO DE LAUNIDAD:
Resolver situaciones empresariales aplicando los conceptos de lmites yderivacin
RESULTADOS DEAPRENDI&A%E DE LAUNIDAD:
Ana7ia 0#$7#. 0a0i#. !7ai#na$#. #n !a#n.$ a06i# C#n.! a67a. C'LCULO DE (ORASDE LA UNIDAD
ESCENARIOS DE
APRENDI&A%E
N)# ("ras a*ren+i,a-e Te.ricas1%
N)# ("ras Prcticas/la0"rat"ri" 1+
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina&P!"#$# 2%1& - 2%1'
-
7/26/2019 Limites, derivadas
9/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
TUTOR1ASN)# ("ras Presenciales
+
N)# ("ras A*ren+i,a-e AulaVirtual
2
TRABA%OAUT2NOMO
("ras +e Tra0a-" Aut.n"3"2+
PRO4RAMACI2N CURRICULAR
CONTENIDOS
ACTIVIDADES DETRABA%O AUT2NOMO5
ACTIVIDADES DEINVESTI4ACI2N Y DEVINCULACI2N CON LA
SOCIEDAD
MECANISMOS DEEVALUACI2N
$.1 a "erivada! "efinicin e ;nterpretacin geomtrica. Reglas de diferenciacin
$.$ Razn de cambio. Regla de la cadena.$.* "erivadas de! funciones
ogartmicas y e&ponenciales."erivadas implcitas."erivadas de orden superior
$.2 Aplicaciones
.Ca77a! in!
RECURSOS DID'CTICOS: Li6!#.9 !i.a. @#77#.(C#0BIBLIO4RAF1A:/aeussler.4r., %. 5., 6 7ood, R. . 0$83. Matemticas para Administracin y Economa. 9&ico! #rentice /all :"ecimosegunda edicin.BIBLIOGRAFA COMPLEMENTARIA/offmann aurence, et al,0$123, Matemticas, Aplicadas a la Administracin y los Negocios, Mxico, Mc Graw HillEducation.
OBRAS F1SICAS
DISPONIBILIDAD EN
BIBLIOTECA VIRTUALNOMBRE
BIBLIOTECAVIRTUAL
SI NOB'SICA 7 http://www.epulibre.o
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina'P!"#$# 2%1& - 2%1'
-
7/26/2019 Limites, derivadas
10/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
rg/
COMPLEMENTARIA
7
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina3P!"#$# 2%1& - 2%1'
-
7/26/2019 Limites, derivadas
11/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
DATOS INFORMATIVOS DE LA UNIDAD CURRICULAR N"# 8NOMBRE DE LAUNIDAD:
DIFERENCIACIN DOS
OB%ETIVO DE LAUNIDAD:
Ana7ia! .iai#n. 0
RESULTADOS DEAPRENDI&A%E DE LAUNIDAD:
I$nia 7#. in!a7#. $ ganania. E
-
7/26/2019 Limites, derivadas
12/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
METODOLO41AS DEAPRENDI&A%E:
"eterminar algoritmos y elaborar grficas, es I$nia a!ia67. n aii$a$. 0
A
-
7/26/2019 Limites, derivadas
13/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
INVESTI4ACI2N Y DEVINCULACI2N CON LA
SOCIEDAD
2.1 a ;ntegral ;ndefinida! concepto eintegracin de formas elementales.
2.$ ;ntegrales indefinidas concondiciones iniciales.(cnicas de;ntegracin! sustitucin eintegracin por partes.
2.* ;ntegral definida, interpretacingeomtrica.
2.2 "eterminacin de >reas ba)o lacurva y entre curvas
2.? Aplicaciones
R.#7!
$ ing!ain$ni$a i7ian$#@!07a.
Ca77a! ing!a7.in$ni$a.9 #n
D!0ina! 7 !a6a# na !a n! !a.(
I$nia! !.#7!
Pa!iiBIBLIO4RAF1A:>a..7!(!(9 E( F(9 Q##$9 R( S( /2%%5( Matemticas para Administracin y Economa.MHi#: P!ni >a77 - Di0#.gn$a $iin(BIBLIOGRAFA COMPLEMENTARIA
>#0ann La!n9 a79/2%1+9 Matemticas, Aplicadas a la Administracin y los
Negocios, M!ico, Mc "raw #ill Education.
OBRAS F1SICAS
DISPONIBILIDAD EN
BIBLIOTECA VIRTUALNOMBRE
BIBLIOTECAVIRTUAL
SI NO
B'SICA http://www.epulibre.
org/
COMPLEMENTARIA
3( RELACI%N DE LA ASI&NATURA CON LOS RESULTADOS DEL PERFIL DE E&RESODE LA CARRERA
VICERRECTORADO ACADMICO DE INVESTIGACIN Y POSGRADODIRECCIN GENERAL ACADMICA Pgina1%P!"#$# 2%1& - 2%1'
-
7/26/2019 Limites, derivadas
14/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
RESULTADOS O LO4ROS DEAPRENDI&A%E DEL PERFIL DE E4RESO
DE LA CARRERAEL ESTUDIANTE DEBE
Ana7ia #0 R.7
Ana7ia 0#$7#. 0a0i#.!7ai#na$#. #n !a#n. $a06i# C#n.! a67a. I$nia 7#. in!a7#. $ganania. I$nia a!ia67. n
aii$a$. 0
A
!na6i7i$a$9 0$ian 7a
A7 na7ia! 7 .0.!9 7 .$ian7g# $ a
A
-
7/26/2019 Limites, derivadas
15/214
UNIVERSIDAD CENTRAL DEL ECUADORSYLLABUS
!$in $ #.#. 7a#5( EVALUACI%N DEL ESTUDIANTE POR RESULTADOS DE APRENDI*A'E
TCNICASPRIMER
(EMISEMESTRE;PUNTOS g (0 )=2
ashP 2=2
asiP La :3n,i6n es ,*ntin3a
as;P Gr4:i,a
as0P @ as(P 9asP 5 asnP !1as*P ! as+P -7
as
-
7/26/2019 Limites, derivadas
104/214
2P limx 4
lim
x 4
xlimx 4
4
limx 4
x+ limx 4
4=
444+4
=0
8=0
-Plim
x 4
x4
x+4 =0 > 6 (4 )=0
atdP $ J $
ateP La :3n,i6n es ,*ntin3a
at:P Gr4:i,a
at.P@ athP9atiP at;P -at0P# at(P 5
atP - atnP7at*P2 at+P-at
-
7/26/2019 Limites, derivadas
105/214
x+4x2
*(2, 0)
1P f(2)=2+ 422
=2
8=0.5 1P f(0)=
0+402
= 4
2=2
2Plim
2 x+ lim2 4
lim2
x lim2
2=
2+422
=0,5 2Plim
0 x+lim0 4
lim0
xlim0
4=
4
2=2
-P La :3n,i6n n* es ,*ntin3a en x=2 -P La :3n,i6n es ,*ntin3a en x=0atPat@P Gr4:i,aat9P
atP @ a3aP 9
a3)P 2 a3,P $5
a3dP 1 a3eP 1
a3:P$ a3.P 2
a3hP 1 a3iP5
a3;P- a30P 7
a3(P! a3P !
a3nPa3*Pa3+P
a3
-
7/26/2019 Limites, derivadas
106/214
a3&P Gr4:i,aa3P
a3@P @ a39P 9
a3P 5 a&aP $$!
a&)P ! a&,P $
a&dP 2 a&eP 1
a&:P1 a&.P -
a&hP $ a&iP2
a&;P
a&0P
a&(Pa&P
a&nP
{x+2 si x7 2x2 si x
-
7/26/2019 Limites, derivadas
107/214
adP
aeP
-
7/26/2019 Limites, derivadas
108/214
a:P En,3entre t*d*s (*s +3nt*s de dis,*ntin3idad
1( f(x )=3x23
a.P Es +*(in6i,a ent*n,es es ,*ntin3a
2( f(x )= 3
x+ 4
ahP x+4=0
aiP x=4
a;P Es ,*ntin3a en @ J !
*(
2x23
3
g (x )=
a0P Es +*(in6i,a ent*n,es es ,*ntin3a
+( f(x )= x
2+6x+9
x2+2x15
a(P x2+2x15=0
aP (x+5)(x3)=0
anP x+5=0x3=0
a*P x=5x=3
a+P La :3n,i6n es dis,*ntin3a en @ J 5 9 @ J -
&( 6 (x )=x7
x3x
aB x3x=0
a! x=3x
asP La :3n,i6n es dis,*ntin3a en x=3x
atP
'( % (x )= x
x2+ 1
a3P x2+1=0
-
7/26/2019 Limites, derivadas
109/214
a&P x2=21
aP La :3n,i6n es dis,*ntin3a en x2=21
a@P
a9P
3( f(x)={1 s ix 01 six 1
a@)P La :3n,i6n es dis,*ntin3a en @ J 1
a@,P
-
7/26/2019 Limites, derivadas
110/214
'6% De=e$ N 5
lim
x 2
16=16
a@eP
a@:P Gr4:i,a
a@hP
a@iP
a
-
7/26/2019 Limites, derivadas
111/214
lim
x 5(x25)
axk) limx 5
(x25) JXx
limx 5
2 lim
x 5 5 J 5P - 5 = 20
axl) Gr4:i,a
a0a@nP
a@*P
a@+P
a@
-
7/26/2019 Limites, derivadas
112/214
a@@P
a@9P
a@P
a9aP
a9)P
a
a9dP
a9eP
-
7/26/2019 Limites, derivadas
113/214
limt 3 (
t2t+5 )
a9:P (t2
t+5)=
(limn 3
tlimn 3
2
limn3
t+limn3 5
)=
(3 )2
(3 )+5=5
2=2, 5
limt 3
a9.Pa9hP Gr4:i,aa9iP
a9;P @ a90P 9a.l)-/ a.0) -12
a9nP 1 a9*P $75a9+P $ a9
-
7/26/2019 Limites, derivadas
114/214
lim% 4
%2+%+5
aiP
lim% 4
%2+%+5=[ lim% 4%2+lim
%4
%+ lim%4
5]1
2=[ (4 )2+4+5 ]1
2=25=25
a;P Gr4:i,a
a0P
a(PaPanPa*P
a+Pa
-
7/26/2019 Limites, derivadas
115/214
aP limx 3
x2x2x3
=[ limx 3
x ]2 limx 3
x limx 3
2
limx 3
xlimx 3
3 =
(3 )2(3 )633
=0
0=I
)aaP x2x2
x3 =
(x3)(x+2)
(x3) =x+2
bab)lim
x 3
(x2x2 )
x3 =lim
x 3
x+ limx 3
2=3+2=5
)a,P)adPGr4:i,a
bae%
ba&%
,iP @ ,;P 9,0P ! ,(P 2,P- ,nP 1,*P 2 ,+P $,
-
7/26/2019 Limites, derivadas
116/214
limx 4
x29x+20
x23x4
)a.P limx 4
x29x+20
x2
3x
4=
(4 )29 (4 )+20
(4 )
2
3 (4 )4
=0
0=I
)ahP x
29x+20
x23x4
=(x5 ) (x4 )(x4 ) (x+1 )
=lim
x 4
xlimx 4
5
limx 4
x+ limx 4
1=
454+1
=0,2
)aiP)a;PGr4:i,a)a0P
)a(P
)aP)anP)a*P)a+P)a
-
7/26/2019 Limites, derivadas
117/214
bbi%
bb'%
-
7/26/2019 Limites, derivadas
118/214
limx
3
x
66X
limx
x
12
3
x=
limx
3
limx
))(P Gr4:i,a
))P))nP))*P))+P))
-
7/26/2019 Limites, derivadas
119/214
),ePGr4:i,a
),:P),.P),hP),iP),;P),0P
),(P),P),nP),*P),+P),
-
7/26/2019 Limites, derivadas
120/214
limt
3 t2+2 t2+9t1
5 t25
),sP limt
3 t2+2 t2+9t1
5 t25
=
=I
),tP 3t
2+2t2+9 t1
5 t25
=
3 t3
t3 +
2t2
t3+
9 t
t3
1
t3
5 t2
t3
5
t3
=
3+2
t+
9
t2
1
t3
5
t
5
t3
),3P
limt
3 t2+2 t2+9t1
5 t25
=
limt
3+limt
2
limt
t+
limt
9
[ limt
t]2
limt
1
[ limtt
]3
limt
5
limt
t
limt
5
[limtt
]3
=
3+2
+
9
2
1
3
5
5
3
=3+0+00
00 =
),&P
),P Gr4:i,a),@P),9P),P)daP)d)P)d,P)ddP)deP
)d:P)d.P)dhP)diP)d;P)d0P)d(P)dP)dnP
:P @ :@P 9:9P - :P 22.aP 2 .)P 2--
.,P $5 .dP 117.eP $ .:P $2..P 1 .hP -27.iP 2 .;P -1-.0P - .(P -!5.P! .nP -"1
-
7/26/2019 Limites, derivadas
121/214
limx
7
2x+1
)d*P
limx
7
2x+1=
7
2 ( )+1=0
)d+P)d
-
7/26/2019 Limites, derivadas
122/214
limx 1
x23x+1
x2+1
)dtP limx 1
x23x+1
x2
+1
=(1)23(1)+1
(1)
2
+1
=13+1
2 =0,5
)d3P)d&PGr4:i,a
)dP
)d@P)d9P
limx 7
x2+1
x249
)dP limx 7
x2+1
x249=
(7)2+ 1
(7)249=
49+18 0
=50
0=
)eaP
)e)PGr4:i,a)e,P
)edP
h.P @ hhP 9hiP - h;P 17h0P 2 h(P 1"hP1 hnP 25h*P $ h+P 1h
-
7/26/2019 Limites, derivadas
123/214
lim
x 0
lxl
)eePlimx 0
lxl=0=0
)e:P)e.P Gr4:i,a)ehP
)eiP@ )e;P9)e0P $ )e(P$)eP 1 )enP 1)e*P 2 )e+P 2)e
-
7/26/2019 Limites, derivadas
124/214
>aga 7 6#.B# $ 7a g!a $
):.P f(x){ 100x+600 si0 x< 5100x +1100 si5 x100x+1600 si10 x):;P : es ,*ntin3a en 2 es ,*ntin3a en 5 es ,*ntin3a en 1$):0P
):(P):P):nP
1( Si . #nina n 22( N# . #nina n &*( N# . #nina n 1%
3")
34)
35)
3r)
3s)
36)
-
7/26/2019 Limites, derivadas
125/214
=@8 CORRECCIN DE LA .RUEBA1( En#n!a! 7 7"0i $ 7a @nin ana7i . #nini$a$
37) limx 0
1
x+2
1
2x
=
1
0+2
1
20
00
=I
):P1
x+2
1
2
x =
2 (x+2 )2 (x2 )
x =
2x2(2x+4 )x
= x
(2x+4 )x=
12x+4
):@Plim
x 0
1
2x+4=
12 (0)+4
=14
):9P
):P -oescontinuaen
(0,
1
4
)).aP).)P Gr4:i,a).,P
).dP @ ).eP 9).:P 5 )..P $1#).hP ! ).iP$25
).;P- ).0P $!$).(P1 ).P $5
).nP $ ).*P $25).+P 1 ).
-
7/26/2019 Limites, derivadas
126/214
3;)
2( Ca77a! 7 .igin 7i0i limx 5
2x213x+15
x2x20
( >aga 7a g!a #0
#nini$a$(
bhg) limx 5
2x213x+15
x2x20
=2(5)213(5)+15
(5)2(5)20=
0
0=I
bhh) 2x
213x+15
x2x20
=(2x3)(x5)
(x5 )(x+4)=
2x3x+4
)hiPlim
x 5
2x3
x+4=
2(5)35+4
=7
9)h;P
)h0P -oescontinuaen
(5,
7
9
))h(P)hP Gr4:i,a)hnP
)h*P @ )h+P 9)h
-
7/26/2019 Limites, derivadas
127/214
*( E7 g!.a . a.an a 7a @nin @/W2x4x+ 2
$#n$ .#n 7# a#. $ i$a $ 7a
0
a R
)i0P f(x)2x4x+2
)i(P)iP Gr4:i,a)inP
)i*P@ )i+P9)i
-
7/26/2019 Limites, derivadas
128/214
6 En BH a# 7a 0
);0P La e+resa 9a n* tiene +erdida en e( se.3nd* a*'
E.n 7i0ia$#. .. 6ni#. Si . a." 7 . . 7"0i
b'l% limx
2x4x+2 = 2( )4()+2 = =I
b'm% 2x4
x+ 2 =
2xx
4
xxx
+2
x
=2
4
x
1+2
x
);nP
limx
24
x
1+2
x
=2
4
1+2
=2
);*P);+P E( (8ite de (*s in.res*s ser4 de 2'
);
x+1$7a!.(
aP C34( ser4 e( +re,i* dentr* de 5 eses
);rP %(5)=4+ 30
5+1=45
);sP E( +re,i* dentr* de 5 eses ser4 de !5 d6(ares'
)P En ,34nt* )a;ara e( +re,i* en e( ,P C34nd* e( +re,i* ser4 !-
);P 43=4 + 30x +1
);@P 4340= 30
x+1);9P (x+1)3=30);P 3x+3=30)0aP x=9)0)P)0,P En e( n*&en* es e( +re,i* ser4 de !- d6(ares
)0dP
-
7/26/2019 Limites, derivadas
129/214
dP /3= (e s3,eder4 a( +re,i* a (ar.* +(a* Gra:i
-
7/26/2019 Limites, derivadas
130/214
3l#)
)(dP
-
7/26/2019 Limites, derivadas
131/214
=e De=e$ N /
3l)Pr"3le0as $$
-
7/26/2019 Limites, derivadas
132/214
60 m%/=e
2( 0,1)10,10
=2,2140
60 m%/=e
2( 0,01)10,010
=2,0201
6na m%/=e
2( 0,001)10,0010=2,0020
)n)P En (*s +r*)(eas - a 1 e+(ee (a de:ini,i6n de (a deri&ada +ara en,*ntrar(a en ,ada,as*
)n,P-' f(x )=x
)ndP df(x)
dx =
d
dxx=x11=1
6n+( f(x )=4x1
6n@ df(x)dx
=4 ddx
xddx
1=(4 ) (1 )0=4
6ng&( y=3x+5
6n8 dy
dx=3
ddx
x +ddx
5=3.1+0=3
6ni'( y=5x
6n dy
dx=5
d
dxx=5.1=5
6nX3( y=(54x )
6n7 dy
dx=
ddx
54 ddx
x=04.1:4
6n0
5( y=(1x2 )6nn
dydx
=ddx
1ddx
x2
=01
2=
12
6n#,( f(x )=3
6n
-
7/26/2019 Limites, derivadas
133/214
6n
6n
-
7/26/2019 Limites, derivadas
134/214
12( y=x2+5x+1
6n dydx
=d
dxx
2
+5d
dxx+
d
dx1=2x21+5.1+0
6n dy
dx=2x+5
1*( %=3 &2+2&+1
6n d%
d&=3
dd&
&2 +2
dd&
&+dd&
1=3 .2 &21+2.1+0
6#a d%
d&=6 & +2
6#61+( y=(x2x3 )
6# dydx
=ddx
x2
ddx
xddx
3=2x2110
6#$
dy
dx =2x16#
1&( y=6
x
6#@ dydx
=(x )
d
dx(6 )(6 )
d
dx(x )
(x )2 =
x (0 )(6 ) (1 )
x2
6#g dy
dx=
06
x2
=6
x2
6#8
1'( c=7+2&3 &2
6#i dc
d&=
dd&
7+2dd&
&3dd&
&2=0+ (2 ) (1 ) (3 ) (2 ) &21
6# dc
d&=26 &
6#X
13( f(x )=x+2=(x+2 )1
2
6#7 df(x )
dx =
ddx
(x+2 )1
2 .ddx
(x +2 )=1
2(x+2 )
1
21
. (1+0 )
6#0df(x )dx =
1
2 (x+2 )
1
2 . (1 )=1
2
[ 1
(x+2)1
2]6#n
df(x )dx
= 1
2x+26##
6#
-
7/26/2019 Limites, derivadas
135/214
15( ;(x )= 3
x2
6#B d;(x )dx
=(x2 )
d
dx(3 )(3 )
d
dx(x2 )
(x2 )2 =
(x2) (0 )(3 )(x110 )(x2 )2
6#! d;(x )
dx = (
0 )(3 ) (1 )(x2 )2
= 03
(x2 )2=
3
(x2 )2
6#.1,( Enn! 7a
-
7/26/2019 Limites, derivadas
136/214
22( Enn! 7a
-
7/26/2019 Limites, derivadas
137/214
25' y=x2+2x+3 * (1,6 )
)
-
7/26/2019 Limites, derivadas
138/214
=$ De=e$ N
)riP Res*(&er (*s si.3ientes e;er,i,i*s +*r (a re.(a de (a ,adena)r;P
1( y=4! * !=x5x4+3
6!X6!7 y=4!=!
1
4
6!0 dyd!=
d
dx!
1
4=1
4!
1
41
6!n dyd!
=1
4!
34
6!#6!
-
7/26/2019 Limites, derivadas
139/214
2( ==u3*u=
t1t+1
cuandot=1
6.6
6. ==u3
6.$ d=du
= ddu
u3=3 u31
6. d=
du=3 u2
6.@
6.g u=t1t+1
6.8 dudt
=(t+1 )
ddt
( t1 )(t1)ddt
(t+1)
(t+1)2
6.i dudt= (t+1 ) (10 )(t1 ) (1+0 )
( t+1)2
6. du
dt=
(t+1 )( t1 )
(t+1)2
6.X
6.7 d=
dt=(3 u2 )[(t+1 )( t1 )( t+1 )2 ]
6.0 d=
dt=[3( t1t+1 )
2
] [( t+ 1 )( t1)( t+1)2 ]6.n d=
dt=[3(
111+1)
2
][(1+1 ) (11)
(1+1 )2 ]6.#
d=dt
=[3( 02 )2
][(2 )(0 )(2 )2 ]6.
-
7/26/2019 Limites, derivadas
140/214
*( !=u2 +u+9 *u=2 s2 1cuando s=16..
6. !=u2+u+9=u2+u1
2+9
6.
u
ddu
( 2+u12+ 9)= d!du
u2
+d!du
u1
2+d!du
9
d!
du=
6. d!du
=2u21+1
2u
1
21
+0
6. d!
du=2u+
1
2u1
2 =2u+1
2 ( 1u )6.
d!
du=2u+
1
2u
6.6. u=2 s21
6a du
ds=
dds
( 2 s2 1 )=2dds
s2
dds
1
66 du
ds=(2 ) (2 ) s210
6 du
ds=4 s
6$
6 d!
ds
=
(2 u+
1
2u )(4 s )
6@ d!
ds=[2 ( 2 s21 )+ 122 s2 1 ] (4 s )
6g d!
ds={2 [2 (1 )21 ]+ 122 s21 }[4 (1 )]
68 d!
ds=[2 (21 )+ 1221 ] (4 )
6i d!
ds=(2+ 121 ) (4 )=(2+
1
2 ) (4 )6 d!ds=(
52 ) (
4 )=10
6X=t
-
7/26/2019 Limites, derivadas
141/214
=t9 De=e$ N
)tnP)t*P Res*(&er +*r d*s =t*d*s distint*s (a si.3iente :3n,i6n)t+P
6 (! )= 62!!
24
)t
-
7/26/2019 Limites, derivadas
142/214
=8' De=e$ N
3=3) Pr"3le0as $$
-
7/26/2019 Limites, derivadas
143/214
)3@P
6 df(x )
dx =
ddx
(9x2 )= (9 ) ddx
x2=9.2x21=18x
5( y=4x3
)3P6a
dydx
=ddx
(4x3 )=4 ddx
x3=4.3x31=12x2
)&)P)&,P
,( g (= )=8 =7
)&dP
6 dg (= )
d= =
d
d=(8 =7 )= (8 ) d
d==
7=8.7x71=5x6
1%( v (x )=xe
)&:P
6g dv (x )
dx =
d
dx(xe)=xe1
)&hP11( y=2/3x4
)&iP
6 dydx =ddx(23x4)=
23
ddxx4= 23
.4x41= 83x3
)&0P12( f(%)=3%4
)&(P
60 df(%)d%
=dd%
( (3 )1
2%4)= (3 )1
2 dd%
%4=4. (3)1
2%41=43%3
)&nP
1*( f( t)=t7
256#
6 x7)=17
ddx
x7= 17> 7
1x
71=x6
-
7/26/2019 Limites, derivadas
144/214
6
1&( f(x )=x+36
6 df(x)
dx =ddx (x+3)=
ddxx +
ddx3=1+ 0=1
6
1'( f(x )=3x26
6 df(x)
dx =
ddx
( 3x2)=3 ddx
xddx
2=3.1+ 0=3
6
13( f(x )=4x22x+36a
66 df(x)
dx =
ddx
(4x22x +3 )=4 ddx
x22ddx
x+ddx
3=4.2x12.1+0=8x2
6 df(x)
dx =2( 4x1)
6$
15( f(x )=5x2
9x6
6@ df(x)
dx =5
ddx
x29ddx
x=5.2x19.1=10x9
6g
1,( g (% )=%43%3168
6i
dg(%)d%
=dd%
(%43%31)=dd%
%4 3 d
d%%
3ddx
1=4%33.3%20=4%39%2
6
2%( f( t)=13 t2+14 t+16X
67df(t)
dt =
d
dt(13 t2+14 t+1)=13
d
dtt
2+14d
dtt+
d
dt1=13.2 t1+14.1+0=26 t+14
60
-
7/26/2019 Limites, derivadas
145/214
21( y=x3x6n
6# dydx
=ddx
(x3x1
2 )=ddx
x3ddx
x1
2=3x21
2x
12
6
-
7/26/2019 Limites, derivadas
146/214
2*( y=13x3+14x22x+36
6 dy
dx=
ddx
(13x3+14x22x+3)=13ddx
x3+14ddx
x22ddx
x +ddx
3
6 dy
dx=13(3)x2+14 (2)x12(1)+0=39x2+28x2
6
2+( ?(r )=r87 r6+3 r2+16
6 d?(r )
dr =
d
dr(r87 r6+3 r 2+1 )=d
drr
87d
drr
6+3d
drr
2+d
dr1
6a d?(r )dr =8 r77( 6)r5+3( 2)r1+ 0=8 r742 r5+6 r
662&( f(x )=2 (13x 4 )
6
6$ f(x )=ddx
(262x4 )=ddx
262ddx
x4=02 ( 4 )x3=8x3
6
2'(12
y=x +3=(x+3 )
6@
1
2
=( 12 ) (x+3 )1
21 d
dx(x+ 3)
dydx
=ddx
(x+3 )
6g dy
dx=( 12 ) (x+3 )
12 .[ddxx11 ddx3]
68
12
dy
dx=
1
2(x+3 )
6i
23( y=x
2+15
=( 15 ) (x2+1 )6
dydx
=ddx( 15 ) (x2+1 )=( 15 ) .(2 ddxx21 ddx1)
6X dydx
=(15 )(2x)=
2x5
-
7/26/2019 Limites, derivadas
147/214
67
60
-
7/26/2019 Limites, derivadas
148/214
25( y=2
2x2
6n y=2
2 x
2=x2
6#
dy
dx=
d
dxx2
6
-
7/26/2019 Limites, derivadas
149/214
*1( f(x )=(2x2+4x )100
6g d f(x )
dx =
dydx
(2x2+ 4x )100
68 d f(x )
dx
=100d
dx
(2x2+4x )1001 d
dx
(2x2+ 4x)
6i d f(x )
dx =100 (2x2+4x )99(2 (2 )ddxx21+ 4 ddxx)
6 d f(x )
dx =100 (2x2+4x )99 ( 4x+4 )
6X d f(x )
dx =100 (4x+4 )(2x2+4x )99
67*2( f(= )===+=2== ( = )1/ 2+=2== 3 /2+=2
60 df( = )
d=
=d
d=
= 3/ 2+
d
d=
= 2
6n df( = )
d= =( 32 ) dd==
3
21
+2 dd=
=21
6# df( = )
d= =
3
2=
1
2 +2 =
6
-
7/26/2019 Limites, derivadas
150/214
*+( y=5x
28x2x
)9P dy
dx=
ddx
5x28x2x
)9@P
2x
dydx
=(2x )d
dx(5x28x )(5x28x ) d
dx(2x )
)99P
2x
dydx
= (2x
)(5
d
dxx
2
8
d
dxx
)(5x2
8x )
(2
d
dxx
)
)9P
2x
dy
dx=
(2x )[5 (2 )x218 (1)](5x28x )[2 (1 )]
)aP
2x
dydx
= (2x )(10x8)(5x28x )(2)
))P
2x
dydx
=20x216x10x2+16x
),P dy
dx=10x
2
4x2=
5
2
)dP
*&( y=(8x+ 2)(x2+1)4
)eP dy
dx=
d
dx(8+2x)(x2+1)4
):P dy
dx=(8+2x )
ddx
(x2+1 )4+(x2+ 1 )4 ddx
(8+2x )
).P dydx
=(8+2x )( ddx (x2+1)4d
dx(x2+1 ))+(x2+1)4(ddx8+2 ddxx)
)hP dy
dx=(8+2x ) [4 (x2+1 )3(2x )]+ (x2+1)4(2)
)iP dy
dx=(8+2x ) [8x (x2+1 )3 ]+( 2x2+2 )4
-
7/26/2019 Limites, derivadas
151/214
);P dy
dx=(8+2x )(8x3+ 8x )3+( 2x2+2 )4
)0P
*'( (! )= (2! )3
5 +5
g
67 d g(! )
d! =
d
d!(2! )
3
5+5
60 d g(! )d!
=d
d!(2! )
3
5+d
d!5=
d
d!
(2! )3
5dd!
(2! )+0
6n d g(! )
d! =
3
5(2! )
3
51
2=6
5(2! )
25
6#
6
-
7/26/2019 Limites, derivadas
152/214
*3( y=11x=11x1
2
6B dydx
=11d
dxx
1
2=11.1
2x
12
6!
dy
dx =
11
2 x
1
2
=
11
2x
6.
*5( y=x7=x7
2
6 dydx
=d
dxx
7
2=7
2x
7
21
6 dydx
=7
2x
5
2
6
*,( y=6 3
r=6 r
1
3
6 dy
dx=6
d
dxr
1
3=6( 13 )r1
31
=6
3r
23
6dy
dx=2( 1r23)=
22r2
6
+%( y=48x2=4x
2
8
6 dy
dx
=4 d
dx
x2
8=( 4 )
(
2
8
)x
2
81
aa dy
dx=
8
8x
34 =x
34 =
14x3
a6+1( f(x )=x4
a df(x )
dx =
d
dxx
4=4x41
a$ df(x )
dx =4x5
a
+2( f( s)=2 s3
a@ df( s )
dx =2
ddx
s3=(2)(3)s31
ag df( s )
dx =6 s4
a8+*( f(x )=x3+x52x6
ai df(x )
dx =
d
dxx
3+d
dxx
52d
dxx
6=3x315x51(2 ) (6 )x61
a df(x )
dx =3x45x6+12x7
aX
-
7/26/2019 Limites, derivadas
153/214
++( f(x )=100x3+10x1
2
a7 df(x )
dx =100
d
dxx
3+10d
dxx
1
2=(3 ) (100)x31+(12 )(10 )x1
21
a0
df(x )dx =300x
3
+5x
12
an
+&( y=1
x=x1
a# dy
dx=
ddx
x1=1x11
a
-
7/26/2019 Limites, derivadas
154/214
6 dy
dx=2x3
&1( f( t)=1
2t=
1
2t
1
6@
df(t)
dx =
1
2
d
dxt
1
=(1
2
)t
11
6g df(t)
dx =
12
t2
,)hP
&2( g (x )=7
9x
1
6i dg (x )
dx =
7
9
ddx
x1=(79 ) (1)x116
dg (x )
dx =
7
9 x2
6X
&*( f(x )=1
7x+7x1
67 df(x )
dx =
1
7
d
dxx+7
d
dxx
1=1
7+(7 ) (1 )x11
60 df(x )
dx =
1
77x2
6n
&+( (x )=1
3
x33x3
6# d (x )
dx =
1
3
ddx
x3
3ddx
x3=( 13 )(3)x31(3 ) (3 )x31
6
-
7/26/2019 Limites, derivadas
155/214
c=8 De=e$ N 10
#37) Pr"3le0as $$
-
7/26/2019 Limites, derivadas
156/214
,,rP
. df(x )
dx =
ddx
(9x2 )= (9 ) ddx
x2=9.2x21=18x
5( y=4x3
,,tP
dydx
=ddx
(4x3 )=4 ddx
x3=4.3x31=12x2
,,&P,,P
,( g (= )=8 =7
,,@P
dg (= )
d= =
d
d=(8 =7 )= (8 ) d
d==
7=8.7x71=5x6
,,P1%( v (x )=xe
,daP
$6 dv (x )
dx =
d
dx(xe)=xe1
,d,P11( y=2/3x4
,ddP
$ dy
dx=
ddx( 23x4)= 23 ddxx4= 23.4x41= 83x3
,d:P12( f(%)=3%4
,d.P
$8 df(%)d%
=dd%
( (3 )1
2%4)= (3 )
1
2 dd%
%4=4. (3)
1
2%41=43%
3
,diP
1*( f( t)=t7
25$
$X df(t)
dt =
ddt(t
7
25 )= 125(7 )(t71 )=725t6
,d(P,dP
,dnP,d*P,d+P
-
7/26/2019 Limites, derivadas
157/214
1+( y=x
7
7$B
$! dy
dx=
d
dx(1
7> x
7
)=
1
7
d
dxx
7=1
7>
7
1x
71
=x6
$.
1&( f(x )=x+3$
$ df(x)
dx =
ddx
(x+3)=ddx
x +ddx
3=1+ 0=1
$
1'( f(x )=3x2$
$ df(x)
dx =
ddx
( 3x2)=3 ddx
xddx
2=3.1+ 0=3
$
13( f(x )=4x22x+3$
a
df(x)
dx =
d
dx(4x2
2x +3 )=4 d
dxx
2
2d
dxx+
d
dx 3=4.2x1
2.1+0=8x2
6 df(x)
dx =2( 4x1)
15( f(x )=5x29x$
df(x)
dx
=5d
dx
x29d
dx
x=5.2x19.1=10x9
@
1,( g (% )=%43%31g
8
dg(%)d%
=dd%
(%43%31)=dd%
%4 3
dd%
%3
ddx
1=4%33.3%20=4%39%2
i
-
7/26/2019 Limites, derivadas
158/214
X
2%( f( t)=13 t2+14 t+17
0df(t)
dt =
d
dt(13 t2+14 t+1)=13
d
dtt
2+14d
dtt+
d
dt1=13.2 t1+14.1+0=26 t+14
n
21( y=x3x#
-
7/26/2019 Limites, derivadas
159/214
@$ f(x )=ddx
(262x4)=ddx
262ddx
x4=02(4)x3=8x3
,:eP
-
7/26/2019 Limites, derivadas
160/214
&&( f(x )=9x1
3+5x2
5
,:.P df(x)
dx =
d
dx
(9x1
3 +5x2
5 )=9 ddx
x1
3 +5d
dxx
25 =9
(1
3
)(x
1
31)+5
(2
5
)(x
25
1)
,:hP df(x)
dx =3x
23 2x
75
,:iP
&'( f(! )=3!1
41228!3
4
@
,:0P
df(!)
d!
=d
d!
(3!1
41228!34 )=3d
d!
!
1
4d
d!
1448d
d!
!34 =3
(
1
4
)(!
1
41)08
(
3
4
)(!
341)
,:(P df(!)
d! =
3
4!
34 +6!
74
,:P
&3( & (x )= 138x2
= 1
38
3x2
= 1
23x2
= 1
2x2
3
=1
2x
23
@n
,:*P d&(x )
dx =
1
2
ddx
x2
3 =1
2 (23)x2
3 1
=1
3 x
53
,:+P
&5( f(x )= 34x3
=3x3
4
@B df(x )
dx =
d
dx(3x
34 )=3 d
dxx
34 =3 (34)x
34 1
=94
x
74
@!#s) Pr"3le0a $$#6),:3P Di:eren,ie (*s +r*)(eas de( 1- a( #!#7)
1-' y=(x21 ) (3x36x+5 )4 ( 4x2 +2x+1 )=(3x59x311x22x9)
,:P dy
dx=
ddx
(3x59x311x22x9 )
,:@P dy
dx=[3 ddx (x5 )9 ddx(x3 )11dydx(x2 )2 ddx(x )ddx ( 9)]
,:9P dy
dx=[3 (5x 4 )9 (3x2 )11 (2x )2 (1 )(0)]
,:P dy
dx=15x427x222x2
,.aP,.)P
-
7/26/2019 Limites, derivadas
161/214
1!' 6 (x )=4 (x53 )+3 (8x25 )(2x+2 )=(4x5+48x3+48x230x42)
,.,P d6 (x )
dx =
d
dx( 4x5+48x3+48x230x42 )
d6 (x )
dx
=
(4dy
dx
x5+48
dy
dx
x3+48
dy
dx
x230
dy
dx
xdy
dx
42
)
d6 (x )
dx
=(20x4+144x2+96x30 )
,.dP
1&( f(%)=3
2(5%2)(3%1)
,.eP df(%)
d% =
3
2
d
d%[(5%2)(3%1 )]
df(%)d%
=3
2 [ (5%2) dd%(3%1 )+(3%1 ) dd%(5%2 )]df(%)
d%
=3
2
[(5%2)
(3
d
d%
%d
d%
1
)+ (3%1)
(5
d
d%
%d
d%
2
)]df(%)d%
=3
2 [ (5%2 ) (3 )+ (3%1 )(52%12 )] df(%)d% =32 [ (15%6 )+(3%1 )(152 %
1
25
2%
12 )]
,.:P
1'( g (x )=(x+5x2 )( 3x3x ) dg (x )
dx =
ddx
[ (x+5x2 ) (3x3x ) ]
dg (x)dx
=[ (x +5x2 ) ddx(3x3 x )+(3x3x ) ddx(x +5x2 )]dg (x)
dx = (x+5x2 )(d
dx
3
x3d
dxx )+(3
x3x ) (d
dxx+5d
dxxd
dx2)dg (x )
dx =[ (x+5x2 )(13x
23
3
2x
12 )+(3x3x )(12x
12 +5)]
dg(x )dx
=1
6(135x
1
2 +40x1
3+5x1
6 +18x1
3 4x2
3 18)
,..P
13( y=7
(
2
3
)=
14
3
dy
dx
=d
dx[14
3
]=
d
dx
14
3
=0
15( y=(x1 ) (x2 ) (x3 )=(x36x2+11x +6) dy
dx=
ddx
(x36x2+11x+6)
dydx
=ddx
x36ddx
x2+11ddx
x+ddx
6=3x212x +11
g8c!i De=e$ N 11
,+aP#43) Pr"3le0as $$#4#),+dP Di:eren,ie (*s +r*)(eas de( 1 a( !#4e)
$# f(x )=(4x+1)(6x+3)
,+:P df(x)
dx =( 4x+1 )
d
dx(6x+3 )+(6x+3)
d
dx(4x+1)
,+.P df(x)
dx =( 4x+1 )(6 ddxx+ ddx3)+(6x+3)(4 ddxx+ddx1)
,+hP df(x )
dx =(4x+1 ) (6+0 )+ (6x+3 ) ( 4+0 )
,+iP df(x)
dx =( 4x+1) (6 )+(6x+3)(4)
,+;P df(x)
dx =24x+6+24x+12=48x+18=6 (8x+3 )
,+0P2( f(x )=(3x1 ) (7x+2)
,+(P
df(x)
dx =(3x1 )
d
dx(7x+2 )+(7x+2)
d
dx(3x1)
,+P df(x )
dx =(3x1 )(7 ddxx+ ddx2)+(7x+2 )(3 ddxxddx1)
,+nP df(x)
dx =(3x1 ) (7+0 )+(7x+2)(30)
,+*P df(x)
dx =(3x1 ) (7 )+(7x +2)(3)=21x7+21x+6=42x1
,++P*( s (r )=(53 r )(r32 r2 )
,+
-
7/26/2019 Limites, derivadas
183/214
-
7/26/2019 Limites, derivadas
184/214
&( / (x )=(3+x )(5x22)
,+@P d/(x)
dx =(3+x ) d
dx( 5x22 )+(5x22)d
dx(3+x)
,+9P
d/(x)dx =(3+x )(
5
d
dxx
2
d
dx2
)+(5x
2
2)(
d
dx3+
d
dxx )
,+P d/ (x )
dx =(3+x )(5 (2x )0 )+ (5x22 ) ( 0+1 )
,
-
7/26/2019 Limites, derivadas
185/214
5( f(x )=x2(2x25)
,
-
7/26/2019 Limites, derivadas
186/214
12( (x )= (35x+2x2 )(2+x4x2)
,rP d(x )
dx = (35x+2x2 ) d
dx(2+x4x2 )+( 2+x4x2 ) d
dx(35x+2x2 )
,rnP d(x )
dx = (35x+2x2
)(d
dx2+d
dxx4d
dxx
2
)+(2+x4x
2
)(d
dx 35d
dxx+2d
dxx
2
),r*P
d(x )dx
= (35x+2x2 )(0+14 (2x ))+(2+x4x2) (05+2(2x))
,r+P d(x )
dx = (35x+2x2 )(18x )+(2+x4x2) (5+4x )
,r
-
7/26/2019 Limites, derivadas
187/214
2%( y=2x34x+1
dy
dx=
( 4x+1 )d
dx(2x3 )(2x3)
d
dx(4x+1)
(4x+1 )2
dydx
=
( 4x+1 )
(2
d
dxx
d
dx3
)(2x3 )
(4
d
dxx+
d
dx1
)(4x+1 )2dydx =
( 4x+1 ) (2 ) (2x3 ) (4 )
(4x+1 )2
.i dy
dx=
(8x+2 )(8x12 )
( 4x+1 )2
. dy
dx=
14
( 4x+1 )2
#s?)
21' f(x )= 5x
x1df(x )
dx =
(x1 )ddx
(5x )(5x )ddx
(x1 )
(x1 )2
,s(P df(x )dx
=
(x1 )(5 ddxx )(5x )(ddxxddx1)(x 1 )2
df(x )dx
=(x1 ) (5 )(5x ) (1 )
(x1 )2
,sP df(x )
dx =
(5x5)(5x )
(x1 )2
,snP df(x )
dx =
5
(x1 )2
,s*P
22' ;(x )=5x5x
d;(x )dx
=(5x )
d
dx(5x )(5x )
d
dx(5x )
(5x )2
,s+P d;(x )dx
=
(5x )(5 ddxx )(5x )( ddx5ddxx)(5x )2
d;(x )dx
=(5x ) (5 )(5x ) (1 )
(5x )2
,s
-
7/26/2019 Limites, derivadas
188/214
2*( f(x )=13
3x5=
133 (1x5 ) df(x )dx =133[ddx ( 1x5 )]=133(
x5ddx
(1)(1 )ddx
(x5 )
x10 ).
df(x )
dx =
13
3
(x
5 (0 )(1 )(5x4 )
x10
) df(x )
dx =
13
3
(5x4
x10
). df(x)dx
=13
3(5x6 )! 65
3x6
,sP
2!' f(x )=5 (x22 )
7 =
5
7(x22 )
df(x )dx
=5
7 [ddx(x22 ) ]= 57 (ddxx2ddx2),s@P
df(x )dx
=5
7(2x )=10
7 (x )
,s9P
df(x)dx =
10
7 x,sP
25' y=x +2x1
a dydx
=(x1 )
dydx
(x+ 2 )(x +2)dydx
(x1)
(x1)2
6 dydx =
(x1 )( dydxx+ dydx2)(x+2 )( dydxx dydx1)(x1 )2
dy
dx=
(x1 ) (1)(x+2)(1)
(x1)2
$ dy
dx=
x1x2
(x 1 )2 =
3
(x1 )2
@
-
7/26/2019 Limites, derivadas
189/214
2'( 6 (= )=3 =
2+5 =1=3
ctg% d6(=)
d= =
dd=( 3 =
2+5 =1=3 )
cth% d6(=)d=
=(=3) d
d=(3 =2+5 =1 )(3 =2+5 =1) d
d=(=3 )
(=3)2
cti% d6(=)d=
=
(=3)(3 dd==2
+5 dd=
= dd=
1)(3 =2+5 =1 )( dd== dd=3)(=3)2
ct'% d6(=)
d= =
(=3) (6 =+5 )(3 =2+5 =1 )(1 )(=3)2
ct%
d6(=)
d= =
6 =2+5 =18 =153 =25 =+1
(=3)2
ctl% d6(=)
d= =
3 =218 =14
(=3)2
,tP,tnP,t*P
23( 6 (! )=62!
!24
,t+P
!
( 24)2
d6(!)d!
=(!24 ) d
d!(2!+6 )(2!+6)
d
d!(!24)
,t
-
7/26/2019 Limites, derivadas
190/214
,t3P,t&P
-
7/26/2019 Limites, derivadas
191/214
25( !=2x
2+5x2
3x2+5x+3
ctw% dy
dx=
ddx
(
2x2+5x2
3x
2
+5x+3 )ct!% dydx
=(3x2+5x+3 ) d
dx(2x2+5x2)(2x2+5x2)d
dx(3x2+5x+3)
( 3x2+5x+3 )2
cty% dydx
=
(3x2+5x+3 )(2 ddxx2
+5d
dxx
d
dx2) (2x2+5x2)(3 ddxx
2
+5 d
dxx+
d
dx3)
(3x2+5x+3 )2
ct$% dy
dx=
(3x2+5x+3 ) ( 4x+5 )(2x2+5x2) (6x+5 )
(3x2+5x+3)2
cua% dydx
=12x3
+15x2
+20x2
+25x+12x+1512x3
+10x2
+30x2
25x+12x+10(3x2+5x+3 )
2
cub% dy
dx=
55x2+ 24x+25
(3x2+5x+3 )2
cuc%cud%
02. y=3x
2x13x
=3x
2x1
x
1
3
,3eP dydx
=(x
13 )ddx ( 3x2x1 )(3x2x1) d
dx (x13)
(x1
3)2
,3:P dydx
=
(x1
3 )(3 ddxx2
x d
dxx
d
dx1)(3x2x1)( ddxx
1
3)(x
1
3)2
,3.P dydx =
(x1
3 ) (6x10 )(3x2x1)( 13
x
23 )
3
x2
,3hP dydx
=
63x7 3x
3x4+
1
3
3x+
1
33x2
3
x2
,3iP dydx
=
63x7
2
3
3x
3x+
1
33x2
3
x2
cu'%cu%
-
7/26/2019 Limites, derivadas
192/214
*5( f(x )=x
0,32
2x2,1+1
,3(P df(x)
dx =
(2x2,1+1 ) ddx
(x0,32 )(x0,32) ddx
(2x2,1+1)
(2x2,1+1 )2
,3P df(x)dx
=
(2x2,1+1 )( ddxx0,3
d
dx2)(x0,32)(2 ddxx
2,1
+d
dx1)
(2x2,1+1 )2
,3nP(2x2,1+1 )
df(x)dx
=(2x2,1+1 ) (0, 3x0,70 )(x0,32)(4,2x1,1+ 0)
2
,3*P df(x)
dx =
0,6x1,4+0,3x0,74,2x1,3+8,4x1,1
(2x2,1
+1 )
2
,3+P df(x )
dx =
0,3 (1+28x1,812x2,1)
(2x2,1+1)2
cu3%#=r)#=s) Pr"3le0a $$2#=6),33P Uti(i,e (a re.(a de (a ,adena en (*s +r*)(eas de( 1 a( cu4%
1( y=u2 2u y u=x2x encuentre
dy
dx
cuw%
u
( 22 u)=ddu
u22ddu
u=2 u2
dydu
=ddu
cu!%
cuy% du
dx=
ddx
(x2x )=ddx
x2
ddx
x=2x1
cu$%
c4a% dydx
=(2u2 ) (2x1 )=(2(x2x)2)(2x1 )=(2x22x2 )(2x1 )
c4b% dy
dx=4x32x24x2+2x4x+2=4x36x22x+2
c4c%
(. y=2 u38u y u=x3+7xencuentre
dy
dx
c4d% dy
du=
ddu
(2u38 u)=2ddu
u38
ddu
u=6 u28
c4e%
c4&% dudx
=ddx
(x3+7x )=ddx
x3
+7 ddx
x=3x2+7
-
7/26/2019 Limites, derivadas
193/214
c4g%
c4h% dy
dx=(6 u28 )(3x2+7)
c4i% dy
dx=(6(x3+7x)28 )(3x2+7)=2(147126x327x64)(73x2)
c4'%c4%
-
7/26/2019 Limites, derivadas
194/214
*( y=1
u3y u=2x encuentre
dydx
c4l%
c4m%
u
( 3)2
dy
du=
d
du (1u3 )=(u3 ) d
du(1 )(1 )
d
du(u3 )
c4n%
u
( 3)2
dy
du=
(u3 )( ddu 1) (1 )( dduu3)
c4o%
u
( 3)2
dydu
=(u3 )(0 )(1 )(3 u2 )
c4p% dy
du=
3 u2
u6
c43%
c4r% du
dx=
ddx
(2x )=ddx
2ddx
x=1
c4s%
c4t% dydx =(3
u4
) (1 )=3u4 = 3(2x ) 4
c4u%c44%
+( y=4! y !=x
3x4+3 encuentredydx
c4w% y=!1 /4=dd!
(! )1
4
c4!% dy
d!=
1
4!
34
c4y%
c4$% d!
dx=
ddx
(x3x4 +3 )=ddx
x3
ddx
x4 +
ddx
3
cwa% d!
dx=4x3+3x2
cwb%
cwc% dy
dx=( 14!
34 ) (4x3+3x2 )=[14(x3x4+3)43] (4x3+3x2 )
cwd%
cwe%
-
7/26/2019 Limites, derivadas
195/214
&( c==3 +=+ 9y ==2x
21 encuentred=dx
@ df( c )
d= =
d
d=(=3+=2
1+9 )=( dd==3
+ d
d==
2
1
+ d
d=9)
cwg% df( c )
d= =3 =2+
1
2=2
1
cwh%
cwi% d=
dx=
ddx
(2x21 )=(2 ddxx2
ddx
1)=4xcw'%
cw% dydx
=(3 =2+ 12=21) ( 4x )=[3 (2x21)2+ 12 ( 2x21 )1
2 ] (4x ),(P
dy
dx=
(6x
43+x11
2
12
)(4x)
,P dy
dx=24x512x+4+
1
2
1 /2
cwn%
'( !=u2 +u+9y u=2x
21 encuentred!ds
cuando s=1
# d!du
=d
du(u2+u
1
2+9)=(ddu u2+ ddu u1
2+d
du9)
cwp% d!
du=2u +
1
2u
1/ 2
cw3%
cwr% du
dx=
d
dx(2x21)=(2 ddxx
2
d
dx1)
cws% du
dx=(2
ddx
x2
ddx
1)=4x
cwt%
cwu% dy
dx=(2u+ 12 u
12 ) (4x )
cw4%
cww% du
dx=[2 (2x21 )+ 12(2x21 )
12 ] (4x )
cw!% du
dx=(4x22+x11)(4x )
cwy% dy
dx=16x38x+3
cw$% du
dx=16 (1 )38 (1 )+3=10
c!a%
c!b%
-
7/26/2019 Limites, derivadas
196/214
3( y=3 u28u +4y u=2x2 +1 encuentre
dydx
cuando x=0
dy
du=
ddu
(3 u2 8 u+4 )
$ dydu
=(3 dduu28 d
duu+ d
du4)
c!e% dy
du=6 u8
c!&%
c!g% du
dx=
d
dx(2x2+1)=(2
d
dxx
2
+d
dx1)=4x
c!h%
c!i% dy
du=(6 u8 ) (4x )=[6 ( 2x2+ 1 )8 ] ( 4x )
c!'% dydu=(12x2+68 )( 4x )
c!% dy
du=48x28x
c!l% dy
dx=48( 0)28 (0 )=0
c!m%
5( y=3 u3u2+ 7 u y u=5x 2 encuentre
dydx
cuando x=1
n
dy
du=
d
du (3 u
2 u2+ 7 u)
# dy
du=(3 dduu2ddu u2+7 dduu)
c!p% dy
du=9 u22 u+7
c!3%
c!r% du
dx=
d
dx(5x2)=5
d
dxx
d
dx2=5
c!s%
c!t%
5x2
9 (5 )
dydx
=(9 u22 u+7 )(5 )=
,@3P dy
dx=(225x23210x+4+7 )(5 )
,@&P dy
dx=1125x250x105=1125(1)250 (1)105
,@P#99)
-
7/26/2019 Limites, derivadas
197/214
#9.) Pr"3le0a $1
-
7/26/2019 Limites, derivadas
198/214
,9tP,93P
-
7/26/2019 Limites, derivadas
199/214
13(x
2
(+4 )y=x2+log2
,9&P dydx
=x2+
dy
dx(x2
+4 )(x2+4 )
. 1
ln2
dy
dx=x2+
2x
x2+ 4
. 1
ln 2
,9@P dy
dx=x2+
2x
(x2+4 )(ln 2)
15( y=x2 log2x
,9P dy
dx = 1
ln 2 (x2
lnx )
a dy
dx=
1
lnx[x2(1x )+lnx (2x ) ],)P
dydx
= xln 2
(1+2 lnx )
,,P
1,( f(!)=ln!
!
,dP
df(! )d! =
!
!
d
d!ln!=
!
!.
1
!.1
,eP df(! )
d! =
!
!2
,:P
2%( y=x
2
lnx
g dydx
=ddx( x
2
lnx )
8
x
ln
dy
dx=
(lnx ) ddx
(x2)(x2 ) ddx
( lnx )
i
xln
dydx =
(lnx ) (2x )(x2)1
x
-
7/26/2019 Limites, derivadas
200/214
,;P
x
ln
dy
dx=2xlnxx
,0P
-
7/26/2019 Limites, derivadas
201/214
!$' y=ln2 (2x+11 )=[ ln (2x+11)]2
,(P dydx
=ddx
[ ln (2x+11)]2
.ddx
[ln (2x+11 )]
,P dy
dx=2 [ ln (2x+11)]
2. [ ln (2x+11 )]
,nP dy
dx=2 ln (2x+11) [ 2+0(2x+11)]
,*P dy
dx=2 ln (2x+ 11)( 22x+11 )
,+P dy
dx=
4 ln (2x +11 )2x+11
,
-
7/26/2019 Limites, derivadas
202/214
!!' y= ln (x+1+x2 )=ln [x+(1+x2 )1
2 ]= lnx+12ln (1+x2 )
da.P dy
dx=
ddx
lnx +1
2
ddx
ln (1+x2 )
dahP dy
dx =1
xddx1+
1
2 ( 1
1+x2 )ddx(1+x
2
)
daiP dy
dx=
1
x+
2
2(1+x2)
da;P
-
7/26/2019 Limites, derivadas
203/214
%' De=e$ N 12
da(PdaP En,3entre 3na e,3a,i6n de (a re,ta tan.ente a (a ,3r&a en e( +3nt* dad*danP
y=x+ 5x
2 * (1,6)
da*P y=x+5
x2
da+P
xx
( 2)2
(x2 ) d
dx
(x+5 ) (x+5 )
ddx
( 2)
dydx
=
da
-
7/26/2019 Limites, derivadas
204/214
d).P Gr4:i,ad)hP
d)iP y=x+5
x2
y=11x+17
d);P
d)0P @ d)(P 9d)P - d)nP $22d)*P 2 d)+P $75d)
-
7/26/2019 Limites, derivadas
205/214
y=( x+1x2 (x4 ))*(2,3
8 )d,,P
d,dP y=
(
x+1
x
3
4x2
)d,eP dy
dx=
(x34x2 ) ddx
(x+1 )(x+1 )ddx
(x34x2 )
(x34x2 )2
d,:P dy
dx=
(x34x2 )(1 )(x+1 )(3x28x )
(x34x2 )2
d,.P dy
dx=
(x34x2 )(3x38x2+3x28x)
(x34x2 )2
d,hP
dy
dx=
(x34x2 )(3x35x28x )
(x34x2 )2
d,iP dy
dx=
x34x23x3+5x2+8x
(x34x2 )2
d,;P dy
dx=
2x3+x2+ 8x
(x34x2 )2
d,0Pd,(P
d,P y=m=
2x3+x2+8x
(x34x2 )2 =
2 (2 )3+(2 )2+ 8 (2 )
[ (2 )34 (2)2 ]2
d,nP m= 4
64=
1
16
d,*Pd,+Pd,
-
7/26/2019 Limites, derivadas
206/214
d,&P Gr4:i,ad,P
d,@P y=( x+1x2 (x4 ))y= 1
16x
1
2
d,9P @ d,P 9ddaP 1 dd)P $
dd,P $5 dddP $!!ddeP $5 dd:P 171
dd.P 1 ddhP $#7ddiP 2 dd;P $-#
dd0P -5 dd(P $7-ddP !- ddnP $"#
dd*P # dd+P $$"
dd
-
7/26/2019 Limites, derivadas
207/214
f(x )=2x13x+5
* x=1
ddtP
dd3P f(x )=2x13x+5
dd&P dydx
=(3x+5 ) d
dx(2x1 )(2x1 ) d
dx(3x+5)
(3x+5 )2
ddP dy
dx=
(3x+5 ) (2 )(2x1 ) (3 )
(3x+5 )2
dd@P dy
dx=
(6x+10 )(6x3 )
(3x+5 )2
dd9P dy
dx=
6x+106x+3
(3x+5 )2
ddPdeaP
de)P y=2x13x+5
=2 (1 )13 (1 )+5
=1
8 (1, 18 )de,PdedPdeeP yy 1=m (xx1 )
de:P y1
8=
13
64(x1 )
de.P y=13
64x
13
64+
1
8
dehP y=13
64x 5
64
deiP
-
7/26/2019 Limites, derivadas
208/214
de;P Gr4:i,a
de0P f(x )=2x13x+5
y=13
64x
5
64
de(P
deP @ denP 9de*P - de+P 175de
-
7/26/2019 Limites, derivadas
209/214
f(x )=3+ 0
24x* x=0
d:dP
d:eP y=3+ 0
24x=3 (0,3 )
d::Pd:.P
d:hP f(x )=3+ 0
24x
d:iP dydx
=d
dx3+
(24x )d
dx( 0 )(0 )
d
dx(24x )
(24x )2
d:;P dy
dx=
(24x ) (0 )(0 ) (4 )
(24x )2
d:0P
dy
dx =
0
(24x )2
d:(P dy
dx=0
d:Pd:nPd:*P f(x )=0d:+Pd:
-
7/26/2019 Limites, derivadas
210/214
d:@P Gr4:i,ad:9P
d:P y=3+ 0
24xy=3
d.aP
d.)P
d.,P
-
7/26/2019 Limites, derivadas
211/214
x3+xy+y2=1 * (1,1 )
d.dPd.eP x3+xy+y2=1
d.:P
x
d
dx (3+xy+y2
)=
d
dx1
d..P 3x2+x
dydx
+y+2y+dydx
=0
d.hP dy
dx=3x2y
x+2yd.iPd.;P
d.0P y =m=3 (1 )2(1 )
1+2 (1 )
d.(P m=311d.P m=4d.nPd.*Pd.+P yy 1=m (xx 1 )d.
-
7/26/2019 Limites, derivadas
212/214
d.&P Gr4:i,ad.Pd.@P x3+xy+y2=1y=4x3d.9P
d.P @ dhaP 9dh)P - dh,P #2> -2dhdP 2 dheP 12> -2dh:P 1 dh.P $> 1
dhhP
;P @ ;nP 9;*P $ ;+P -
;
-
7/26/2019 Limites, derivadas
213/214
dhiP
dh;P
-
7/26/2019 Limites, derivadas
214/214
+< BIBLIO4RAF1A
dh(P Ar*((* ' 2- de $ de 2$12P' economia W..O)tenid* de htt+%%'e,*n*ia's%de)eha)er'+h+dhP Car(*s ' 2- de 3ni* de 2$1!P'!"#.O)tenid* de
htt+%%'eh3'e3s%;3an,ar(*s'.*r*stia.a%a+*9*%(iites'htdhnP :,a'3n('ed3'a' 2- de O,t3)re de 2$$7P' #$%' O)tenid* de htt+%%':,a'3n('ed3'ar%Liite%2'2Z2$L
ZEDites(atera(es'htdh*P air* L' $5 de Di,ie)re de 2$1$P' Wikipendia.O)tenid* de htt+s%%es'i0i+edia'*r.%i0i%L
ZC-ZADite[ateZC-ZA1ti,*dh+P Leana ?' 25 de A.*st* de 2$1!P' &it'tor' O)tenid* de htt+%%'&it3t*r',*%:3n%!%)[1'ht(dh