CÁLCULO DE DERIVADAS

1 Halla la función derivada de las siguientes funciones: 2

3

3 2

32

3 2

2

2

2

a) y 6x 61 1b) y +

2 x 3 x3c

y 3x 6

) y2x x

d) y 6x 6x1e) y33x 1f)

x 5

y x x

1yx x

y 2x 3

y 3x2

x 6xy 23x 3 xy x2 2 22

= − +

= +

=

= +

′ = −

′ =

−′ =

′ = +

′ =

′ = + −

−

= +

= + −

Sol.

Sol.

Sol.

Sol.

Sol.

Sol.

2 Halla la función derivada de las siguientes funciones:

( )( )

( )

( ) ( ) ( )

( )

2 2

x

x

2

x

x

2

52

2

2

3

4 52

2

22

a) y cos x sen x cos 2xb) y e 1 x

c) y 2 1 x ln 2

x 1d) y 2x ln xx

1e) y

y sen xcos

15x 1 log x

xy x e

y x 2

y x 1 ln x

y 3x x l 3x xx ln2

4xf) y

og x

x 1yx 1

x 3x 5x 3yx

x 1

g)

′ = − =

′ = +

′ = +

+′ = +

′ = − +

=

=

=

= +

= − −

−′ =+

=−

+

−

+ −=

Sol.Sol.

Sol.

Sol.

Sol.

Sol.

S3 2

2

2 2 2

2x 3x 3yx

1 logx loge logx 1 eln10h) y llo ogxx x x

gxyx

+ −′ =

− −′ == = =

ol.

Sol.

3 Halla la función derivada de las siguientes funciones: 3

3 2

1 5a) y2x 3 25x

y 2x 5x ′ = += + Sol.

( ) ( ) ( )22b) y 2x 5 cos x 5x 7y sen x 5x 7 ′ = −− + += −Sol.

( )233

10c) y 5 y3 5 3

x 3x

′ =+

= + Sol.

( ) ( ) ( )y sen 3x 1 cod s 3x 1) y 3cos 6x 2= + + ′ = +Sol.

2

2 2

2

2

2 2logxlogxy2logx 2 eln10 ln10e) y lo

xx x xxg

− −= ′ = = =Sol.

( ) ( )f) y 3sen 3xy cos 3x ′ = −− −= ππ Sol.

y 1 2x 1g) y1 2x

+ ′= =+

Sol.

( )22x 1 x 1h) y e 1 2e xy x + +′ = += Sol.

1/7

( ) ( ) ( ) ( )( )

2 22

2

2

2 2

2x 1 x cos x 1 xsen x 1i) y

1 x 1 x

sen x 1y

1 x

− + + +′ =

−

+=

−−Sol.

( ) ( ) ( )j) y sen x x cosx sen x xy ′ = − −= −− π ππ Sol.

3 23

2

2k) 3x x

1yx

y −′ == Sol.

4 33xl) yx 52

6 52

3y ⎛ ⎞′ = −⎜ ⎟⎝ ⎠

⎛ ⎞= −⎜ ⎟⎝ ⎠

Sol.

( )2

3x 7 13m) yx

yx 2 2−

=+

′ =+

Sol.

( ) ( )x xn) y e ln cos x tgy e ln cos x x−− ′ ⎡ ⎤= − +⎣ ⎦= Sol.

2

1ñ) y2cos x tg

tgxx

y ′ == Sol.

2

x 1y lnx

1o) yx x

+⎛ ⎞= ⎜ ⎟⎝ ⎠

−′ =+

Sol.

( )

5 4

6

5xp) yxx 1

yx 1

′ =+

⎛ ⎞= ⎜ ⎟+⎝ ⎠Sol.

( ) ( ) ( )

23

23

4q) y3 1

1 x

x 1 x 1 xy

1 x−⎛ ⎞=

−′ =+

⎟+

⎜⎠ −+⎝

Sol.

2 2

2x 4yx

4r) yx x 4

′ =−

−= Sol.

( )2

x 1yx 1

1s) yx x 1

′−+

==+

Sol.

( ) ( )7 6y 2 x 3 7t) y 2 x 3

x′ = −= − Sol.

4 (Ejercicio resuelto) Halla la función derivada de las siguientes funciones:

a) 1 cy cos x; y2 sen x 2 se

sn

y xx

en ′ ′= → = ⋅ =os x

b) ( )22

2

1y cos xy sen x 5x 5x 1 2x 52 x 5x 1

1 ′→ = + − ⋅ ⋅+ −

= − ++

( ) 2

2

2x 5 cos x 5x 1y

2 x 5x 1

+ +′ =

+ −

−

c) ( ) ( )2 2 2x +3x x + xx 3x +3y y 5e 2x 3 ; y 5 25e x 3 e′ ′→ = ⋅ + = +=

d) ( ) 1 1 1y ; yln x x

y ln ln xx ln x

′ ′→ = ⋅ ==

e) ( ) ( )24 3

2 1y 4 x x sen x 2y x co2

x x senx

x ⎛ ⎞′→ = − + − += − ⎜⎝

+ ⎟s x⎠

2/7

f) 1 1y 1 x 1 x2 x 12 x

y xx 1

x 1 ⎛ ⎞′→ = ⋅ ⋅= + + + ⋅⎜ ⎟+⎝ ⎠+

( )( ) ( )4 3 2 2 243 2 4

2 x 1 x1 1 3x 2 1 3x 2y ; y2 x 1 2 x 12 x x 2 x x 1 2 x 12 x x

⎛ ⎞+ + + +′ ′= ⋅ = ⋅ = ⋅⎜ ⎟⎜ ⎟+ ++ +⎝ ⎠ ++

( )324

3x 2y4 x x 1

+′ =+

g) ( ) ( )

( ) ( ) ( )2 2

1 1 x 1 x 1 1 x 1 x 2y ;1 xy1 x 1 x 1 x 1 x

− ⋅ + − − ⋅ − − − + −′ ′→ = = =+ ++ +

−= 2y

h) ( ) ( )( )2

1 1 x 1 x 11 1y1 x

1 xy1 x 1 x2 2

1 x

− ⋅ + − − ⋅′→ = ⋅

− ++

+=

−=

21 x1 x

−⋅

−+

( )21 x+

( ) ( ) ( ) ( ) ( )4 32

1 1y1 x 1 x 1 x 1 x 1 x1 x1 x 1 x

− − −′ == = =− + − + −+ ⋅+ +

1

( ) ( ) ( ) ( ) 2

1 1y ; y1 x 1 x 1 x 1 x 1 x

− −′ ′= =+ + − + −

( ) ( ) ( )2 2

1 2 2y 1 x 1 x 1 x 1 x1 x

1 xy ln1 x

− −′→ = ⋅ =− + − ++

−=

+

1 x+

( ) ( ) 2

2 2; y1 x 1 x 1 x

− −′= =− + −

i)

De otra forma: Desarrollando el logaritmo previamente:

( ) ( ) ( )1 1 1 1y 1 .11 x 1 x 1

y ln 1 x ln 1 xx 1 x

−′→ = ⋅= − − + − − = −+ − +

−

( )( ) ( ) ( ) ( ) 2

1 x 1 x 1 x 1 x 2y ;1 x 1 x 1 x 1 x 1 x

− − − − − − − + −′ ′= = =− + − + −

y

j) ( ) ( )

( )2 2

2

1 tgx1 11 tgx 1 tgxcos x cos xy1 tgxy

1 tgx 1 tgx

− −−⋅ + − − ⋅

′→ = =+

−=

+

1 tgx− +

( )2

2cos x

1 tgx+

( )22

2ycos x 1 tgx

−′ =+

1 1 tgy D1 tgx1 tgx2

1

1 t

t

gxy1 tg

gxx

⎛ ⎞−−= ′→ = ⋅ ⎜ ⎟+− ⎝+

x⎠

+

k)

( )22

1 tgx 2D1 tgx cos x 1 tgx⎛ ⎞− −

=⎜ ⎟+ +⎝ ⎠ En el ejercicio anterior se calculó:

1y2

′ =2

1 tgx1 tgx

−⋅

−+

( ) ( ) ( )22 4

2

1cos x 1 tgx 1 tgx 1 tgx

cos x1 tgx

−=

+ + −+

3/7

( ) ( ) ( ) ( ) ( )232

1 1ycos x 1 tgx 1 tgx 1 tgxcos x 1 tgx 1 tgx

− −′ == =+ + −+ −

( )2 2

1ycos x 1 tgx 1 tg x

−′ =+ −

l) ( ) x 1x 1 x 1 x 1x 1 2

x 1x

x 1

1 x 12

1 1 3 ln3 3 ln3y 3 ln3 ln3 32 2 22 3 3

y3

3++ + + −+

++ +

+ ′→ = ⋅ ⋅ = ⋅ ⋅ = ⋅ = ⋅=

x 1x 12ln3 ln3y 3 ; y 3

2 2

++′ ′= ⋅ =

m) ( ) ( )2y log sen x cos x ; y 2 log sen x cos x= ⋅ = ⋅ ⋅

( )( )1 1y 2 cos x cos x sen x -sen xsen x cos x ln10

′→ = ⋅ ⋅ ⋅ ⋅ +⋅

( ) ( )2 22 2

2 cos x sen x2y cos x sen xln10 sen x cos x ln10 sen x cos x

−′ = ⋅ − =

⋅ ⋅ ⋅ ⋅

Teniendo en cuenta que: y que 2cos 2x = cos x - sen x2

→1sen 2x = 2 sen x cos x sen x cos x = sen 2x2

2 cos2x 2 cos2x 4 cos2x 4 cos2xy ;sen2x ln10 sen2x ln10 sen2x ln10 sen2xln10

2 2

⋅ ⋅ ⋅′ ′= = = = ⋅⋅ ⋅⋅

y

Teniendo en cuenta que: cos 2x 1= cotg 2x =sen 2x tg 2x

, se tiene:

4 4y cotg2x; yln10 ln10 tg2x

′ ′= ⋅ =⋅

De otra forma:

( ) ( )2 sen2xy log sen x cos x ; y 2 log sen x cos x ; y 2 log2

⎛ ⎞= ⋅ = ⋅ ⋅ = ⋅ ⎜ ⎟⎝ ⎠

1 1 1y 2sen2x ln10 2

2

′→ = ⋅ ⋅ ⋅ ( )cos2x 2⋅ ⋅4cos2x 4 cotg2x

ln10sen2x ln10= = ⋅

n) ( )2 2 y 2sen xcy sen x co os 2cos x s ns x x e x 1′→ = + −+ + += y 2sen x cos x 2sen xcos x 1; y 1′ ′= − + =

ñ) y sen x 1 cos x 1= + ⋅ −

( )1 1y cos x 1 cos x 1 sen x 1 sen x 12 x 1 2 x 1

′→ = + ⋅ ⋅ − + + ⋅ − − ⋅+ −

cos x 1 cos x 1 sen x 1 sen x 1y2 x 1 2 x 1+ ⋅ − + ⋅ −′ = −

+ −

o) ( ) ( )5 53 3y sen 3x 2 x 2x ; y sen 3x 2 x 2 x= − + = − + ⋅ 3

( )5 43 33 2

2 1y cos 3x 2 x 2x 15x 22 x 3 x

⎛ ⎞′→ = − + ⋅ − + ⋅⎜ ⎟

⎝ ⎠

( )3

4 5 33 2

1 2y 15x cos 3x 2 x 2xx 3 x

⎛ ⎞′ = − + ⋅ − +⎜ ⎟⎜ ⎟

⎝ ⎠

4/7

( )2

2 2

1 cy cos x 2x ; y2 sen x x 1 2 sen x x

y sen x1

x 1 +′ ′→ = ⋅ + =++ +

= +p) os x 2x+ +

q) ( ) ( )2 32 2 2 23 3y cos x 3 x ; y cos x 9 6x x ; y cos x 5x 9= + − = + − + = − 2 +

( )( )

( )3 32 2

223

1y 2cos x 5x 9 sen x 5x 9 2x 53 x 5x 9

′→ = − + ⋅ − − + ⋅ ⋅ −− +

( )

( )

3 32 2

223

2 2x 5 cos x 5x 9 sen x 5x 9y

3 x 5x 9

− − ⋅ − + ⋅ − +′ =

− +

( ) ( )( )

3 32 2

223

2x 5 2 sen x 5x 9 cos x 5x 9y

3 x 5x 9

− − ⋅ ⋅ − + ⋅ − +′ =

− +

( ) ( )( )

3 2

223

2x 5 sen 2 x 5x 9y

3 x 5x 9

− − ⋅ − +′= → =

− +sen2a 2 senacosa

( ) ( )( )

3 2

223

5 2x sen 2 x 5x 9y

3 x 5x 9

− ⋅ − +′ =

− +

r) ( )2 2

1 1 1 1 1y ;y arc y ; y2 x 1 x 2 x 2 x

ex

sx

n1

x ′ ′ ′→ = ⋅ = ⋅ =− −−

=

s) ( )2 2

1 1 1 1 1y ;y arc c y y2 x 1 x 2 x 2 x

o1

sx

xx

− −′ ′ ′→ = ⋅ = ⋅ =− −−

= −

t) ( ) ( )2

1 1 1 1 1y ;y arc tg x y ; y1 x2 x 2 x 2 x 1 x1 x

′ ′ ′→ = ⋅ = ⋅ =+

=++

5 Halla la función derivada de las siguientes funciones:

a) ( )

45 5xxy y⎛ ⎞= ′ =Sol. 6x 11 x⎜ ⎟++⎝ ⎠

2 3 2

2 2x 8yx 4x x 4

yx− +

=−

− ′ =Sol. b)

( )y ln sec x tgx y sec x′= + =Sol. c) 2 2y x sen x 2x cos x 2sen x y x cos′− == + Sol. d) x

e) g3 x 3 x2 x3x 2 3y 6e t ey tg e sec e′ == Sol. 2 2y x arc cos

x= f)

222

2 1y 2x arc cos xx 21

x

− −′ = +⎛ ⎞− ⎜ ⎟⎝ ⎠

2x

;

5/7

2

2

2 2 2 2xy 2x arc cos ; y 2x arc cos ;x x4 x 41

x

′ ′= + = +−−

2

2 1y 2x arc cosx x 4

⎛ ⎞′ = +⎜ ⎟

−⎝ ⎠

( ) ( )2y x 1 2x x arcsen x 1= − − + − g)

( )( )

2

2 2

2 2x 1y 1 2x x x 12 2x x 1 x 1

;−′ = ⋅ − + − ⋅ +− − −

( )2 2y 2x x x 1′ = − + −

( )1 x

2

−2

12x x 1

+− 2x 2x 1− + −

;

22

2 2

x x 1 x 1y 2x x2x x 2x x

− − +′ = − + +− −

;

( )22 2 2

2 2

2 2x x2x x x x 1 x 1 4x 2xy ;2x x 2x x 2x x

−− + − − + + −′ = = =− −

2−

( )( )

( )22 2

22

2 2x x2 2x x 2x xy

2x x

−− −′ = =

−

2

2

2x x

2x x

−

−;

2y 2 2x x′ = −

( )2 xy x ln 4 x 4arc tg 2x2

= + + − h)

( )22 2

1 4 1y 1 ln 4 x x 2x 224 x x1

4

′ = ⋅ + + ⋅ ⋅ + ⋅ −+

+ ;

( )2

22 2

2x 16 1y ln 4 x 224 x 4 x

′ = + + + ⋅ −+ +

;

( )2

22 2

2x 8y ln 4 x 24 x 4 x

′ = + + + −+ +

;

( ) ( )2 2

2 22 2

2x 8 8 2x 0y ln 4 x ln 4 x4 x 4 x+ − −′ = + + = + +

+ + ;

6 Aplica la derivación logarítmica para derivar: a) Se toma logaritmo neperiano en cada miembro:

Se deri a en cada miembro:

( )2y ln 4 x′ = +

3xy x=

3xln y ln x ; ln y 3x ln x= = v

( ) ( ) 1D ln y D 3x ln x ; y 3 ln x 3 x′= ⋅ = ⋅ +y

1⋅

xy; 3 ln x 3y′= ⋅ +

a de y: ( ) Se despeja la función derivad y 3ln x 3 y′ = + Se sustituye la función y por su expresión:

( ) ( )3x 3xy 3ln x 3 x ; y 3x ln x 1′ ′= + = +

6/7

7/7

) b x 1x 1n x xx

1xy x + +y l +⎛ ⎞′ = +⎜ ⎟⎝ ⎠

)

= Sol.

x x1 1 1y ln 11y 1 11x x x x

⎡ ⎤⎛ ⎞ ⎛ ⎞′ = + − +⎜ ⎟ ⎜ ⎟⎢ ⎥+⎝⎛ ⎞= +⎜ ⎟⎝ ⎠ ⎠ ⎝ ⎠⎣ ⎦

Soc l.

d) ( ) ( ) ( )x 1x

x 1 x 1y ln ln xy ln x ln xln x

++ +⎡ ⎤′ = += ⎢ ⎥⎣ ⎦Sol.

x xsen x sen xysen x ln xcotgx 1x x

yx

⎡ ⎤⎛ ⎞ ⎛ ⎞′ = + −⎜⎛ ⎞= ⎜ ⎟⎝ ⎠

⎟ ⎜⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Sol. e) ⎟

f) ttgx gx2

ln x tgxy xxcos

y xx

⎛ ⎞′ = +⎜ ⎟⎝ ⎠

= Sol.