2
2
2
2
x
x +2x
x +2x 0
41. ( 1+x
)0 = (1+x)·2x−1·x
= (1+x)
2 . ( (1+x)2 ) = (1 −
(1+x)2
1)−3 ⇒ f 00 (1) = 2 · 2−3 = 41 .
1
)0
(x+1)2
50. (a) P 0 (2) = F 0 (2)G(2) + F (2)G0 (2) = 0 · 2 + 3 ·
1
2
(b) Q0 (7) =
55. R0 (0) =
G(7)F 0 (7)−F (7)G0 (7)
G(7)2
g(0)f 0 (0)−f (0)g 0 (0)
g(0)2
=
=
1·1−0·0
12
1· 14 −5· −2
3
12
=
2
·0−2(x+1)·1
= − (x+1) (x+1)
= 2(x +
4
= 32 .
43
.
12
= 1.
59. (a) (f gh)0 = ((f g)h)0 = (f g)0 h + (f g)h0 = (f 0 g + f g 0 )h + f gh0 .
(b) f = g = h ⇒ (f 3 )0 = 3f 2 f 0 by (a).
(c) f = ex ⇒ (e3 x)0 = 3e2x ex = 3e3x by (b).
7. (c cos t + t2 sin t)0 = −c sin t + (2t sin t + t2 cos t).
31. (a)
(
tan x − 1 0 sec x(tan x − 1)0 − (sec x)0 (tan x − 1)
) =
sec x
sec2 x
2
sec x(sec x) − (tan x sec x)(tan x − 1)
=
sec2 x
sec2 x − tan x(tan x − 1)
1 + tan x
=
=
.
sec x
sec x
sin x
x−1 0
(b) ( tan
) = ( cos x1
sec x
−1 0
) = (sin x − cos x)0 = cos x + sin x.
cos x
(c)
1+tan x
sec x
=
sin x
1+ cos
x
1
cos x
= cos x + sin x.
. We know cos x = 12 for x = π3 . Hence, for
33. f 0 (x) = 1 + 2 cos x = 0 ⇔ cos x = −1
2
. So, we should consider 2kπ + (π ± π3 ), where k ∈ Z.
x = π ± π3 , we’ll have cos x = −1
2
35. (a) velocity = v(t) = (8 sin t)0 = 8 cos t cm/s; acceleration = a(t) = (8 cos t)0 =
−8 sin t cm/s2 .
√
2π
π
(b) x( 2π
)
=
8
sin
=
8
sin
=
4
3. v( 2π
) = 8 cos 2π
= −8 cos π3 = −4. a( 2π
)=
3
3√
3
3
3
3
2π
−8 sin 3 = −4 3.
37. x = 10 sin θ.
dx
dθ
= 10 cos θ. θ =
π
3
⇒ 5ft/rad.
45.
sin θ
lim sinθ θ
sin θ
1
1
θ→0
θ
=
lim
= lim
=
=
sin
θ
sin
θ
1
θ→0 θ + tan θ
θ→0 1 +
1+1
2
1 + lim θ cos θ
θ cos θ
θ→0
1
49. (sin x)0 = cos x; (cos x)0 = − sin x; (− sin x)0 = − cos x; (− cos x)0 = sin x. Hence,
d99
d3
(sin x) = dx
3 (sin x) = − cos x.
dx99
51. (−A sin x−B cos x)+(A cos x−B sin x)−2(A sin x+B cos x) = sin x ⇒ (A−3B) cos x =
if A − 3B 6= 0, which is impossible because we
(3A + B + 1) sin x ⇒ tan x = 3A+B+1
A−3B
get a constant value for tan x. Hence, A − 3B = 0 and 3A + B + 1 = 0. And we can
and B = −1
.
solve A = −3
10
10
cos x cos x−sin x(− sin x)
cos2 x
sin x)
( cos1 x )0 = cos x·0−1·(−
cos2 x
sin x 0
53. (a) (tan x)0 = sec2 x and ( cos
) =
x
(b) (sec x)0 = tan x sec x and
=
1
cos2 x
= sec2 x.
= tan x sec x.
(c) (sin x + cos x)0 = cos x − sin x and
(
54. A(θ) = 21 π(10 sin
lim+
θ→0
π(10 sin( θ2 ))
2(10 cos(
θ
2
))
1 + cot x 0 csc x(− csc2 x) − (1 + cot x)(− csc x cot x)
) =
csc x
csc2 x
2
2
csc x(cot x − csc x) + csc x cot x
=
csc2 x
1
cot x
=−
+
= − sin x + cos x
csc x csc x
θ
2
)2 and B(θ) = 2 · 12 (10 sin
= lim+
θ→0
π(10 tan( θ2 ))
2
= 0.
2
θ
2
)(10 cos
θ
2
). Hence, lim+
θ→0
A(θ)
B(θ)
=