Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 93
15. y =
t 2 +2
t 4 − 3t 2 +1
QR
⇒
y 0 = (t4 − 3t 2 +1)(2t) − (t 2 +2)(4t 3 − 6t)
= 2t[(t4 − 3t 2 +1)− (t 2 + 2)(2t 2 − 3)]
(t 4 − 3t 2 +1) 2 (t 4 − 3t 2 +1) 2
= 2t(t4 − 3t 2 +1− 2t 4 − 4t 2 +3t 2 +6)
= 2t(−t4 − 4t 2 +7)
(t 4 − 3t 2 +1) 2 (t 4 − 3t 2 +1) 2
17. y =(r 2 − 2r)e r PR
⇒ y 0 =(r 2 − 2r)(e r )+e r (2r − 2) = e r (r 2 − 2r +2r − 2) = e r (r 2 − 2)
19. y = v3 − 2v √ v
v
= v 2 − 2 √ v = v 2 − 2v 1/2 ⇒ y 0 =2v − 2 1
2
v −1/2 =2v − v −1/2 .
We can change the form of the answer as follows: 2v − v −1/2 =2v − 1 √ v
= 2v √ v − 1
√ v
= 2v3/2 − 1
√ v
21. f(t) = 2t
2+ √ t
QR
⇒
(2 + t 1/2 1
)(2) − 2t
f 0 (t) =
(2 + √ t ) 2 = 4+2t1/2 − t 1/2
(2 + √ t ) 2 = 4+t1/2
(2 + √ t ) 2 or 4+ √ t
(2 + √ t ) 2
2 t−1/2
23. f(x) =
A
B + Ce x
QR
⇒ f 0 (x) = (B + Cex ) · 0 − A(Ce x )
= − ACex
(B + Ce x ) 2 (B + Ce x ) 2
x
25. f(x) =
x + c/x ⇒ f 0 (x) = (x + c/x)(1) − x(1 − c/x2 ) x + c/x − x + c/x
x + c 2
= x 2 2
= 2c/x
+ c (x 2 + c) 2
x
x
x 2
· x2
x = 2cx
2 (x 2 + c) 2
27. f(x) =x 4 e x ⇒ f 0 (x) =x 4 e x + e x · 4x 3 = x 4 +4x 3 e x or x 3 e x (x +4) ⇒
f 00 (x)=(x 4 +4x 3 )e x + e x (4x 3 +12x 2 )=(x 4 +4x 3 +4x 3 +12x 2 )e x
=(x 4 +8x 3 +12x 2 )e x
or x 2 e x (x +2)(x +6)
29. f(x) = x2
1+2x ⇒ f 0 (x) = (1 + 2x)(2x) − x2 (2)
(1 + 2x) 2 = 2x +4x2 − 2x 2
(1 + 2x) 2 = 2x2 +2x
(1 + 2x) 2 ⇒
f 00 (x)= (1 + 2x)2 (4x +2)− (2x 2 +2x)(1 + 4x +4x 2 ) 0
= 2(1 + 2x)2 (2x +1)− 2x(x +1)(4+8x)
[(1 + 2x) 2 ] 2 (1 + 2x) 4
= 2(1 + 2x)[(1 + 2x)2 − 4x(x +1)]
= 2(1 + 4x +4x2 − 4x 2 − 4x) 2
=
(1 + 2x) 4 (1 + 2x) 3 (1 + 2x) 3
31. y = 2x
x +1
⇒
y 0 =
(x +1)(2)− (2x)(1) 2
=
(x +1) 2 (x +1) . 2
At (1, 1), y 0 = 1 2 , and an equation of the tangent line is y − 1= 1 2 (x − 1),ory = 1 2 x + 1 2 .
33. y =2xe x ⇒ y 0 =2(x · e x + e x · 1) = 2e x (x +1).
At (0, 0), y 0 =2e 0 (0 + 1) = 2 · 1 · 1=2, and an equation of the tangent line is y − 0=2(x − 0),ory =2x. Theslopeof
the normal line is − 1 2 , so an equation of the normal line is y − 0=− 1 2 (x − 0),ory = − 1 2 x.