Derivadas

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EXERCÍCIOS DERIVADAS – PARTE II

1) Dada a função do 1º grau f ( x ) = 2 x – 3, calcule:

a) f ′(2) = c) f ′( -1) =

b) f ′(5) = d) f ′(- 3) =

2) Calcule o valor numérico dasderivadas:

a) y = 2 x 3 + 3 x 2 – 4 x + 1 y ′ ( 1 ) =

b) y = 5 x 4 – 2 y ′ ( - 1 ) =

c) f ( x ) = x 2 – 2 x f ′ ( 2 ) =

d) f ( x ) = [pic] f ′ ( 1 ) =

e) y = 8 x ( 3 x + 2 )y ′ ( 0 ) =

f) f ( x ) = ( 2 x + 4 ) ( 3 x – 4 ) f ′ ( - 1 ) =

g) y = 3 x 2 – 1 y ′ ( - 1 ) =

h) y = x 3 + x 2 – x + 5 y ′ ( - 2 ) =

i) f ( x ) = ( 3 x + 5 ) 3 f ′ ( - 1) =

j) y = ( 2 x 2 – x + 1 ) 2 y ′ ( 3 ) =

l) f ( x ) = [pic] f ′ ( - 1 ) =

m) y = ( 5 x 4 – 8 ) 4 y ′ ( 1 ) =
n) ) y = [pic] y ′ ( 0 ) =
o) f ( x ) = 3 x 4 . ( 7 x 2 – 5 )f ′ ( 1 ) =

p) f ( x ) = ( 7 – 10 x 2 ) ( - 8 x + 4 x 3 ) 3 f ′ ( 0 ) =

q) f ( x ) = - 8 x 4 . ( 4 x + 1 ) 5 f ′ ( -1 ) =

3) Calcule a derivada nos pontos:

a) f ( x ) = x 2 + 1no ponto x = 5

b) f ( x ) = 3 x 2 no ponto x = 2

c) f ( x ) = 2 x 3 no ponto x = 1

d) f ( x ) = 2 x 3 – 2 no ponto x = 3

e) f ( x ) = x 3 + 4 x no ponto x = 2

f) f ( x ) =[pic] no ponto x = 1
g) f ( x ) = [pic] no ponto x = - 1

h) f ( x ) = ( 4 x 3 – 3 ) ( - 6 x 2 + 4 x ) 6 no ponto x = 1


4) Considere f ( x ) = 2 x 3 + 15 x 2 + 12 x, determine f ′ ( 1) =



5) Sendo f ( x ) = 2 x 3 – 15 x 2 + 36 x – 7 e g ( x ) = x 3 – 6 x 2 + 11 x – 6 , determine :
f ′ ( 0 ) - 2 g ′ ( 1 ) =



6) Dada a função f ( x ) = 3 x 2 – 1, calcule f ′ (- 4 ) =



7) Calcule a derivada de f ( x ) = ( x 2 – 3 x ) 3 e encontre f ′ ( 2 ) =



8) Calcule a derivada de f ( x ) = ( x 3 – 2 x ) 2 e encontre f ′ ( - 2 ) =






RESPOSTAS:
1)a) 2 b) 2 c) 2 d) 2

2) a) 8 b) – 20 c) 2 d) - [pic]
e) 16 f) -8 g) – 6 h) 7

i) 36 j) 352 l) - [pic] m) 2160

n) -30 o) 66 p) - 168 q) - 20736...
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