• Nenhum resultado encontrado

Prof. M.Sc. Carlos Alberto Bezerra e Silva

N/A
N/A
Protected

Academic year: 2022

Share "Prof. M.Sc. Carlos Alberto Bezerra e Silva"

Copied!
2
0
0

Texto

(1)

Material de apoio - Exercícios – Derivadas - Respostas

a ) Respostas

( )

x = 0

'

f f'

( )

x = 3

( )

3

= 1 x '

f f'

( )

x = 2x f

( )

x = x6

( )

12

x x '

f = −

( )

34

x x '

f = − f'

( )

x = 1 f'

( )

x = 100x99 f'

( )

x = 3x4 g'

( )

x = 12 g'

( )

x = 5

( )

92

x x '

g = g'

( )

x = 3x g'

( )

x = 0 g'

( )

x = 0 g'

( )

x = 9 g'

( )

x = 13x14

( )

x x

'

g = 8 g'

( )

x = x6 g'

( )

x = 3xln3 g'

( )

x = 4xln4 g'

( )

x = 100xln100

( )

10 10

10

1 . ln x

'

g = x g'

( )

x = ex f'

( )

x = 88x10 f'

( )

x = 7x2

( )

x x

'

f = 1+ 6 f'

( )

x = 10x + 5x4 f'

( )

x = 3x2 + 2x + 1

( )

x = 2nxn1

'

f f'

( )

x = 8x + cosx f'

( )

x = 3x2 6x+ 3

( )

x cosx senx

'

f = −1+ + f'

( )

x = a 4x3 f'

( )

x = 32x3 3xln3

( )

x senx ex

'

f = − − f'

( )

x = 3x2 4xln4 f'

( )

x = 20x3 9

( )

x ln e cosx

'

f = 5x 5+ x +

b ) Respostas

( )

2 4

2

' 3 3

3 2

+

= + x

x x x

f

( ) ( )

3 3 3

1 1

' 2 2

3 2

+ +

+ +

= +

x x

x x x x

f

( )

x x

(

x x

)

f' = (−2 − 1)sen 2 + f'

( )

x = 2xcos

( )

x2

( )

cot ( )

' x g x

f = f'

( )

x = 18x

(

3x2 +1

)

2

( )

x

( )

x

f' = −3sen 3

( )

3 ' 22

= + x x x f

( )

x e x

f' = 3 3 f'

( )

x = cos

( )

x.esenx

( )

x

( ) (

x x

)

g' = −sen .coscos g'

( )

x = exsen

( )

ex

( )

x e x

g' = −5 5 g'

( )

x = sec

( )

x.etg( )x

( )

8

(

2 3

)

3

' t = t t +

g g'

( )

x = 3sec

( ) ( )

3x.tg 3x

( )

2 cossec2

( )

2

' x x x

g = − g'

( )

x = 2cossec

( )

2x cotg

( )

2x

( )

x

( ) ( ) ( )

x tgx tg tgx

g' = sec2 .sec .

( ) ( )

x x

x x x x

e e

e e e x e

g

+ +

= −

2 ' 2

c ) Respostas

( )

x = 15x4 + 4x3+ 9x2 + 8x+ 1

'

g'

( )

x = x2

(

9x6 + 7x4 + 12x3 + 4x+ 3

)

g'

( )

x x e

(

x

)

g = 2 x 3+

( )

x e

(

x

)

senx

'

g = x + 1 −

( )

x x a

(

xlnx

)

'

g = 3 x 4+

( )

x = acosx bsenx

(

a,b

)

'

g'

( )

x e

(

cosx xcosx xsenx

)

g = x + −

( )

x x ex

'

g = 2 + 2

( )

x e .

(

senx.cosx cos x

)

'

g = x + 2

( )

x xe

(

x

)

'

g = 5 x 2+

Prof. M.Sc. Carlos Alberto Bezerra e Silva

(2)

Material de apoio - Exercícios – Derivadas - Respostas

( )

148

x x '

f = −

( )

x =

(

x2+2xx+11

)

2

' f

( )

x =

(

x21

)

2

' f

( )

x =

(

3x23+ 2

)

2

' f

( ) ( )

2

2

9 2

70 12 27

x x x x

'

f

+ +

= −

( ) (

3

)

2

2

1 6

= + x x x ' f

( ) (

2 3

)

2

2

1

3 2 1

x x x

x x x

'

f + + +

= −

( )

2 23

x x x '

f = −

( )

7

2 6 +

= x

x ' f

( ) ( )

2

2

1

3 2

x x x x

'

f +

= +

d ) Respostas

( )

2.3 ln3 3

' x = 2x + g

( )

2 1

2 2

' 2

+ +

=

xe x x

g x

( )

x e x e x

g' = − + 2 2

( )

2 .2 .ln2 2.3 .ln3

' x x x2 2x

g = +

( )

x x

(

x

) (

x

)

g' = −sen .3+ cos x.ln3+ cos

( ) (

10 10

)

ln10

' x x x

f = +

( ) [

sen

( )

6 sen

( )

4 sen

( )

2 1

]

4

' x = 1 x + x + x +

f

( )

x

( )

x e x ex

( )

ex

f' = − sen . cos − sen

( )

x e

(

x

)

f' = 3x 1+ 3

( )

x x e

(

x

)

f' = 3 2 3x 1−

( ) ( )

1 2 1 2 2 ln

' = + + +

x x x

x f

( )

x e

( ( )

x

( )

x

)

f' = x cos 2 − 2sen 2

( ) (

2

) (

3

) (

14 36

)

' x = x+ 7 x+ 5 x+ f

( ) ( )

4 4

1

' 21 4

3

+

= +

x x x x

f

( )

x

x x f' = 2

( ) ( ) ( ) ( )

4

sen cos

sen

' 3 2 x 2 x 4 x

x

f = −

( )

x

[ ( )

x

( )

x

]

f cos2 sen2

8

' = 3 −

( )

x x

f' = sen4

( )

x

( )

x

( )

x

f' = −2sen + 2sen3

( )

x x

f' = sen5

Prof. M.Sc. Carlos Alberto Bezerra e Silva

Referências

Documentos relacionados