ejerciciosderivadasresueltos
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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.
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