derivadas de funções trigonométricas (ex resolvidos)

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Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 1 de 7 
Lista de Exercícios - Derivadas de Funções Trigonométricas 
 
 
1) Nos exercícios abaixo, ache a derivada da função. 
 
a) 2 cosy x x= − 
 
2 cosy x x= − 
' 2 ( sen )y x x= − − 
' 2 seny x x= + 
 
b) 1 3senx
2
y = − 
 
1 3senx
2
y = − 
' 0 3cos xy = − 
' 3cos xy = − 
 
c) 2( ) cosf t t t= 
 
2( ) cosf t t t= 
[ ]' 2 2( ) cos cosd df t t t t tdt dt  = ⋅ + ⋅   
' 2( ) sen 2 cosf t t t t t= − ⋅ + ⋅ 
( )'( ) 2 cos senf t t t t t= ⋅ ⋅ − ⋅ 
 
d) cos( ) tg t
t
= 
 
cos( ) tg t
t
= 
[ ] [ ]
'
2
cos cos
( )
d dt t t t
dt dtg t
t
⋅ − ⋅
= 
'
2
( sen ) cos( ) t t tg t
t
⋅ − −
= 
'
2
sen cos( ) t t tg t
t
⋅ +
= − 
 
e) 2tgy x x= + 
 
2tgy x x= + 
' 2sec 2y x x= + 
Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 2 de 7 
f) 5 secy x x= 
 
5 secy x x= 
[ ] [ ]' 5 sec sec 5d dy x x x xdx dx= ⋅ + ⋅ 
' 5 sec tg 5 secy x x x x= ⋅ ⋅ + ⋅ 
( )' 5sec x tg 1y x x= ⋅ + 
 
g) sen4y x= 
 
sen4y x= 
[ ]' cos4 4dy x xdx= ⋅ 
' 4cos4y x= 
 
h) 2cossecy x= 
 
2cossecy x= 
' 2 2 2cossec co tg dy x x x
dx
 = − ⋅ ⋅   
' 2 22 cossec co tgy x x x= − ⋅ ⋅ 
 
2) Nos exercícios abaixo, ache a derivada da função e simplifique a 
resposta utilizando as identidades trigonométricas. 
 
a) 2cosy x= 
 
2cosy x= 
( )2cosy x= 
[ ]' 2 cos cosdy x xdx= ⋅ ⋅ 
( )' 2 cos senxy x= ⋅ ⋅ − 
' 2 sen cosy x x= − ⋅ ⋅ 
' sen2y x= − 
 
b) 2 2cos seny x x= − 
 
( ) ( )2 2cos seny x x= − 
( )' 2 cos sen x 2 sen cosy x x x= ⋅ ⋅ − − ⋅ ⋅ 
' 2 cos sen x 2 sen cosy x x x= − ⋅ ⋅ − ⋅ ⋅ 
' sen 2 sen 2y x x= − − 
' 2sen 2y x= − 
Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 3 de 7 
c) cos
sen
xy
x
= 
 
cos
sen
xy
x
= 
[ ] [ ]
( )
'
2
sen cos cos sen
sen
d d
x x x x
dx dxy
x
⋅ − ⋅
= 
( )
'
2
sen sen cos cos
sen
x x x xy
x
⋅ − − ⋅
= 
2 2
'
2
sen x cos
sen
xy
x
− −
= 
2 2
'
2
sen x cos
sen
xy
x
+
= − 
'
2
1
sen
y
x
= − 
' 2cossecy x= − 
 
d) ln seny x= 
 
ln seny x= 
[ ]' 1 sen
sen
dy x
x dx
= ⋅ 
'
cos
sen
xy
x
= 
' cotgy x= 
 
e) 2 2ln cossec co tgy x x= − 
 
2 2ln cossec co tgy x x= − 
' 2 2
2 2
1
cossec co tg
cossec co tg
dy x x
x x dx
 = ⋅ − 
−
 
' 2 2 2 2
2 2
1
cossec cotgx 2 ( cossec ) 2
cossec co tg
y x x x x
x x
 = ⋅ − ⋅ ⋅ − − ⋅ 
−
 
' 2 2 2 2
2 2
1
cossec cotgx 2 cossec 2
cossec co tg
y x x x x
x x
 = ⋅ − ⋅ ⋅ + ⋅ 
−
 
'
2 2
1
cossec co tg
y
x x
=
−
2 2 22 cossec cossec cotgxx x x ⋅ ⋅ −  
' 22 cossecy x x= ⋅ 
 
 
Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 4 de 7 
f) tgy x x= − 
 
tgy x x= − 
' 2sec 1y x= − 
' 2tg 1y x= − 
 
g) seny x= 
 
seny x= 
( )12seny x= 
( ) [ ]1' 21 sen sen2
dy x x
dx
−
= ⋅ ⋅ 
( ) 1' 21 sen cos2y x x
−
= ⋅ ⋅ 
'
cos
2 sen
xy
x
= 
 
h) 1( tg sec )
2
y x x x= − 
 
1( tg sec )
2
y x x x= − 
[ ] [ ]' 1 tg tg sec tg2
d dy x x x x x x
dx dx
 
= ⋅ + ⋅ − ⋅ 
 
 
( )' 21 sec tg sec tg2y x x x x x= ⋅ + − ⋅ 
 
3) Utilize a diferenciação implícita para achar dy/dx e calcule a 
derivada no ponto de coordenadas ,
2 4
pi pi 
 
 
. 
a) sen cos2 1x y+ = 
 
sen cos2 1x y+ = 
cos sen2 2 0dyx y
dx
 
+ − ⋅ ⋅ = 
 
 
cos sen2 2 0dyx y
dx
− ⋅ = 
sen2 2 cosdyy x
dx
⋅ = 
cos
2 sen2
dy x
dx y
=
⋅
 
Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 5 de 7 
( )
( )
cos 02 0
2 12 sen 2 4
dy
dx
pi
pi
= = =
⋅
⋅ ⋅
 
 
4) Ache uma equação da tangente ao gráfico da função ( ) tgf x x= no 
ponto de coordenadas , 1
4
pi 
− − 
 
. 
 
( ) tgf x x= 
 
' 2
22
2 2
1 1 1 1 1( ) sec 21cos 2cos cos 24 4 2
f x x
x pi pi
= = = = = = =
     
−          
 
 
 
 0 0( )y y m x x− = − 
( 1) 2
4
y x pi  − − = ⋅ − −  
  
 
1 2
2
y x pi+ = + 
2 1
2
y x pi= + − 
 
5) Nos exercícios abaixo, determine os extremos relativos da função 
no intervalo (0, 2 )pi . 
 
a) 2sen sen2y x x= + 
 
2sen sen2y x x= + 
' 2cos 2 cos2y x x= + ⋅ 
2cos 2 cos2 0 ( 2)x x+ ⋅ = ÷ 
cos cos2 0x x+ = 
2cos 2cos 1 0x x+ − = 
22cos cos 1 0x x+ − = 
 
Façamos cos x y= 
 
22y 1 0y+ − = 
 
2
2
4
( 1) 4 2 ( 1)
1 8
9
b ac∆ = −
∆ = + − ⋅ ⋅ −
∆ = +
∆ =
 
 
Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 6 de 7 
' ''
1 3 1
 e 1
4 2
y y y− ±= ⇒ = = − 
 
Assim sendo: 
 
1 5
cos ou 
2 3 3
x x x
pi pi
= ⇒ = = 
 
cos 1x x pi= − ⇒ = 
 
Intervalo (0, pi/3) (pi/3, pi) (pi, 5pi/3) (5pi/3, 2pi) 
Valor de teste pi/4 pi/2 5pi/4 7pi/4 
Sinal de f’(x) + - - + 
Conclusão Crescente Decrescente Decrescente Crescente 
 
Extremos relativos: 
 
Mínimo: 3 3,
3 2
pi 
  
 
 
 
Máximo: 3 3,
3 2
pi 
−  
 
 
 
b) 2seny x x= − 
 
2seny x x= − 
' 1 2cosy x= − 
1 2cos 0x− = 
1 5
cos ou 
2 3 3
x x x
pi pi
= ⇒ = = 
 
 
Intervalo (0, pi/3) (pi/3, 5pi/3) (5pi/3, 2pi) 
Valor de teste pi/4 pi/2 7pi/4 
Sinal de f’(x) - + - 
Conclusão Decrescente Crescente Decrescente 
 
Extremos relativos: 
 
Mínimo: , 3
3 3
pi pi 
− 
 
 
 
Máximo: 5 5, 3
3 3
pi pi 
+ 
 
 
 
Disciplina: Cálculo I Prof. Rogério Dias Dalla Riva 
Página 7 de 7 
c) cosxy e x−= 
 
cosxy e x−= 
[ ]' cos cosx xd dy e x x edx dx
− − = ⋅ + ⋅   
' ( ) cos ( )x xy e sen x x e− −= ⋅ − + ⋅ − 
' ( cos )xy e sen x x−= − ⋅ + 
( cos ) 0xe sen x x−− ⋅ + = 
 
Como 0xe−− ≠ , temos que: 
 
 cos 0sen x x+ = 
cossen x x= − 
 
3 7(2 e 4 ) Quadrantes ou 
4 4
o ox x x
pi pi
∈ ⇒ = = 
 
Intervalo (0, 3pi/4) (3pi/4, 7pi/4) (7pi/4, 2pi) 
Valor de teste pi/4 pi 11pi/6 
Sinal de f’(x) - + - 
Conclusão Decrescente Crescente Decrescente 
 
Extremos relativos: 
 
Mínimo: 3 , 0,06694
4
pi 
− 
 
 
 
Máximo: 7 , 0,002893
4
pi 
 
 