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


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(',*


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('

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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


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'


5/214


$' DATOS INFORMATIVOS


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


6/214


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


7/214


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'


8/214


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


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'


9/214


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'


10/214


rg/

COMPLEMENTARIA

7



11/214


DATOS INFORMATIVOS DE LA UNIDAD CURRICULAR N"# 8NOMBRE DE LAUNIDAD:

DIFERENCIACIN DOS


Ana7ia! .iai#n. 0

RESULTADOS DEAPRENDI&A%E DE LAUNIDAD:

I$nia 7#. in!a7#. $ ganania. E


12/214


METODOLO41AS DEAPRENDI&A%E:

"eterminar algoritmos y elaborar grficas, es I$nia a!ia67. n aii$a$. 0

A


13/214


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'


14/214


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

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15/214


!$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


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


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


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


107/214

adP

aeP


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


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


110/214

'6% De=e$ N 5

lim

x 2

16=16

a@eP

a@:P Gr4:i,a

[email protected]

a@hP

a@iP

a


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@


112/214

a@@P

a@9P

a@P

a9aP

a9)P

a

a9dP

a9eP


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


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


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 $,


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


117/214

bbi%

bb'%


118/214

limx

3

x

66X

limx

x

12

3

x=

limx

3

limx

))(P Gr4:i,a

))P))nP))*P))+P))


119/214

),ePGr4:i,a

),:P),.P),hP),iP),;P),0P

),(P),P),nP),*P),+P),


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


121/214

limx

7

2x+1

)d*P

limx

7

2x+1=

7

2 ( )+1=0

)d+P)d


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


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


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)


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 ).


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


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


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


129/214

dP /3= (e s3,eder4 a( +re,i* a (ar.* +(a* Gra:i


130/214

3l#)

)(dP


131/214

=e De=e$ N /

3l)Pr"3le0as $$


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


133/214

6n

6n


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#


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


136/214

22( Enn! 7a


137/214

25' y=x2+2x+3 * (1,6 )

)


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!


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.


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


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


142/214

=8' De=e$ N

3=3) Pr"3le0as $$


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


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


145/214

21( y=x3x6n

6# dydx

=ddx

(x3x1

2 )=ddx

x3ddx

x1

2=3x21

2x

12

6


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


147/214

67

60


148/214

25( y=2

2x2

6n y=2

2 x

2=x2

6#

dy

dx=

d

dxx2

6


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


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


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


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


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


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


155/214

c=8 De=e$ N 10

#37) Pr"3le0as $$


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


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


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#


159/214

@$ f(x )=ddx

(262x4)=ddx

262ddx

x4=02(4)x3=8x3

,:eP


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


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 )

,+


183/214


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 )

,


185/214

5( f(x )=x2(2x25)

,


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


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


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

@


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


190/214

,t3P,t&P


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%


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


193/214

c4g%

c4h% dy

dx=(6 u28 )(3x2+7)

c4i% dy

dx=(6(x3+7x)28 )(3x2+7)=2(147126x327x64)(73x2)

c4'%c4%


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%


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%


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)


197/214

#9.) Pr"3le0a $1


198/214

,9tP,93P


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


200/214

,;P

x

ln

dy

dx=2xlnxx

,0P


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

,


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


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


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)


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,


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


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


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


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:


210/214

d:@P Gr4:i,ad:9P

d:P y=3+ 0

24xy=3

d.aP

d.)P

d.,P


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.


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 -

;


213/214

dhiP

dh;P


214/214

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Limites, derivadas

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