Resolução parte 1 Cap 4 Calculo 1, Rubens Starke

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1
RESOLUÇÕES DOS EXERCÍCIOS: CÁLCULO 1 
CAPÍTULO 4 
 
Professores Inder Jeet Taneja e Rubens Starke. 
 
1. Em cada item devemos examinar o 0 0
0
( ) ( )lim
x
f x x f x
xΔ →
+ Δ −
Δ . 
1.1 ( ) 3f x x= − ; 0 4x = . 
0 0
(4 ) (4) 4 3 1lim lim
x x
f x f x
x xΔ → Δ →
+ Δ − + Δ − +=Δ Δ ( ) ( )
( )
( )
0
0
4 2 4 2
lim
4 2
1lim .
44 2
x
x
x x
x x
x
x x
Δ →
Δ →
+ Δ − + Δ += Δ + Δ +
Δ= =Δ + Δ +
 
 
Portanto o gráfico admite reta tangente no ponto (4, 1)− e seu coeficiente angular é 
1
4
. 
Achar a equação: 1
4
y x b= + . Para achar b substituímos x por 4 e y por 1− , 
obtendo 2b = − . A equação da reta tangente é: 2
4
xy = − . 
1.2 
, se 0
( )
, se 0
x x
f x
x x
≤⎧⎪= ⎨ >⎪⎩
, (0,0)P ; 0 0x = . 
 
Como a função está definida através de 2 sentenças, uma à esquerda e outra a direita 
de 0, devemos calcular os limites laterais: 
 
0 0
(0 ) (0) 0lim lim 1
x x
f x f x
x x− −Δ → Δ →
+ Δ − Δ −= =Δ Δ . 
e 
0 0
(0 ) (0) 0 1lim lim
x x
f x f x
x x x+ +Δ → Δ →
+ Δ − Δ −= = = +∞Δ Δ Δ . 
 
Como os limites laterais são diferentes e não são ambos infinitos, o gráfico não 
possui reta tangente nesse ponto. 
 
1.3 2/3( )f x x= ; 0 0x = . 
( )
( )
2/3
1/3 30 0 0 0
(0 ) (0) 1 1lim lim lim lim
x x x x
xf x f
x x xxΔ → Δ → Δ → Δ →
Δ+ Δ − = = =Δ Δ ΔΔ . 
 
Agora note que este limite não é finito. Mas os limites laterais são ambos infinitos, 
ou seja, 
30
1lim
x x−Δ →
= −∞Δ e 30
1lim
x x+Δ →
= +∞Δ . 
 2
 
Concluímos que a reta tangente é 0x = que é o eixo Y . 
 
2) a) 
0
(1 ) (1)(1) lim
x
f x ff
xΔ →
+ Δ −′ = Δ 
 
( )
2 3
0
2 3
0
2
0
1 3 3 5 5 1 3lim
2 3lim
lim 2 3 2.
x
x
x
x x x x
x
x x x
x
x x
Δ →
Δ →
Δ →
+ Δ + Δ + Δ − − Δ + += Δ
− Δ + Δ + Δ= Δ
= − + Δ + Δ = −
 
 
 b) 
0
( ) ( )( ) lim
x
f x x f xf x
xΔ →
+ Δ −′ = Δ 
 
3 2 2 3 3
0
3 2 3
2
0
3 3 5 5 1 5 1lim
3lim 3 5.
x
x
x x x x x x x x x x
x
x x x x x
x
Δ →
Δ →
+ Δ + Δ + Δ − − Δ + − + −= Δ
Δ + Δ + Δ= = −Δ
 
 
3. Devemos calcular as derivadas laterais 
 
a) 
0
(3 ) (3)(3) lim
x
f x ff
x−− Δ →
+ Δ −′ = Δ 
 
2
0
2(3 ) (3 ) 21lim 13
x
x x
x−Δ →
+ Δ + + Δ −= =Δ . 
 
 
0
(3 ) (3)(3) lim
x
f x ff
x++ Δ →
+ Δ −′ = Δ 
 
2
0
(3 ) 7(3 ) 9 21lim 13
x
x x
x+Δ →
+ Δ + + Δ − −= =Δ . 
 
 Logo, f é derivável no ponto 3 e (3) 13f ′ = . 
 
 b) 
0
4(2 ) 9 1(2) lim 4
x
xh
x−− Δ →
+ Δ − +′ = =Δ 
 
0
2(2 ) 7 1(2) lim 8.
x
xh
x++ Δ →
− + Δ + +′ = = −Δ 
 
 Como as derivadas laterais são diferentes, h não é derivável em 2. 
 
4. a) 2 3 3
22y x y x
x
− −′= ⇒ = − = − . 
 1/3 2/3
23
1 1
3 3
y x y x
x
−′= ⇒ = = . 
 
1
22 5/2 3/25 5
2 2
y x x y x x x+ ′= = ⇒ = = . 
 
 3
5. 3 2 4 2 5 2( ) (2 4 )(15 2 ) (6 8 )(3 )f x x x x x x x x x′ = − + + − + 
 7 6 4 348 84 10 16 .x x x x= − + − 
 
 (1) 48 84 10 16 42.f ′ + − + − = − 
 
6. 
2 2 3
2 2
( 4 1)6 (2 4)(2 4)( )
( 4 1)
x x x x xg x
x x
− + − + −′ = − + 
 
4 3 2
2 2
2 16 6 8 16
( 4 1)
x x x x
x x
− + − += − + . 
 
7. 
( ) ( )
( )2
1 11 1
2 2( )
1
t t
t tf t
t
+ − −
′ =
+ ( ) ( )2 2
2 1 1
2 1 1t t t t
= =
+ +
. 
 
8. 
2 2 2 2 2
2 2 2 2 2 2
( )2 ( )2 4( )
( ) ( )
R a R R a R a Rh R
R a R a
+ − −′ = =+ + 
 
2
3
2 42
2
4 16 322 2 .
2 25 25
4
aaah a
a aa a
⋅⎛ ⎞′ = = =⎜ ⎟⎝ ⎠ ⎛ ⎞+⎜ ⎟⎝ ⎠
 
 
9. 2 4 3 4( ) 5(8 3) 16 80 (8 3) .f x x x x x′ = − = − 
 
10. 4 1/3( ) ( 2 )g t t t= − ( )
3
4 2/3 3
243
1 4 2( ) ( 2 ) (4 2)
3 3 2
tg t t t t
t t
− −′⇒ = − − =
−
. 
 
11. 2 3 2 2 2( ) ( 1) 2(2 5) 2 3( 1) 2 (2 5)f r r r r r r′ = + ⋅ + ⋅ + + ⋅ + 
 
2 2 2
2 2 2
2( 1) (2 5) 2( 1) 3 (2 5)
2( 1) (2 5)(8 15 2).
r r r r r
r r r r
⎡ ⎤= + + + + +⎣ ⎦
= + + + +
 
 
12. 
2 2 2 2 2 3 2
2 4
( 4) 3( 5) 2 ( 5) 2( 4)2( )
( 4)
y y y y y yf y
y
+ − − − +′ = + 
2 2 2 2 2
2 4
2 2 2
2 3
2 ( 5) ( 4) 3( 4) 4( 5)2
( 4)
2 ( 5) ( 22) .
( 4)
y y y y y
y
y y y
y
⎡ ⎤− + + − −⎣ ⎦= +
− += +
 
 
13. 1 1 1( ) .
( ) 9 9
g x
f x
′ = = = −′ − 
 A inversa de f é 4( )
9
xg x −= . Agora você pode ver que ( )g x′ é de fato 1
9
− . 
 
 4
14. Vamos primeiro calcular ( )f x′ . Como 1/3( ) (2 1)f x x= − , temos 
2/3
23
1 2( ) (2 1) 2
3 3 (2 1)
f x x
x
−′ = − ⋅ = − . Chamamos de 
1f − de g , temos: 
233 (2 1)1( )
( ) 2
xg y
f x
−′ = =′ . Mas sendo 
3 2 1y x= − , temos que 
23( )
2
yg y′ = e 
23( )
2
xg x′ = . 
 
15. Note que (0) 6f = − e, portanto, ( 6) 0g − = . Então se (0) 0f ′ ≠ , temos 1( 6) .
(0)
g
f
′ − = ′ 
Sendo 6( ) 21 2f t t′ = + , temos (0) 2.f ′ = Logo, 1( 6)
2
g′ − = . 
 Como (1) 1f = − , temos ( 1) 1g − = e se (1) 0f ′ ≠ , temos 1( 1)
(1)
g
f
′ − = ′ . Mas (1) 23f ′ = e 
assim, 1( 1)
23
g′ − = . 
 
16. 
2 2 5( ) x x xf x e e− −= + 
 
2
2
2 5
2 5
( ) (2 2) ( 5)
( ) (2 2) 5 .
x x x
x x x
f x e x e
f x e x e
− −
− −
′⇒ = − + −
′⇒ = − −
 
 
 Logo, 
 10 10
5(2) 2 5 2f e
e
−′ = − = − . 
 
17. 
3 31 2 1
2
3 5 ( 3 ) ln5 5 3( )
9
x xx xg x
x
− −⋅ − − ⋅′ = 
 
3
3
1 3
2
1 3
2
3 5 ( 3 ln5 1)
9
5 (3 ln5 1) .
3
x
x
x
x
x
x
−
−
⋅ − −=
− +=
 
18. 
5
6
6 1 .
( ) ln 2
xy
x x
−′ = − (aplicação direta da fórmula). 
 
19. 2
4
2 5
xy
x
′ = − . (idem) 
 
20. [ ]
2
2
3ln(3 4) 2
3 4( )
ln(3 4)
x x x
xu x
x
− ⋅ − ⋅ −′ = − 
 Logo, 
 2
4 ln 2 6(2)
(ln 2)
u −′ = . 
 
 5
21. ( )5 5 1( ) 2 1 ln 3 5x xv x e x e x⎛ ⎞′ = + − −⎜ ⎟⎝ ⎠ . 
 
22. 2( ) 2 cos3 ( sen 3 ) 3 2(cos3 ) 3f x x x x x x′ = + − ⋅ − ⋅ 
 
2
2
2 cos3 3 sen 3 6cos3
(2 6)cos3 3 sen 3 .
x x x x x
x x x x
= − −
= − − 
 
23. 2( ) 2sen cos 3cos 4 ( sen 4 ) 4g x x x x x′ = − − ⋅ 
 22sen cos 12cos 4 sen x x x x= ⋅ + ⋅ . 
 Logo, 
 
21 3 1 3 3 3( ) 2 12 3 2 3.
2 2 2 2 2 2 2
g π ⎛ ⎞′ = + − = + =⎜ ⎟⎝ ⎠ 
 
24. ( ) 1/2 2 21( ) tg 3 sec 3 3 2 cotg 3 ( cossec 3 ) 3
2
f x x x x x−′ = ⋅ + ⋅ − ⋅ 
 
2
23sec 3 6 cotg 3 cossec 3
2 tg 3
x x x
x
= ⋅ − ⋅ ⋅ . 
 
25. cotg3 2 cos( ) 2 ( cossec 3 ) 3ln 2
sen 
x xf x x
x
′ = − ⋅ + 
 cotg3 23 2 cossec 3 ln 2 cotg x x x= − ⋅ ⋅ ⋅ + . 
 
 1
2
1 1 13 2 ln 2 1 3 ln 2 1 1 3ln 2.34 2 1 / 2sen
4
f π π
−⎛ ⎞′ = − ⋅ ⋅ + = − ⋅ ⋅ + = −⎜ ⎟⎝ ⎠ 
 
26. a) cos sen ( ) sen cos ln(sen )
sen cos
x xf x x x x
x x
′ = + ⋅ − 
 cos cos ln(sen ) tg x x x x= + ⋅ − . 
 
b) 2( ) sec 1f α α′ = − e como 2 21 tg secα α+ = temos que 2( ) tgg α α′ = . 
 
c) 2cos( ) (1 cotg cossec )
sen 
xg x x x x x
x
′ = − ⋅ − − 
 2cotg cot cossec 1x x x x= − + ⋅ − 
 2(cossec 1)x x= − 
 2cotgx x= ⋅ , pois 2 21 cotg cossecx x+ = . 
d) 2 5( ) 6 (cossec 2 1) 2cossec 2 ( cotg 2 cossec2 ) 2f x x x x x′ = ⋅ + ⋅ ⋅ − ⋅ ⋅ 
 2 5 224(cossec 2 1) cossec 2 cotg 2 .x x x= − + ⋅ ⋅ 
 
e) 
2 2
2
2 2 2 2 4
tg sec( ) sec 3
3
sec (1 tg ) sec sec sec .
f θ θθ θ
θ θ θ θ θ
′ = +
= + = =
 
 
 6
27. 2
(1 cos )( sen ) (1 cos )sen ( )
(1 cos )
x x x xf x
x
− − − +′ = − 
 2
sen sen cos sen sen cos
(1 cos )
x x x x x x
x
− + − −= − . 
 2
2 sen 
(1 cos )
x
x
= − − . 
 
2
32 32 4 3.
3 1 / 411
2
f π
− ⋅ −⎛ ⎞′ = = = −⎜ ⎟⎝ ⎠ ⎛⎞−⎜ ⎟⎝ ⎠
 
 
28. Antes de começar a derivar, vamos escrever ( )f x de outro forma, usando a propriedade do 
logaritmo. 
 
 1 tg ( ) ln ln(1 tg ) ln(1 tg )
1 tg 
xf x x x
x
⎛ ⎞+= = + − −⎜ ⎟−⎝ ⎠ 
 Então: 
 
2 2 2
2
sec sec 2sec( )
1 tg 1 tg 1 tg
x x xf x
x x x
′ = + =+ − − . 
 
 
4 82
3 3 4.1 26 1
3 3
f π
⋅⎛ ⎞′ = = =⎜ ⎟⎝ ⎠ −
 
 
29. 
arcsen arccos
arcsen arccos
2 2 2
1 ( 1)( ) .
1 1 1
x x
x x e ef x e e
x x x
− −′ = − =− − − 
 
30. 2 2 2
1 1 1 1 1( ) 2 arctg 
2 1 2 1 2
f x x x x
x x
⎛ ⎞′ = ⋅ + + −⎜ ⎟+ +⎝ ⎠ 
 
2
2 2
2
1 1arctg 
2(1 ) 2(1 ) 2
1arctg 
xx x
x x
xx x
= ⋅ + + −+ +
+= ⋅ +
22(1 x+
1
2)
1 1arctg arctg .
2 2
x x x x
−
= ⋅ + − = ⋅
 
 
31. ( ) 2/3
2
1 ( 2)( ) arccossec2
3 | 2 | 4 1
f x x
x x
− −′ = ⋅ − 
 
2 23
2
3 (arccossec2 ) 2 | | 4 1x x x
−= ⋅ − 
 
2 23
1
3| | (arccossec2 ) 4 1x x x
−= ⋅ − . 
 7
 
32. Aqui usamos a derivada da função arctg u : 
 2 2
1 / 1( ) .
1 (ln ) 1 (ln )
xf x
x x x
′ = =+ ⎡ ⎤+⎣ ⎦
 
33. Deve-se usar a fórmula: 2(arccotg ) 1
uu
u
′−′ = + . 
 Para facilitar, vamos neste caso calcular separadamente u′ e 21 u+ . 
 
 2 2
( ) 1 ( ) 1 2 .
( ) ( )
x a x a au
x a x a
− ⋅ − + ⋅ −′ = =− − 
 
 
2 2 2 2
2
2 2
( ) ( ) ( )1 1 1
( ) ( )
x a x a x a x au
x a x a x a
+ + − + +⎛ ⎞+ = + = + =⎜ ⎟− − −⎝ ⎠ 
 
 
2 2 2 2 2 2 2 2
2 2 2
2 2 2 2 2( )
( ) ( ) ( )
x ax a x ax a x a x a
x a x a x a
− + + + + + += = =− − − . 
 
 Então, 
 
22
2 2 2 2 2 2 2
2
2
2 ( )( )( ) .
2( ) ( ) 2( )
( )
a
a x a ax af x
x a x a x a x a
x a
−
−−′ = − = ⋅ =+ − + +
−
 
 
34. 4 4 1/2 3
4
1 2( ) 1 1 (1 ) ( 4 ) .
2 1
xf x x x x x
x
−′ = ⋅ − + ⋅ ⋅ − − − − 
 
4
4
4 4
2 21
1 1
x xx
x x
= − − −− − 
 
4 4 4
4 4
1 2 2 1 2 3 .
1 1
x x x x x
x x
− − − − −= =− − 
 
35. Devemos calcular ( )g t′ e depois aplicar em K− . 
Para calcular ( )g t′ , observe que temos uma função do tipo 2( )g t u= , cuja derivada é 
( ) 2 .g t u u′ ′= ⋅ 
 
Então: 
22
2 2 2 2
2
( ) 2arctg 2arctg
1
K
K K K ttg t
Kt t t t K
t
− ⎛ ⎞⎛ ⎞ ⎛ ⎞′ = = − ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟+⎝ ⎠ ⎝ ⎠⎝ ⎠+
 
 2 22arctg
K K
t t K
⎛ ⎞= − ⎜ ⎟ +⎝ ⎠ . 
 
Logo, 
2
1( ) 2 arctg( 1) ( 2) .
2 4 2 4
Kg K
K K K
π π⎛ ⎞′ − = − ⋅ − ⋅ = − − ⋅ =⎜ ⎟⎝ ⎠ 
 8
36. a) 22 2 0 3x yy y y′ ′− = − 
 22 2 3 0x yy y y′ ′⇒ − + = 
 ( )22 3 2y y y x′⇒ − + = − 
 
2
2
2
2 3
2 .
2 3
xy
y y
xy
y y
−′⇒ = − +
′⇒ = −
 
 
 b) 2 2 2 21 2 (2 2 ) 1 0y ye x e y x y x yy′ ′⋅ + ⋅ ⋅ − ⋅ + ⋅ + = 
2 2 2 2(2 2 ) 2 1y yy xe x y e xy′⇒ − = − + − 
( )
2 2
2 2
2 2
2
2 1
2 2
2 1.
2
y
y
y
y
e xyy
xe x y
xy ey
x e xy
− + −′⇒ = −
− −′⇒ = −
 
37. xy x= 
 ln lny x x⇒ = 
11 lny x x
y x
′⇒ = ⋅ + ⋅ 
(ln 1)
(ln 1).x
y y x
y x x
′⇒ = +
′⇒ = + 
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