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Mart´ D´ V´squez ın ıaz a Aixa Villadiego Universidad de Cartagena 4 de marzo de 2011

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1. Calcular las derivadas de las funciones a) f (x) = 5 b) f (x) = 2x c) f (x) = −2x + 2 d ) f (x) = −2x2 − 5 e) f (x) = 2x4 + x3 − x2 + 4 x3 + 2 3 1 g) f (x) = 2 3x x+1 h) f (x) = x−1 i ) f (x) = (5x2 − 3)(x2 + x + 4) f ) f (x) = 2. Calcula mediantela f´rmula de la derivada de una potencia o a) f (x) = b) 5 x5

5 3 + 2 5 x x √ c) f (x) = x

1 d ) f (x) = √ x 1 e) f (x) = √ x x √ √ 3 f ) f (x) = x2 + x g) f (x) = (x2 + 3x − 4)4 3. Calcular mediante la f´rmula de la derivada de una ra´ o ız √ a) f (x) = x2 − 2x + 3 √ 4 b) f (x) = x5 − x3 − 2 √ 2 3 x + 1 c) f (x) = x2 − 1 4. Deriva las funciones exponenciales a) f (x) = f (x) = 10
√ x
2b) f (x) = f (x) = e3−x c) f (x) = f (x) =

ex + e−x 2 √ 2x d ) f (x) = f (x) = 3 − x 2

e) f (x) = f (x) =

e2x x2

5. Calcular la derivada de las funciones logaritmicas ( ) a) f (x) = f (x) = ln 2x4 − x3 + 3x2 − 3x ) ( x e +1 b) f (x) = f (x) = ln x e −1 √ 1+x c) f (x) = log 1−x √ d ) f (x) = ln x(1 − x) √ 3x e) f (x) = ln 3 x+2 6. Calcular la derivada de la funci´n logar´ o ıtmica(x − 2)3 f (x) = ln √ 2x − 1 7. Derivar la funci´n o x f (x) = arctan √ 1 − x2 8. Derivar a) f (x) = cos (cos (cos x))
2 b) f (x) = arc sen xcos x

c) f (x) = logx tan x 9. Hallar las ecuaciones de la rectas que pasan por el punto (1, −3) y son tangentes a la funci´n definida como f (x) = x2 . o 10. Hallar la ecuaci´n de una recta que sea tangente a la gr´fica de y = x3 y o a sea paralela a larecta 3x − y + 1 = 0. 11. Hallar la ecuaci´n de la recta tangente y la recta normal a y = x3 en x = 2 o

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RESPUESTAS 1. f (x) = 5 f (x) = −2x f (x) = −2x + 2 f (x) = −2x2 − 5 f (x) = 2x4 + x3 − x2 + 4 f (x) = f (x) = f (x) = x3 + 2 3 1 3x2 x+1 x−1 f ′ (x) = 0 f ′ (x) = −2x f ′ (x) = −2 f ′ (x) = −4x f ′ (x) = 8x3 + 3x2 − 2x f ′ (x) = f ′ (x) = f ′ (x) = 2 2 x 3 −1 x3 −2 (x − 1)2

f (x) =(5x2 − 3)(x2 + x + 4) 2. a) f (x) =


f ′ (x) = 25x4 + 20x3 + 60x2 − 6x − 3

5 x5(

f (x) = 5 b) f (x) =

1 x5

)′

( ) = 5 −5x−6 = −25x−6

c)

d)

e)

f)

( )′ ( ) 1 = −25x−6 + 3 −2x−3 = −25x−6 − 6x−3 f ′ (x) = −25x−6 + 3 2 x √ f (x) = f (x) = x 1 1 1 1 1 f ′ (x) = x 2 −1 = x− 2 = √ 2 2 2 x 1 f (x) = f (x) = √ x 1 − 1 −1 1 3 1 f ′ (x) = − x 2 = − x− 2 = − √ 2 2 2 x3 1 f (x) =√ x x )′ ( )′ ( 1 3 3 1 3 5 ′ = = − x− 2 −1 = − x− 2 f (x) = 3 1+ 1 2 2 2 2 x x √ √ 3 f (x) = f (x) = x2 + x f ′ (x) = 2 2 −1 1 2 1 1 x 3 + √ = x− + √ 3 3 3 2 x 2 x

5 3 + 2 x5 x

4

3.

g) f (x) = (x2 + 3x − 4)4 ( )4−1 ( 2 )′ ( )3 f ′ (x) = 4 x2 + 3x − 4 x + 3x − 4 = 4 x2 + 3x − 4 (2x + 3) √ a) f (x) = x2 − 2x + 3 ) 1 −1 ( 2 )′ )− 1 1( 2 1( 2 x − 2x + 3 2 x − 2x + 3 = x − 2x + 3 2 (2x −2) f ′ (x) = 2 2 b) f (x) = f ′ (x) = √ 4 x5 − x3 − 2 ) 1 −1 ( 5 )− 3 ( )′ ) 1( 5 1( 5 x − x3 − 2 4 x − x3 − 2 4 5x4 − 3x2 x − x3 − 2 = 4 4 √
3

c) f (x) =

x2 + 1 x2 − 1 2 ( ) 1 −1 ( 2 )′ ( )− 3 [ 2 ] 1 x2 + 1 3 1 x2 + 1 x +1 (x + 1)′ (x2 − 1) − (x2 − 1)′ (x2 + 1) ′ f (x) = = 3 x2 − 1 x2 − 1 3 x2 − 1 (x2 − 1)2 2 ( 2 )− 3 [ ] 1 x +1 2x(x2 − 1) − 2x(x2 + 1) = 2−1 3 x (x2 − 1)2


4.

a)f (x) = f (x) = 10 x 1 1 1 1 2 −1 √ f ′ (x) = (x 2 )′ 10x = √ 10x x 2 x b) f (x) = f (x) = e3−x
2

f ′ (x) = (3 − x2 )′ (e3−x ) = −2xe3−x
2

2

ex + e−x 2 (ex + e−x ) 2 (ex + e−x ) = f ′ (x) = 2 2 2 √ d ) f (x) = f (x) = 32x − x c) f (x) = f (x) = 1 1 (2x)′ 32x−1 − √ = 2 · 32x−1 − √ 2 x 2 x e) f (x) = f (x) = e2x x2

5.

( )′ 2xe2x−1 − 2e2x f ′ (x) = e2x x2 − 2xe2x = x3 ( 4 ) a) f (x)= f (x) = ln 2x − x3 + 3x2 − 3x ( )′ 8x3 − 3x2 + 6x − 3 1 · 2x4 − x3 + 3x2 − 3x = 4 2x4 − x3 + 3x2 − 3x 2x − x3 + 3x2 − 3x ( x ) e +1 b) f (x) = f (x) = ln x e −1 ] ( x )−1 ( x )′ ( x )[ x ′ ′ e +1 (e + 1) · (ex − 1) − (ex − 1) (ex + 1) e +1 e −1 ′ f (x) = · = 2 ex − 1 ex − 1 ex + 1 (ex − 1) f ′ (x) = 5

[ ] ex − 1 ex (ex − 1) − ex (ex + 1) ex (ex − 1) − ex (ex + 1) · = 2 ex + 1 (ex + 1) (ex...
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