Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.4 THE CHAIN RULE ¤ 99
(c)
d
d 1+cotx
(sin x +cosx) =
dx dx csc x
⇒
cos x − sin x = csc x (− csc2 x) − (1 + cot x)(− csc x cot x)
csc 2 x
= − csc2 x +cot 2 x +cotx
csc x
So cos x − sin x = cot x − 1
csc x .
= −1+cotx
csc x
= csc x [− csc2 x +(1+cotx) cotx]
csc 2 x
51. By the definition of radian measure, s = rθ,wherer is the radius of the circle. By drawing the bisector of the angle θ,wecan
see that sin θ 2 = d/2
r
⇒ d =2r sin θ s
.So lim
2 θ→0 + d = lim
θ→0 +
rθ
2r sin(θ/2) = lim
θ→0 +
2 · (θ/2)
2sin(θ/2) =lim
θ→0
θ/2
sin(θ/2) =1.
sin x
[This is just the reciprocal of the limit lim =1combined with the fact that as θ → 0, θ → 0 also.]
x→0 x
2
3.4 The Chain Rule
1. Let u = g(x) =4x and y = f(u) =sinu. Then dy
dx = dy du
=(cosu)(4) = 4 cos 4x.
du dx
3. Let u = g(x) =1− x 2 and y = f(u) =u 10 . Then dy
dx = dy du
du dx =(10u9 )(−2x) =−20x(1 − x 2 ) 9 .
5. Let u = g(x) = √ x and y = f(u) =e u .Then dy
dx = dy du
du dx =(eu )
7. F (x) =(x 4 +3x 2 − 2) 5 ⇒ F 0 (x) =5(x 4 +3x 2 − 2) 4 ·
or 10x(x 4 +3x 2 − 2) 4 (2x 2 +3)
1
2 x−1/2
= e √x ·
1
2 √ x = e√ x
2 √ x .
d
x 4 +3x 2 − 2 =5(x 4 +3x 2 − 2) 4 (4x 3 +6x)
dx
9. F (x) = 4√ 1+2x + x 3 =(1+2x + x 3 ) 1/4 ⇒
F 0 (x) = 1 4 (1 + 2x + x3 ) −3/4 ·
11. g(t) =
2+3x 2
=
4 4 (1 + 2x + x 3 ) 3
d
dx (1 + 2x + x3 )=
1
4(1 + 2x + x 3 ) 3/4 · (2 + 3x2 )=
2+3x 2
4(1 + 2x + x 3 ) 3/4
1
(t 4 +1) 3 =(t4 +1) −3 ⇒ g 0 (t) =−3(t 4 +1) −4 (4t 3 )=−12t 3 (t 4 +1) −4 = −12t3
(t 4 +1) 4
13. y =cos(a 3 + x 3 ) ⇒ y 0 = − sin(a 3 + x 3 ) · 3x 2 [a 3 is just a constant] = −3x 2 sin(a 3 + x 3 )
15. y = xe −kx ⇒ y 0 = x e −kx (−k) + e −kx · 1=e −kx (−kx +1)
or (1 − kx)e
−kx
17. g(x) =(1+4x) 5 (3 + x − x 2 ) 8 ⇒
g 0 (x) =(1+4x) 5 · 8(3 + x − x 2 ) 7 (1 − 2x)+(3+x − x 2 ) 8 · 5(1 + 4x) 4 · 4
=4(1+4x) 4 (3 + x − x 2 ) 7 2(1 + 4x)(1 − 2x)+5(3+x − x 2 )
=4(1+4x) 4 (3 + x − x 2 ) 7 (2 + 4x − 16x 2 )+(15+5x − 5x 2 ) =4(1+4x) 4 (3 + x − x 2 ) 7 (17 + 9x − 21x 2 )