⇥sin 26 =or27(cos
+1
3), or y = 27(cos 26)x +6 81 cos
26 sin 26.⇤
yy=+
= 426)(x
1 7 )⇤4+⇥2x
5 x=sin0,2x
+ytan
(x
cos 2x + sin 2x + 28x tan3 (x7 ) sec2 (x7 ) .
dy
dy
pdy = 27
p the equation of
p the tangent line is
d x
✓22 (4x
✓3 );2 if=x8x
◆=sec23!
44. dy
= sec
3x
cos(1
+
then
sin =
26,
cos 26,8so
2 ◆
2 ydy=
dy =
p8dx
3 1) 2 (4x ) p
(4x
),x = ⇡/2
sec2!(4⇡)
⇡. the
When
x = of⇡,
y =tangent
tan(4⇡)
= 2is
0, so
7047.
Chapter
p then
dx
1
27
dy⇡
9 y5 == 135
45. dy
=
3
sec
(⇡/2
x)
tan(⇡/2
x);
if
1, so
equation
the
line
2
2 +=5 0,
dy
(7
x)
3x
+
5
(7
x)
3x
dx
dx
dx
x=
⇡
3 27(cos 26)(x
2p
46.
=
3
x
1
+
;
if
x
=
2
then
y
=
,
=
3
=
,
so
the
equation
of
the
tangent
line
is
dx
dx
p
p
y
+
sin
26
=
+
3),
or
y
=
27(cos
26)x
+
81
cos
26
sin
26.
2
42. the
==
4 tan
2x=
+
sec 2 +
dx
4
16
3 8+=
ydx
0, or yof
the1tangent
8 x ⇡x 4 8⇡.
dx+ 1equation
x3 +xsinline
x is y = 8 ⇡(x
x⇡)
sin
135
27
!
2 (135/16)(x
p
ydy
27/8
=
or y 3= then xy = . sinp26, 2dydy = 27
dy =
3x✓
+2 ✓
x3x)
); tan(⇡/2
if 2),
x◆=
3 ◆
2 cos 26, so the equation of the tangent line is
2cos(1
16
2
45.
=
3
sec
(⇡/2
x);
if
x
=
⇡/2
then
=
0,
y
=
1,
so the equation of the tangent line is
3x
+
5
(7
x)x
(7
x)
3x
+
5
(3x
+
cos
x)
p44.
123/
43
–
55
odds
dy
d
dy
dx
dx
1
1
27
dy
9
5
135
3 + 3p
3
2
dx
dx
48.
=x326
12
cot
x
cot
x
=
12
cot
x
csc
x,
=
24.
When
x
y = 3,ofsothe
thetangent
equation
46. yy⇥
=
x
1
+
;
if
x
=
2
then
y
=
,
=
3
=
,
so =the⇡/4,
equation
lineofis the
+
sin
=
27(cos
26)(x
+
3),
or
y
=
27(cos
26)x
+
81
cos
26
sin
26.
3
3
2
2
3
sinyx
+ sin x)
2 5 (x + sin x)
1 3x x+
70 dx+ 1 = 0,+or
Chapter 2
dx x=⇡/4
x=dx
8 (xdxp
44
16 p
p
dy
d
dy
2
2
2
2
2
2
135or
3=) =
24(x
y 27
= 24x
+
47. tangent
= secline
(4xis )y◆ ✓
(4x
8x sec⇡/4),
(4x
),
= 38 +⇡6⇡.
sec (4⇡) = 8 ⇡. When x = ⇡, y = tan(4⇡) = 0, so
ydx 27/8✓= (135/16)(x
2),◆or y =
x dx x=
. p⇡
dy
dyp
2
3 1 dx
dy
p
p
dy
1
27
dy
9 0,
5 y =1351, so the equation of the tangent line is
16
2
45.
=
3
sec
(⇡/2
x)
tan(⇡/2
x);
if
x
=
⇡/2
then
=the
dy =
43. the
cos
3x
3xthe
sinx3x;
⇡if then
1y and
y⇡)
=
⇡,
of the tangent
y +⇡
= (xline⇡),
equation
tangent
ythen
=
8=y ⇡(x
8=so
⇡x
8⇡.
46.
=
33xp
13 );+ifif2x2x=line
;=
xis3=
2 then
= sin
, =dy
3 27 equation
=
, so
equationline
of isthe
tangent
is
2x of
dx
dx
44.
=
cos(1
+
=
26,
=
dy
x
dy
dx
dx
x
x
8
dx
4
4 cos 26,
16 so the equation of the tangent line is
y
+
1
=
0,
or
y
=
1
2
dx
dx
p
y==2x x.
49. or
5
x
+
=
4
1/2
=
7/2.
When
x
=
1,
y
=
2,
so
the
equation
of
the
tangent
(
2x),
p
p
p
dy
d
dy
135
27
2= 27(cos
2
2y =
227(cos
2sec
sinsec
26=
26)(x
+
or
26)x
818cos⇡26
sin 26.= 8 ⇡. When x = ⇡, y = tan(4⇡) = 0, so
dx+ =
dx
2 2 )5=
xor
47. yydy
(4x
) d (4x
8x
(4x
),
sec2 (4⇡)
27/8
(135/16)(x
2),3),
y=
x 2x=1
.p
dy⇡+ =
3 ◆
✓ cot
dx = 12
dx
dx
x= p
2 ✓cot x =◆ 12 cot3
16
2
7
3
48. dy
x
x
csc
x,
=
24. When
p
p
11),line
27
135 x = ⇡/4, y = 3, so the equation of the
is 3y x2 =
orif yis
=y==2x8then
.y =
dx
dx tangent
dx⇡)
x=⇡/4
the
equation
of1(7/2)(x
the
⇡(x
=dy8 dy
⇡x3 9 58⇡.
46. line
1
+
;
x
,
=
=
, so the equation of the tangent line is
dy =
2
2
2
dx
xy d 3 =
x24(x
4 0,
4 y =16p1, so the equation
⇡/4),
y== ⇡/2
24x8then
+dx
3 + 6⇡.
45. tangent
sec3is2(⇡/2
x)
tan(⇡/2
x);2 iforxdy
=
p
p of the tangent line is
dy = 3line
2
2
2
2
dx
dx
47.
= sec (4x ) (4x ) = 8x sec (4x
⇡, y = tan(4⇡) = 0, so
135),
27 dy
p = 8 ⇡ sec (4⇡) = 8 ⇡. When x =
dy
d
y
+
1
=
0,
or
y
=
1
dx
dx
dx
x=
⇡
=cot
or12ycot
= 3 x csc
xp2 x, . p
3
dy 27/8
1(135/16)(x
dy
p 24. When x = ⇡/4, y = 3, so the equation of the
48. ydy
=p
12p
x xcot
x x2),
=22 )3/2
=
16
2
50.
=
(1
(
2x),
=
1.
When
x
=
0,
y
=
0,
so
the
equation
of
the tangent line is y = x.
the
equation of2 the
tangent line is y = 8dy ⇡(x dx x=⇡/4
⇡) = 8 ⇡x 8⇡.
dx =
dx
2+
2✓ p
49. dx
2x✓
5 x x◆
( 2x), dx x=0 = 4 1/2 = 7/2. When x = 1, y = 2, so the equation of the tangent
1
◆
2
tangent
line is 1y 3 2= 524(x
= 24x + 3dy
+ 6⇡. 9 5
dx
dxy x=1
dy
1 x2 ⇡/4), or
135
p of the tangent line is
dy3 y = 27 , p
2 d
46. dy
= sec
3 2x
; sec
if 2x(4x
=722), then
= 32 (4⇡) == 8p⇡.
, soWhen
the equation
dy =
d (4x12+
dy
47.
(4x
)
)
=
8x
=
8
⇡ sec
x == 3, ⇡,
y = tan(4⇡) = 0, so
2
3
3
2
p
dx
x
x
8
dx
4
4
16
is
y
2
=
(7/2)(x
1),
or
y
=
x
.
48. line
=
12
cot
x
cot
x
=
12
cot
x
csc
x,
=
24.
When
x
=
⇡/4,
y
so the equation of the
dx
dx
dx x=dx
dy
d x2
dp
⇡
p
p
dy = x(psin(5x))
dy
dx
dx
x=⇡/4
2
2
135
27
51.
(5x)
+
cos(5x)
2
sin
x
(sin
x)
=
5x
sin(5x)
+
cos(5x)
2
sin
x
cos
x = of the tangent
2
equation
ofx
tangent
8xy⇡(x
=
8=6⇡.
⇡x When
8⇡. x = 1, y = 2, so the equation
p 2),line
49. the
= 2x =5(135/16)(x
4 ⇡)
7/2.
2x),
ydx
or( yis
=y =or
.24x
tangent
is
ythe+3dx
= =dx
+1/2
3+
dx 27/8line
x=1
2= 5 24(x
x2 ⇡/4),16dx
2
dy 5x sin(5x)
1
x
=
+ cos(5x)
sin(2x), 7 dy 3
2 3/2
50. dy
2x),
= 1. When x = 0, y = 0, so the equation of the tangent line is y = x.
line =
is p
y p2 3=2 (7/2)(x
x
.
d (1 x21),) or (y =
dy
3 dy 2
dy2 = 121cot
p= 24.
2cot
dx dy
2x=0
48. dx
12 cot
x,
When
x = ⇡/4, y = 3,p so the equation of the
2 5 xx
2 x2d+
2 2x2csc
2 When
d y= 2x
dp2x)x=
d=
d x8p
49.
=dxp
4 x=⇡/4
1/2
=⇡7/2.
=⇡.1, yWhen
= 2, so =
the equation
of the tangent
(sec2x),
dx
47. dx
sec
(4x
)
(4x
=
8x
(4x
),
8
sec
(4⇡)
=
⇡, y 10
= sin(5x)
tan(4⇡)
=2 0,
so
2
=
5x
cos(5x)
(5x)
5
sin(5x)
sin(5x)
(5x)
cos(2x)
(2x)
= 25xxcos(5x)
cos(2x).
dx 2
dx x=1
2
5
x
dx
dx
x=
⇡
tangent
line
is
y
3
=
24(x
⇡/4),
or
y
=
24x
+
3
+
6⇡.
p
p
p
dx
dx
dx
dx
dy equation
1 of the
x tangent
7 dy8 3⇡(x d ⇡) = 8 ⇡x 8⇡.
2 3/2
the
line
dy
d x1),
p
50. line
=
(1
( y is
2x),
When
0, ysin(5x)
= 0, so+the
equation
of the
tangent
is x(
y 2sin(5x))
=2 (7/2)(x
=y =
x sin
. x= 1.(sin
51.
=
(5x))+orcos(5x)
x) =x =5x
cos(5x)
2 sin
x cos
x = line is y = x.
dx
2
2 dx2 x=0
1
x
22
2
p
dx
dx
dx
dy
d
dy
2 d
2 x
2 dy d y
2
2
2
52. dy =
= 2x
cos(3x
(3xp
) = 6x (cos(3x
6x(
sin(3x
)) When
(3x2 x
)+
sin(3xof2 )the
+ 6tangent
cos(3x2 ).
49.
5 3 )x+2d+
=dy
4 1/2
= 7/2.
= 61,cos(3x
y = 2, )so=the36x
equation
2x), ), 2 2 =
5x 12
sin(5x)
sin(2x),
2 12 cot3 xdy
dx
dx
dx
48. =
=
cot
xdxcos(5x)
cot
x5 =
csc
x,
=
24.
When
x
=
⇡/4,
y
=
3,
so
the
equation
of the
dx
dx
dy
1
x
x=1
2
x
dy = p
d x2 )3/2 ( 2x),
dx (1
x=⇡/4
50. dx
=ddx
1.
When
x=
0,sin(5x)
y = 0,+socos(5x)
the equation
of
the xtangent
line is y = x.
7
3
2 = x( sin(5x))
51.
(5x)
+
cos(5x)
2
sin
x
(sin
x)
=
5x
2
sin
x
cos
=
2
dx
2
dx
ddxy is y 1line
d
xis (7/2)(x
tangent
y 3 d= (5x)
24(x
⇡/4),
3 + 6⇡.
line
2 cos(5x)
=
1),
or5ysin(5x)
= xor yx=0
.= dx24xd+(5x)
2 cos(2x).
dy =(1 5xx)
+ (1dx
+ x)
2 2
d2 ycos(2x) (2x) = 25x3cos(5x) 10 sin(5x)
2sin(5x) dx
2
dx2 =
dx
dx
53. =
=
=
2(1
x)
and
=
2(2)(
1)(1 x) = 4(1 x) 3 .
5x sin(5x) + cos(5x)
sin(2x), 2
2
2
dx
(1 x) d x2 (1 x) dy
dx
dy
d
dy2 = x( p
dy
1sin(5x))
dy2 sin
2x
51.
(5x)2 +
cos(5x)
x)
== 7/2.
5x sin(5x)
cos(5x)
x cosequation
x=
2 x = (sin
d
y
d
d
d 0,+x
3/2
p
49.
=
2x
5
x
+
4
1/2
= 1, yequation
= 2,2 sin
so of
the
ofline
theistangent
(
2x),
dsin(5x)
y dx
50. dy
(12 2✓
x◆
) x25( sin(5x)
2x),
= 1. When
=
yWhen
the10
tangent
y = x.
dx
22 d◆ dx
2 0, d
2 so the
2
2 sin(5x)
=p
5x cos(5x)
(5x)
(5x)
=
25x cos(5x)
✓
✓ 2 dx
◆
✓sin(3x
◆xcos(2x)
✓=
◆(2x)
dx2=
dx
x=1=
5
52. dx
=
cos(3x
)
(3x
)
=
6x
cos(3x
),
6x(
))
(3x
)
+
6
cos(3x
)
=
36x
sin(3x2 ) +2 6cos(2x).
cos(3x2 ).
2
x=0
1 2x 1
dx
dx
dx
dx
dy
d
1
1
1
1
1
2
2
7
3
dx
dx
dx
dx
=
5x
sin(5x)
+
cos(5x)
sin(2x),
54. line =
+ tan
=
sec
+ tan
,
is xy sec2 = x(7/2)(x
dx
dx x1), or y = 2xx 2 2 . x
x
x
2
d
y
d
d
d
dy
d
d
y
d
dy =(1 5x x)
d+2x)
d ◆ (5x) d✓22cos(2x)
◆
✓
◆ 2 10
✓sin(5x)
◆2
2 +
2 (2x)
2 cos(5x)
2
2◆ 2 ), 2 sin
y ◆sin(5x)
cos(5x)
51sin(5x)
sin(5x)
=
cos(2x).
52. dy
= x(
cos(3x
) ✓(1
==+
6x✓
cos(3x
=✓6x(
sin(3x
)1+◆cos(5x)
61)(1
cos(3x
) 2=✓
).
3 1x
51.
cos(5x)
x) =2 ))
5x
sin
cos sin(3x
x=
d2 y2=
2sin(5x))
1(3x
d)(5x)
1=dx
1 (sin
1 dx
d dx
2 25x
1 3 ). + 6 2cos(3x
2 2x x)
53.
=
2(1
= (3x
2(2)(
x)
=36x
4(1
x)
dx
dx(5x)
dx2 and
2
dx
dx
dx
dx
2
2
2
dy
1
x
dy
=
sec
sec
+
sec
+
sec
=
sec
tan
.
dx
(1
x)
(1
x)
dx
2 3/2
2
2
3
p x
50. =
( 2x),x
=x 1. When x x
= 0,dx
y = 0,
line is y = x.
dx =
x (1dx x )sin(2x),
x
x so thex equationxof the tangent
x
5x sin(5x)
+ cos(5x)
dx
2
dx2 x=0
1 x)
x✓2+
2
dy
d
dy
(1
(1
+
2
d
y
✓
◆
✓
✓
◆
✓
◆
2 d◆
2x)
2 ◆d y
2
2
2
2
2
2
2 =
52. ddy
) 1 (3x
) ==6x cos(3x =
), 2(12 =x)
sin(3x
)) (3x1d ) +1)(1
6 cos(3x
36x sin(3x
53.
x) )3== 4(1
x) 3 . ) + 6 cos(3x ).
126x(2d 2and
y = cos(3x
3 (1
2 1dx2 =dx 2(2)(
2
2 dd cot13 ✓(1
2 1=
dx
dx
dx
dx
x)
x)
55.
y
=
cot
(⇡
✓)
=
so
dy/dx
3
cot
✓
csc
✓.
54. dy2 ==x sec
tan
, = 25x cos(5x) 10 sin(5x) 2 cos(2x).
5x cos(5x) d (5x) +5 tan
sin(5x) =
sin(5x)sec (5x) +
cos(2x)
(2x)
dx = x( sin(5x))
x dx
x + cos(5x) x 2 sin x xd dx
x= 5x sin(5x)
x + cos(5x) 2 sin x cos x =
dx(5x)
dx
51. dx
(sin
x)
2
✓+✓◆
✓ ◆ ✓ 2◆✓ ◆
◆
✓dx ◆ ✓ ◆d y✓ ◆✓ ◆✓ ◆
✓ ◆
✓ ◆
dx
dx
(1 2◆5x)
2✓
2
dy
1 (1
d x)
1=1 2(1
1 2 =
1
ddy
yau
1+
d 1=
1
1 2(2)(
d
1 1)(1 2x) 3 2= 4(1
1
53. dy
=
x) 31.
2 2 1x)1 2 and
+
b 2secad
2 bc
22
22tand
d
d
y
54.
=
x
sec
+
tan
=
sec
+
,
=
sec
+
sec
+
sec
=
sec
tan
=
5x
sin(5x)
+
cos(5x)
sin(2x),
dx
(1
x)
(1
x)
dx
2
2
2
2
2
2.
2
56. 6dx2= cos(3x
.x 6x cos(3x
52.
)x +
))x (3x
dx
x),x2 2 =
x 6x(
x
x ) x+ 6 cos(3x
x(3x
dx
x
x sin(3x
dx
x3 ) = x36x sin(3x
x ) + 6 cos(3x ).
2) =
cu + dx (cu
d)
dx
dx
dx
dx
2
✓ ◆ d✓ ◆ ✓ ◆ ✓ ✓◆d ◆
✓ ◆
✓ ◆
d y
✓ ✓◆1 ◆ dd
✓ ◆ ✓ 1 ◆ ✓ 1◆
d2 y = 5x
2 cos(5x)
d 1 1 (2x) =
2 25x
1
1 sin(5x) 2 cos(2x).
5 sin(5x)
10
2 1 1 2 (5x)
2 cos(5x)
dy
1 d (5x)
1sec
1 secsin(5x)
1 2 1cos(2x)
2
2
3
3
2
2
=
sec
+
+
sec
=
sec
tan
.
dx
dx
2 tan
54. ydy
x sec
=3 cotxsec
,
55.
=2=
cot
(⇡
✓)(1
= +dx
cot ✓+soxtan
dy/dx
✓ dx
csc ✓.d+
2=
3
d
(1
x)
+
x)
2
y
dx
x
x
dx
x
x
dx
x
x
x
x
2
2
2
3
3
x b sin
dx ⇡!]
x= = 2⇡a x
xx)⇡! +
x sin ⇡!
x ⇡! =
57. dx [a
cos
⇡!
sin
2⇡b
cos
⇡(b
a)(2
sin
⇡!
cos
⇡!)
53.
= cos ⇡! +
=
2(1
and
=
2(2)(
1)(1
x)
=
4(1
x)
. = ⇡(b a) sin 2⇡!.
2◆
✓ ◆x)2
✓2 ◆
✓ ◆
✓ ◆
✓ ◆
dx
(1 ✓x)
(1
dx
d!
2 ✓ ◆
2
✓
◆
dy
d
d
y
d
5
1(3x
1
12 )) d (3x12 ) + 6 cos(3x
2
1 36x2 sin(3x
1
2d 3
21
2
216x( 2 sin(3x
2 =
2
22 ) =
+3 (⇡
b2 2✓)
bccot
52. yd =yau
=
cos(3x
)ad=
)=
6x cos(3x
),=sec
) + 6 cos(3x2 ).
55.
cot
3 cot
✓ csc
✓.◆
+
+✓sec
✓ dx
◆x dx
✓
◆✓ soxdy/dx
✓ x◆
✓ ◆x = x3 sec x tan x .
2 dx2 x
56. 6dx
. sec
dx2 2= x sec
dx
x
dx
2
58. 2dycsccu(⇡/3
y)
1 cot(⇡/3
1
1
1
1
+ d 2 (cu
+dd) 1 y).
54. ✓ = x sec◆
+ tan
=
sec2
+ tan
,
5
2
dx
x
dx
x
x
x
x
x
au
+
b
ad
bc
dy
(1
x)
+
(1
+
x)
2
d
y
3
3
2
2
2
55.
cot (⇡ ✓)
cot
so dy/dx
3 cot
✓◆csc and
✓. ✓ ◆= 2(2)(
56.
. =✓ ✓
53. y6d2= =
✓=
◆
◆ 2 ==2(1
✓ x)
✓ ◆ 1)(1 x) ✓3 =◆4(1 ✓x)◆3 .
2d)22
2
22 (1(cux)
cu
+
d
+
dx
(1
x)
dx
d
y
1
d
1
1
1
1
d
1 = ⇡(b
2
1 ⇡!) = ⇡(b a) sin 2⇡!.
57.
[a cos ⇡! + b sin ⇡!] = 2⇡a cos ⇡!2 sin ⇡! + 2⇡b
sin ⇡! cos ⇡!
a)(21sin ⇡! cos
sec
sec
+ 2 sec
+ sec2
= 3 sec2
tan
.
d! =
dx✓2 au + bx◆5 ✓ad◆
x bcdx
x
x
x
x
dx
x
x
x
x
✓ ◆
✓ ◆
✓ ◆
✓ ◆
d
56. 6dy 2 2 2 1 d22 . 1
1 ⇡! sin1⇡! +2 2⇡b
1
1
57.
[a
+ bcot(⇡/3
2⇡a cos
cu
d ⇡!(cu
+sin
d) ⇡!] =y).
54. 2d!
=cos
x+3sec
+ tan
= 2 sec 2 sin+⇡!
tancos ⇡! ,= ⇡(b a)(2 sin ⇡! cos ⇡!) = ⇡(b a) sin 2⇡!.
58.
csc
(⇡/3
y)
3
dx
x
dx
x
x
x
x
x
55. y = cot (⇡ ✓) = cot ✓ so dy/dx = 3 cot ✓ csc ✓.
✓ ◆
✓ ◆
✓ ◆
✓ ◆
✓ ◆
✓ ◆
✓ ◆
2 2
dd✓
y (⇡/3
1 2d
1
1
1
1 d 1
2
1
1
22◆ y) cot(⇡/3
2
2
2
58.
2
csc
y).
5sec+ b sin ⇡!]
57.
[a=cos ⇡!
cos ⇡!) =
sec= 2⇡a+cos2⇡!
secsin ⇡! ++2⇡b
secsin ⇡! cos ⇡! = ⇡(b
= 3 a)(2
sec sin ⇡!tan
. ⇡(b a) sin 2⇡!.
2
au
+
b
ad
bc
d!
x
x dx
x
x
x
x dx x
x
x
x
56. 6dx
.
cu + d (cu + d)2
2
3
58.
2
csc
(⇡/3
55. y = cot
(⇡ y)
✓)cot(⇡/3
= cot3 ✓y).
so dy/dx = 3 cot2 ✓ csc2 ✓.
d
57. ✓[a cos2 ⇡!
+ b sin2 ⇡!] = 2⇡a cos ⇡! sin ⇡! + 2⇡b sin ⇡! cos ⇡! = ⇡(b a)(2 sin ⇡! cos ⇡!) = ⇡(b a) sin 2⇡!.
d! au + b ◆5 ad bc
56. 6
.
cu + d (cu + d)2
58. 2 csc2 (⇡/3 y) cot(⇡/3 y).
57.
d
[a cos2 ⇡! + b sin2 ⇡!] =
d!
58. 2 csc2 (⇡/3
y) cot(⇡/3
y).
2⇡a cos ⇡! sin ⇡! + 2⇡b sin ⇡! cos ⇡! = ⇡(b
a)(2 sin ⇡! cos ⇡!) = ⇡(b
a) sin 2⇡!.