SECTION 3.3
3.3
DERIVATIVES OFTRIGONOMETRI C FUNCTIONS
D
Derivatives ofTrigonometric Functions
1. f (x ) =3x 2 -2 cos x
=>
2. f (x ) = ~s inx
j' (x ) =
=>
3. f (x ) = sinx + ~ cot x
=>
8. Y = e U (cos u + cu)
=
10 . y
2 - tan x
1 + sin x
x + cos x
=> hi(0) = - esc 0 cot 0 + eO (- csc 2 0) + (cot O)eO = - csc 0 cot 0 + eO(cot 0 - csc'' 0)
=> y' = eU ( - sin u + c) + (cos u + cu) eU = e (cos u - sin u + cu + c)
U
=> Y
I
(2 - tanx)(1) -x( - sec2x )
2 - ta n x + x sec 2 x
-------,;-:!--''---'--- -7-:c-- - --"­
(2 - tan z l?
(2 - tan z )?
= "----
=>
x cos x + cos 2 X - (cos 2 x)
(x + COSX)2
11. f (O ) =
sin;:.
2v x
j' (x ) = cos x - ~ csc 2 X
x cosx + cos 2 X - (1 - sin 2 x)
(x + cos z )?
(x + cos x) (cos x) - (1 + sin x ) (l - sin x )
(x + cos xF
y'
~cosx +
=> g'(t) =4secttant + sec 2t
7. h(0) = esc 0 + eOcot 0
X
=
=> g' (t) = t 3 ( - sin t ) + (cos t) · 3t 2 = 3e cos t - t 3 sin t or t 2 (3 cos t - t sin t )
6. g (t ) = 4 sec t + t an t
9. y =
~cos x + sin x (~ X-l/2)
=> y' = -2csc x cot x -5sin x
4. y =2csc x + 5 cos x
5. g( t) = t 3 cos t
j' (x ) =6x -2(- sin x ) = 6x +2 sin x
xcosx
(x + cos X)2
sec 0
l + secO
j'(0) = (l +sec O)(sec Otan O) -(sec O)(sec O t an O) = (sec O t an O) [(1+ sec O) - sec O] = sec O t a n O
(1 + sec 0)2
(1 + sec OF
(1 + sec 0)2
12. y
•
I
y
=
1 - sec x
t an x
=
tan x (- secx t anx) - (1 - sec x ) (sec 2 x)
(t an x)2
13 = sin x
. y
x2
=>
I
x 2 cos x - (sinx)(2x)
=> y =
14. y = cscO (0 + cot 0)
(X2)2
sec x (- tan 2 x - sec x + sec x )
t a n2 x
2
=
x (x cos x - 2 sin x )
x4
sec x (1 - sec x)
t an" x
xcos x - 2 sin x
x3
=>
y' = cscO (1 - csc 2 0) + (0 + cot O)(-csc O cot 0) = cscO (1- csc' 0 - OcotO - cot 2 0)
= esc 0 (- cot 2 0 - 0 cot 0 - cot 2 0)
[1 + cot 2 0 = csc 2 OJ
= esc 0 (-0 cot 0 ~ 2 cot 2 0) = - csc 0 cot 0 (0 + 2 cot 0)
179
180
CHAPTER 3
0
DIFFERENTIATION RULES
15. Using Exercise 3.2.55(a), f(x) == xe" esc x
f'(X) == (x)'e x cscx + x(e x)' cscx
== eX esc
x (1
f' (x) == (x 2 )' sin x tan x
sin x tan x
17. ~ (cscx)
dx
==
18, ~ (sec x)
dx
==
+ xeX(cscx)' == lex cscx + xe" cscx + xe X(­
cot x cscx)
+ x -x cot x)
16. Using Exercise 3.2.55(a), f(x)
== 2x
::::}
== x
2
sinx tanx
+ x 2 (sin x)
tan x
I
::::}
+ x 2 sin x (tan x)
+ x 2 sin x + x 2 sin x
I
== 2x
sec 2 x == x sin x (2 tan x
~ (_1_)
==
(cosx)(O) -l(-sinx) == sinx == _1_. sinx == secx tanx
cos? X
cos? X
cos X cos x
dx
cos X
== cosx
+ h) h
h~O
.
(
== 1im
h~O
22. y
sirr' x + cos'' X
1
2
--s-in-2-x-- == - -si-n2 -x == - esc x
. cos (x
f (x ) == 11m
+ h) -
h~O
cosx
h
li
cosx cosh - sinx sinh - cosx
h
== 1m - - - - - - - - - - - ­
h~O
cos h - 1
.
sin h)
u cos hh - 1 - sin
.
u sin h
h
- sin z -h- == cosx im
z im -h-
cosx
h~O
h~O
x ) (0) - (sin x) (1) == - sin x
y' == sec x tan x, so y' ( i) == sec
::::}
== eX cos
(i, 2)
x
is y - 2 == 2 V3
::::}
(x - i)
y' == eX(- sin x)
i
tan i == 2 V3.
or y == 2 V3 x + 2 - ~
+ (cos x )e X == eX(cos x
An equation of the tangent line to the curve y == sec x
V3 at.
- sin x)
::::}
the slope of the tangent line at (0, 1) is
eO ( cos 0 - sin 0) == 1(1 - 0) == 1 and an equation is y - 1 == 1 (x - 0) or y == x
23. y == x
24, y
+ cos x
== .
sin z
1
::::}
+ cos x
== 2x
sin x
::::}
y' == 2 (~ cos ~
+ 1.
y' == 1 - sin x. At (0, 1), y' == 1, and an equation of the tangent line is y - 1 == 1 (x - 0), or y == x
::::}
y' ==
cos x - sinx
[Reciprocal Rule].
(sin x + cos x ) 2
of the tangent line is y - 1 == -l(x - 0), or y == -x
25. (a) y
sec ' x
::::}
f' (x) == lim f (x
at the point
+ x 2 sin x
x).
(sin x) (0) -l(cosx) == -cosx == _ _
1_. cosx == -cscx cot x
sin' x
sin" x
sin x sin x
sin x
(sin x ) (- sin x) - (cos x) (cos x )
d
d ( cos x )
(cot x) == - . - == --'----------'--'-----'-------'-----'---'-----...:....
dx
dx SIn x
sin 2 x
21, y == sec x
+ x + x sec '
tan x
==
dx
== (cos
+ x 2 cos x
~ (_1_)
19, -
20. f(x)
sin x tan x
y' == 2( x cos x
+ sin ~)
== 2(0
+ sin x . 1).
1-0.
I
At ( 0,1 ) ,y == - (0 + 1)2 == -1, and an equation
+ 1.
At (~, 1r) ,
+ 1.
(b)
37T
2
+ 1) == 2, and an equation of the
tangent line is y - t: == 2 (x - ~), or y == 2x.
o~---.....L--------
7T
SECTION 3.3
26. (a) y = sec x - 2cos x
=>
sec
i
tan
0
(b)
(i , 1) is
i + 2 sin i = 2 . J3+ 2 . "{( = 3 J3, and an equation
is y - 1 = 3 J3 (x -
i), or y = 3 J3x + 1 -
= sec x - x => 1' (x ) = secx
27. (a) f (x )
+ 2 sin x =>
y' = sec x tan x
the slope of the tangent line at
DERIVATIVESOFTRI GONOMETRIC FUNCTIONS
(b)
-
J3.
tan x - 1
Note that l'
/}
f
1T 1---- 2
tt
= 0 where f
when f is decreasing and
has a minimum. Also note that l' is negative
r is positive when f is increasing.
-t-_ 7"-_ ---i 1T
2
r
28. (a) f (x ) = e" cos x
=> 1'(x ) = eX(- sinx)
f" (x ) = eX(- sin x - cos x)
+ (cos x -
+ (cos x) e" = eX (cos x -
sin x)
sin x) eX = eX(- sin x - cos x
=>
+ cos x -
sin x)
= - 2ex sin x
(b)
Note that
r = 0 where f has a minimum and r = 0 where l' has a
minimum. Also note that l' is negative when f is decreasing and f" is
negative when l' is decreasing.
f
- 36
= OsinO => H' (O) = 0 (cos O) + (sin O) . 1 = OcosO + sin O
H" (0) = 0 (- sin 0) + (cos 0) . 1 + cos 0 = - 0 sin 0 + 2 cos 0
29. H (O)
30. f (x ) = sec x
=> 1' (x ) = sec x tan x
f " (X) = sec x (sec 2 x)
=>
=>
+ t an x (sec x
tan x)
= sec x (sec 2 x + tan 2 x) .
J" ( ~) = v'2 [(v'2 ) 2 + 12 ] = v'2 (2 + 1) = 3 v'2
31 . (a) f (x )
=
1' (x ) =
t an x - 1
=>
secx
2
sec x (sec x ) - (t anx - l ) (sec x t a n x )
= sec x (sec2 x -
(sec z)?
sin x
- - - 1
(b) f (x ) = t an x - 1 = cosx
1
sec x
cos x
(c) From part (a,
)
32. (a) g(x)
g' (i)
tan 2 x
sec ? x
+ t a nx ) = 1 + tan x
sec x
sinx - cosx
cosx
1
=>
r (x ) = cos x -
(- sin x)
= cos x + sin x
cos x
. x , W hiICh iIS the expression
.
x = 1 + tan x = - 1- + -tan-x = cos x + Sill
f ' ()
secx
sec x
sec x
= f(x) sin x => g' (x ) = f (x ) cosx + sin x· 1' (x ), so
= f ( i) cos i + sin i . l' (i) = 4 . ~ + "{( . (-2) = 2 - J3
"lor t ' (x ).In part (b).
181
182
0
CHAPTER 3
DIFFERENTIATION RULES
(b) h(x) = CfO(SXX)
=}
h' ( )
x
f(x )· (- sin x ) - cos x . j' (x)
[f(X) ]2
' so
=
4(- 4 ) - ( ~) ( -2 )
- 2 V3+ 1
16
42
33. f (x ) = x + 2 sin x has a horizontal tangent when j' (x ) = 0
x = 2; + 21m or
1 + 2 cos x =0
{o}
4; + 21m , where n is an integer. Note that 4; and 231r are ± i
{o}
1 -2V3
16
cosx = -~
{o}
units from Jr. This allows us to write the
solutions in the more compact equivalent form (2n + l )Jr ± i, n an integer.
34. y =
cos .x
2 + sm x
- 2 sinx- 1= 0
sinx = -~
{o}
- 2 sinx - sin 2 x - cos'' x
- 2 sinx- 1
------;-::------:-...,..,,---- =
= 0 when
(2 + sin z )?
(2 + sin z )?
(2 + sin x )(- sin x ) ~cosx cos x
(2 +sinx)2
y'
=}
{o}
x = 1~1r + 2 Jrn orx = 7;+ 2Jrn, n an integer.
the points on the curve with horizontal tangents are: ( 1~1r + 2Jrn,
35. (a) x(t) = 8 sint
v( t) =x' (t) =8cost
=}
has position x
(b) The mass at time t = 21r
3
=}
en
and acceleration a (2; ) = - 8 sin 2; = - 8 (
36. (a) s(t) =2cost + 3 sint
+ 2Jrn, -
J:J),n an integer.
a(t ) = x" (t ) =- 8 sin t
= 8 sin 2; = 8 ( ~) = 4 V3, velocity v ( 231r) = 8 cos 2; = 8 (- ~) = - 4,
~ ) = - 4 V3. Since
v( t)= -2sin t +3cos t
=}
J:J), (7;
Soy = J:Jor y = - J:J and
ve3
1r
)
< 0, the particle is moving to the left.
(b)
=}
a(t) = - 2 cos t - 3 sin t
(c) s = 0
=}
t-z ~
2.55.
So the mass passes through the equilibrium
position for the first time when t
(d) v = 0
=}
Li ~
0.9 8, s(t I)
~
~
2.55 s.
3.61 em.
So the mass travels
a maximum of about 3.6 em (upward and downward) from its equilibrium position.
(e) The speed Ivl is greatest when s = 0, that is, when t = t2 +
rut; n
From the diagram we can see that sin 0 = x / 10
37.
a positive integer.
{o}
x = 10 sin O. We want to find the rate
of change of x with respect to 0, that is, dx / dO. Taking the derivative of x = 10 sin 0, we get
dx /dO = 10(cos O). So when 0 = i , ~~ = lO cos i = 10( ~) = 5 ft/rad.
x
38. (a) F
=
(b) dF = 0
dO
j1W
j1sin O+ cos O
{o}
dF
dO
(j1sinO + cos 0)(0) - j1W (j1 cos 0 - sin O)
(j1sinO + COS O)2
j1W(sinO - j1 cos 0) = 0
{o}
sin O = j1 cos 0
{o}
t an 0 = j1
j1W (sinO - j1 cosO)
(j1 sin 0 + cos 0)2
{o}
0 = tan - 1 j1
SECTION 3.3
(c)
From the graph of F ==
120
dF
d() ==
o::::}
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
0.6(50)
. ()
() for 0
0.6 SIn + cos
~
D
183
() ~ 1, we see that
() ~ 0.54. Chec kimg this with
. part (b) and
fL
== 0.6, we
calculate O == tan- 1 0.6 ~ 0.54. So the value from the graph is consistent
with the value in part (b).
o
· sin 3x
1 m - ­ == lim 3 sin 3x
39. 1
x-+O
X
x-+O
3x
· sin 3x
== 3 11 m - 3x-+0
3x
· sin ()
== 3 11m-8-+0
[multiply numerator and denominator by 3]
[as x
--+
0, 3x
--+
0]
[let () == 3x]
()
[Equation 2]
== 3(1)
==3
· sin4x
4sin4x
sin4x
1 l'1m -6x- -_ 4(1) ' -1 (1) -_ ­ 2
- -- l'1m (sin4x
- - , -x
-) -_ li1m - - , li1m - 6x
- - -_ 41'1m - ,40· 11m
x-+O sin 6x
x-+O
x
sin 6x
x-+O
4x
x-+O 6 sin 6x
x-+O
4x
6 x-+O sin 6x
6
3
41 lim tan6t == lim (sin6t ,_1_ ,_t_) == lim 6sin6t -Tim _1_, lim _2_t_
• t-+O sin 2t
t-+O
t e a s 6t sin 2t
t-+O
6t
t-+O cos 6t
t-+O 2 sin 2t
== 6 lim sin6t , lim _1_ , ~ lim ~ == 6(1) , ~ , !(1) == 3
t-+O
6t
t-+O cos 6t
2 t-+O sin 2t
1 2
cos () - 1
42
•
r
li
8~
()
cos () - 1
li
8~ sin f == 8~
sin ()
cos () - 1
()
, sin ()
11m-­
()
8-+0
o
-1-­ 0
()
sin (lim cos ())
43. lim sin(cos ())
8-+0
sec ()
8-+0
lim sec ()
8-+0
44 l'
• t~
23t
sin
_
- t 2-- -
r
t~
(sin 3t sin 3t)
l' sin 3t li sin 3t
- t - ' - t - == t~ - t - , t~ - t - ==
45. Divide numerator and denominator by (),
lim
sin ()
+ tan ()
== lim
8-+0
1
()
sin ()
+ -()-
1
, cos ()
t~
sin 3t) 2
3 lim sin3t) 2 = (3.1)2 = 9
-t­
( t-+O 3t
(sin () also works.)
r
sin ()
8-+0 ()
(1'
sin()
8~ - ( ) -
1+
,
sin () li
11m - - IIfl
8-+0
()
8-+0
u [ Sin(x 2)] u
u sin(x 2) 0
, sin(x2)
46· 11m - - - == 1m x' - - - == 1m x' 1m - -2 - == '
x-+O
X
x-+O
X ' X
x-+O
x-+O
x
x) , x
1 - sin
cos
(
1- tanx
cos x
47. lim,
== lim
X-+7r /4 SIn x - cos X
X-+7r /4 (sin x - cos x) , cos x
lim
X-+7r /4
1
cos ()
u1m
y-+o+
1
1 + 1,1
siny
-­
Y
1
2
==0,1==0
-=- ---=-
cos x - sin x I I
== lim
==
== (sin x - cos x) cos X
X-+7r /4 cos x
1/ V2
48. lim sin(x - 1) == lim sin(x - 1)
== lim _1_ lim sin(x - 1)
x-+l x 2 + X - 2
x-+l (x + 2)(x - 1)
x-+l X + 2 x-+l
x-I
=~
.1
=~
V2
184
D
CHAPTER 3
DIFFERENTIATION RULES
d sinx
tanx == - - dx
dx cos x
49. (a) (
d
d
1
b) d
dx sec x == dx cos x
d
x
.
(c) -d (sin z
sec x tan x ==
==}-
d 1 + cot x
x cscx
+ cos x)
COS X cos x - sin x (- sin x)
cos 2 X + sin'' x
==.
cos? X
cos? X
2
sec x ==
==}-
== -d
( cos x) ( 0) - 1(- sin x )
cos 2 X
.
sin x
So sec x tan x == cos 2 X
X
1
cos
--2-'
X
•
==}­
.
esc x (- csc 2 x) - (1 + cot x) (- esc x cot x)
esc x [- csc 2
cos x - SIn x ==
esc? X
==
- csc 2
2
So sec x ==
+ cot 2 X + cot x
-1
cscx
X
+ (1 + cot x)
esc?
cot x]
X
+ cot x
esc x
.
cot x - I
So cos x - SIn x ==
.
cscx
50. We get the following formulas for rand h in terms of B:
sin
~=
;0
r
=}-
= 10 sin ~
and
cos
~ = 1~
A(O)
==}-
h == lOcos
B
"2
Now A( B) == ~1fr2 and B( B) == ~ (2r)h == rho So
lim A(B) == lim
B( B)
8---+0+
~1fr2
==
rh
8---+0+
lim
!...
8---+0+
h
1.1f
2
==
1.1f
2
lim 10 sin( B/2)
10 cos( B/2)
8---+0+
== ~1f lim tan( B/2) == 0
R
8---+0+
51. By the definition of radian n1easure,
. B
d/2
see t h at SIn - == -
2
r
==}-
S
== rt), where r is the radius of the circle. By drawing the bisector of the angle B, we can
. B S
d == 2 r SIn
-2 . 0
[This is just the reciprocal of the limit lim
x---+O
sin x
x
li
S
li
rB
== 11' ITl 2· (B/ 2 ) == 11' m
B/ 2
== 1
im -d == 8---+0+
im 2 r SIn
. (B/ 2 )
(B/ 2 )
( /)
.
8---+0+ 2 sin
8---+0 sin B 2
8---+0+
== 1 combined with the fact that as B -----7 0,
f!..2
-----7
0 also.]
3.4 The Chain Rule
.
dy
dy du
1. Letu == g(x) == 4x andy == f(u) == sm tz. Then dx == du dx == (cosu)(4) == 4cos4x.
. r:
1/2
dy
dy du
1
1/2 ( )
3
3
2. Letu==g(x ) ==4+3xandY==fu
.Then-==--==-u3 ==--==
.
() ==yu==u
dx
du dx
2
2 yU
2 V4 + 3x
..
3. Letu == g(x) == 1- x 2 andy == f(u) == u
4. Let u
10
.
dy
dy du
2
9)(-2x)
Then dx == du dx == (10u
== -20x(1- x ) 9 .
= g(x) = sin x and y = f( u) = tan u. Then ~~ = ~~ ~~ =
(scc'' u)( cos x)
= sec 2 (sin x) . cos x,
or equivalently, [sec(sin x )]2 cos X.
5. Let u == g(x) ==
VX and y ==
dy
dy du
()
f(u) == e": Then - == - - == (e U ) l.X- 1/ 2 == e vx
dx
du dx
2
vx
1
e
.== - - .
2 VX
2 VX