CalA 4 12

231

4.12 – EXERCÍCIOS – pg. 138
Nos exercícios de 1 à 22 encontrar a derivada das funções dadas. A seguir, comparar os resultados encontrados com os resultados obtidos a partir do uso de um software algébrico. 1-

f (r ) = πr 2 f ′(r ) = 2πr

2-

f ( x) = 3 x 2 + 6 x − 10 f ′( x) = 6 x + 6

3-

f ( w) = aw2 + b f ′( w) = 2aw

4-

f ( x) = 14 −

f ′( x) =
5-

1 −3 x 2

3 −4 x 2

f ( x) = (2 x + 1) (3 x 2 + 6)

f ′ ( x ) = ( 2 x + 1) . 6 x + ( 3 x 2 + 6 ). 2
= 12 x 2 + 6 x + 6 x 2 + 12
= 18 x 2 + 6 x + 12
6-

f ( x) = (7 x − 1) ( x + 4)

f ′( x) = (7 x − 1).1 + ( x + 4).7
= 7 x − 1 + 7 x + 28
= 14 x + 27
7-

f ( x) = (3 x 5 − 1) (2 − x 4 )

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f ′( x) = (3 x 5 − 1) (−4 x 3 ) + (2 − x 4 ) . 15 x 4
= −12 x8 + 4 x 3 + 30 x 4 − 15 x8
= −27 x8 + 30 x 4 + 4 x 3
8-

4 f ( x) = (5 x − 3) −1 (5 x + 3)
6
f ( x) =

4 (5 x + 3)
6 (5 x − 3)

4 (5 x − 3) . 5 − (5 x + 3) . 5
6
(5 x − 3) 2
4 25 x − 15 − 25 x − 15
=
6
(5 x − 3) 2
4 − 30
− 20
=
=
2
6 (5 x − 3)
(5 x − 3) 2

f ′( x) =

9-

f ( x) = ( x − 1) ( x + 1) f ′( x) = ( x − 1).1 + ( x + 1).1
= x −1+ x +1
= 2x

10-

f ( s ) = ( s 2 − 1) (3s − 1) (5s 3 + 2 s )

[

]

f ′( s ) = ( s 2 − 1) (3s − 1) (15s 2 + 2) + (5s 3 + 2 s )3 + (3s − 1) (5s 3 + 2 s )2 s
= ( s 2 − 1) (3s − 1) (15s 2 + 2) + 3( s 2 − 1) (5s 3 + 2 s ) + 2 s (3s − 1) (5s 3 + 2 s )
11-

f ( x) = 7(ax 2 + bx + c) f ′( x) = 7(2ax + b)

12-

f (u ) = (4u 2 − a ) (a − 2u ) f ′(u ) = (4u 2 − a) (− 2 ) + (a − 2u )8u
= −8u 2 + 2a + 8au − 16u 2
= −24u 2 + 8au + 2a

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13-

2x + 4
3x − 1

f ( x) =

(3 x − 1).2 − (2 x + 4).3
(3 x − 1) 2
6 x − 2 − 6 x − 12
=
(3 x − 1) 2
− 14
=
(3 x − 1) 2

f ′( x) =

14-

f (t ) =

t −1 t +1

(t + 1).1 − (t − 1).1
(t + 1) 2
2
t +1− t +1
=
=
2
(t + 1)
(t + 1) 2

f ′(t ) =

15.

3t 2 + 5t − 1 f (t ) = t −1 f ′(t ) =
=

16-

(t − 1) (6t + 5) − (3t 2 + 5t − 1).1
(t − 1) 2
3t 2 − 6t − 4
(t − 1) 2

2 − t2 f (t ) = t−2 f ′(t ) =

(t − 2) (−2t ) − (2 − t 2 ).1
(t − 2) 2

=

− 2t 2 + 4t − 2 + t 2
(t − 2) 2

=

− t 2 + 4t − 2
(t − 2) 2

− t 2 + 4t −