Download 3.6 Solutions

256
CHAPTER 3
DIFFERENTIATION
54. Use the Product Rule twice to find a formula for .fg/00 in terms of f and g and their first and second derivatives.
SOLUTION
Let h D fg. Then h0 D fg 0 C gf 0 D f 0 g C fg 0 and
h00 D f 0 g 0 C gf 00 C fg 00 C g 0 f 0 D f 00 g C 2f 0 g 0 C fg 00 :
55. Use the Product Rule to find a formula for .fg/000 and compare your result with the expansion of .a C b/3 . Then try to guess
the general formula for .fg/.n/.
SOLUTION
Continuing from Exercise 54, we have
h000 D f 00 g 0 C gf 000 C 2.f 0 g 00 C g 0 f 00 / C fg 000 C g 00 f 0 D f 000 g C 3f 00 g 0 C 3f 0 g 00 C fg 000
The binomial theorem gives
.a C b/3 D a3 C 3a2 b C 3ab 2 C b 3 D a3 b 0 C 3a2 b 1 C 3a1 b 2 C a0 b 3
and more generally
.a C b/n D
n
X
kD0
!
n n!k k
a
b ;
k
where the binomial coefficients are given by
!
n
k.k ! 1/ " " " .k ! n C 1/
D
:
k
nŠ
Accordingly, the general formula for .fg/.n/ is given by
.fg/.n/ D
!
n
X
n
f .n!k/ g .k/ ;
k
kD0
where p .k/ is the kth derivative of p (or p itself when k D 0).
56. Compute
f .x C h/ C f .x ! h/ ! 2f .x/
h2
h!0
"f .x/ D lim
for the following functions:
(a) f .x/ D x
(b) f .x/ D x 2
Based on these examples, what do you think the limit "f represents?
SOLUTION
(c) f .x/ D x 3
For f .x/ D x, we have
f .x C h/ C f .x ! h/ ! 2f .x/ D .x C h/ C .x ! h/ ! 2x D 0:
Hence, ".x/ D 0. For f .x/ D x 2 ,
f .x C h/ C f .x ! h/ ! 2f .x/ D .x C h/2 C .x ! h/2 ! 2x 2
D x 2 C 2xh C h2 C x 2 ! 2xh C h2 ! 2x 2 D 2h2 ;
so ".x 2 / D 2. Working in a similar fashion, we find ".x 3 / D 6x. One can prove that for twice differentiable functions, "f D f 00 .
It is an interesting fact of more advanced mathematics that there are functions f for which "f exists at all points, but the function
is not differentiable.
3.6 Trigonometric Functions
Preliminary Questions
1. Determine the sign (C or !) that yields the correct formula for the following:
d
(a)
.sin x C cos x/ D ˙ sin x ˙ cos x
dx
d
(b)
sec x D ˙ sec x tan x
dx
d
(c)
cot x D ˙ csc2 x
dx
S E C T I O N 3.6
Trigonometric Functions
257
The correct formulas are
d
.sin x C cos x/ D ! sin x C cos x
dx
d
sec x D sec x tan x
dx
d
cot x D ! csc2 x
dx
Which of the following functions can be differentiated using the rules we have covered so far?
y D 3 cos x cot x
(b) y D cos.x 2 /
(c) y D e x sin x
SOLUTION
(a)
(b)
(c)
2.
(a)
SOLUTION
(a) 3 cos x cot x is a product of functions whose derivatives are known. This function can therefore be differentiated using the
Product Rule.
(b) cos.x 2 / is a composition of the functions cos x and x 2 . We have not yet discussed how to differentiate composite functions.
(c) x 2 cos x is a product of functions whose derivatives are known. This function can therefore be differentiated using the Product
Rule.
3. Compute
SOLUTION
d
.sin2 x
dx
C cos2 x/ without using the derivative formulas for sin x and cos x.
Recall that sin2 x C cos2 x D 1 for all x. Thus,
d
d
.sin2 x C cos2 x/ D
1 D 0:
dx
dx
4. How is the addition formula used in deriving the formula .sin x/0 D cos x?
SOLUTION The difference quotient for the function sin x involves the expression sin.x C h/. The addition formula for the sine
function is used to expand this expression as sin.x C h/ D sin x cos h C sin h cos x.
Exercises
In Exercises 1–4, find an equation of the tangent line at the point indicated.
1. y D sin x, x D
SOLUTION
!
4
Let f .x/ D sin x. Then f 0 .x/ D cos x and the equation of the tangent line is
y Df0
2. y D cos x, x D
SOLUTION
!
3
!# " !
4
x!
p !
p
p
p !
!# "
#"
2
#"
2
2
2
#"
Cf
D
x!
C
D
xC
1!
:
4
4
2
4
2
2
2
4
Let f .x/ D cos x. Then f 0 .x/ D ! sin x and the equation of the tangent line is
p !
p
p
!# "!
!# "
#"
3
#" 1
3
1
# 3
yDf
x!
Cf
D!
x!
C D!
xC C
:
3
3
3
2
3
2
2
2
6
0
3. y D tan x, x D
!
4
Let f .x/ D tan x. Then f 0 .x/ D sec2 x and the equation of the tangent line is
!# "!
!# "
!
#"
#"
#
y Df0
x!
Cf
D2 x!
C 1 D 2x C 1 ! :
4
4
4
4
2
!
4. y D sec x, x D 6
SOLUTION
SOLUTION
Let f .x/ D sec x. Then f 0 .x/ D sec x tan x and the equation of the tangent line is
yDf
0
!# "!
6
p
!# "
#"
2!
#"
2
2
2 3
#
x!
Cf
D
x!
C p D xC
C :
6
6
3
6
3
3
9
3
In Exercises 5–24, compute the derivative.
5. f .x/ D sin x cos x
SOLUTION
Let f .x/ D sin x cos x. Then
f 0 .x/ D sin x.! sin x/ C cos x.cos x/ D ! sin2 x C cos2 x:
258
DIFFERENTIATION
CHAPTER 3
6. f .x/ D x 2 cos x
SOLUTION
Let f .x/ D x 2 cos x. Then
f 0 .x/ D x 2 .! sin x/ C .cos x/ .2x/ D 2x cos x ! x 2 sin x:
7. f .x/ D sin2 x
SOLUTION
Let f .x/ D sin2 x D sin x sin x. Then
f 0 .x/ D sin x.cos x/ C sin x.cos x/ D 2 sin x cos x:
8. f .x/ D 9 sec x C 12 cot x
SOLUTION
Let f .x/ D 9 sec x C 12 cot x. Then f 0 .x/ D 9 sec x tan x ! 12 csc2 x.
9. H.t/ D sin t sec2 t
SOLUTION
Let H.t/ D sin t sec2 t. Then
d
.sec t " sec t/ C sec2 t.cos t/
dt
D sin t.sec t sec t tan t C sec t sec t tan t/ C sec t
H 0 .t/ D sin t
D 2 sin t sec2 t tan t C sec t:
10. h.t/ D 9 csc t C t cot t
SOLUTION
Let h.t/ D 9 csc t C t cot t. Then
h0 .t/ D 9.! csc t cot t/ C t.! csc2 t/ C cot t D cot t ! 9 csc t cot t ! t csc2 t:
11. f .$/ D tan $ sec $
SOLUTION
Let f .$/ D tan $ sec $. Then
!
"
f 0 .$/ D tan $ sec $ tan $ C sec $ sec2 $ D sec $ tan2 $ C sec3 $ D tan2 $ C sec2 $ sec $:
12. k.$/ D $ 2 sin2 $
SOLUTION
Let k.$/ D $ 2 sin2 $. Then
k 0 .$/ D $ 2 .2 sin $ cos $/ C 2$ sin2 $ D 2$ 2 sin $ cos $ C 2$ sin2 $:
Here we used the result from Exercise 7.
13. f .x/ D .2x 4 ! 4x !1 / sec x
SOLUTION
Let f .x/ D .2x 4 ! 4x !1 / sec x. Then
f 0 .x/ D .2x 4 ! 4x !1 / sec x tan x C sec x.8x 3 C 4x !2 /:
14. f .z/ D z tan z
SOLUTION
15. y D
Let f .z/ D z tan z. Then f 0 .z/ D z.sec2 z/ C tan z.
sec $
$
SOLUTION
Let y D
sec $
. Then
$
y0 D
16. G.z/ D
1
tan z ! cot z
SOLUTION
Let G.z/ D
1
. Then
tan z ! cot z
G 0 .z/ D
17. R.y/ D
3 cos y ! 4
sin y
$ sec $ tan $ ! sec $
:
$2
sec2 z C csc2 z
.tan z ! cot z/.0/ ! 1.sec2 z C csc2 z/
D!
:
2
.tan z ! cot z/
.tan z ! cot z/2
Trigonometric Functions
S E C T I O N 3.6
SOLUTION
Let R.y/ D
R0 .y/ D
18. f .x/ D
SOLUTION
x
sin x C 2
Let f .x/ D
3 cos y ! 4
. Then
sin y
sin y.!3 sin y/ ! .3 cos y ! 4/.cos y/
sin2 y
SOLUTION
1 C tan x
1 ! tan x
Let f .x/ D
sin2 y
D
4 cos y ! 3
sin2 y
.2 C sin x/ .1/ ! x cos x
.2 C sin x/2
D
2 C sin x ! x cos x
.2 C sin x/2
:
1 C tan x
. Then
1 ! tan x
f 0 .x/ D
#
$
.1 ! tan x/ sec2 x ! .1 C tan x/ ! sec2 x
.1 ! tan x/
20. f .$/ D $ tan $ sec $
SOLUTION
4 cos y ! 3.sin2 y C cos2 y/
x
. Then
2 C sin x
f 0 .x/ D
19. f .x/ D
D
2
D
2 sec2 x
.1 ! tan x/2
:
Let f .$/ D $ tan $ sec $. Then
f 0 .$/ D $
d
.tan $ sec $/ C tan $ sec $
d$
D $.tan $ sec $ tan $ C sec $ sec2 $/ C tan $ sec $
D $ tan2 $ sec $ C $ sec3 $ C tan $ sec $:
21. f .x/ D e x sin x
SOLUTION
Let f .x/ D e x sin x. Then f 0 .x/ D e x cos x C sin xe x D e x .cos x C sin x/.
22. h.t/ D e t csc t
SOLUTION
Let h.t/ D e t csc t. Then h0 .t/ D e t .! csc t cot t/ C csc te t D e t csc t.1 ! cot t/.
23. f .$/ D e " .5 sin $ ! 4 tan $/
SOLUTION
Let f .$/ D e " .5 sin $ ! 4 tan $/. Then
f 0 .$/ D e " .5 cos $ ! 4 sec2 $/ C e " .5 sin $ ! 4 tan $/
D e " .5 sin $ C 5 cos $ ! 4 tan $ ! 4 sec2 $/:
24. f .x/ D xe x cos x
SOLUTION
Let f .x/ D xe x cos x. Then
f 0 .x/ D x
d x
.e cos x/ C e x cos x D x.e x .! sin x/ C cos xe x / C e x cos x
dx
D e x .x cos x ! x sin x C cos x/:
In Exercises 25–34, find an equation of the tangent line at the point specified.
25. y D x 3 C cos x, x D 0
SOLUTION
Let f .x/ D x 3 C cos x. Then f 0 .x/ D 3x 2 ! sin x and f 0 .0/ D 0. The tangent line at x D 0 is
26. y D tan $,
SOLUTION
$D
!
6
y D f 0 .0/.x ! 0/ C f .0/ D 0.x/ C 1 D 1:
Let f .$/ D tan $. Then f 0 .$/ D sec2 $ and f 0 . !6 / D
4
3.
The tangent line at x D
p
!
6
is
p
! "!
!# "
#"
4!
#"
3
4
3 2#
0 #
yDf
$!
Cf
D
$!
C
D $C
!
:
6
6
6
3
6
3
3
3
9
:
259
260
DIFFERENTIATION
CHAPTER 3
27. y D sin x C 3 cos x,
SOLUTION
28. y D
Let f .x/ D sin x C 3 cos x. Then f 0 .x/ D cos x ! 3 sin x and f 0 .0/ D 1. The tangent line at x D 0 is
sin t
,
1 C cos t
SOLUTION
xD0
y D f 0 .0/.x ! 0/ C f .0/ D x C 3:
!
3
tD
sin t
1Ccos t .
Let f .t/ D
Then
.1 C cos t/.cos t/ ! sin t.! sin t/
1 C cos t
1
D
D
;
2
2
1 C cos t
.1 C cos t/
.1 C cos t/
f 0 .t/ D
and
f0
The tangent line at x D
!
3
!# "
3
D
1
2
D :
1 C 1=2
3
is
y Df0
29. y D 2.sin $ C cos $/, $ D
!# " !
3
x!
!
3
p
p
!# "
#"
2!
#"
3
2
3 2#
Cf
D
x!
C
D xC
!
:
3
3
3
3
3
3
3
9
p
Let f .$/ D 2.sin $ C cos $/. Then f 0 .$/ D 2.cos $ ! sin $/ and f 0 . !3 / D 1 ! 3. The tangent line at x D
!# "!
!# "
p !
p
#"
#"
y Df0
x!
Cf
D .1 ! 3/ x !
C 1 C 3:
3
3
3
3
30. y D csc x ! cot x, x D !4
SOLUTION
SOLUTION
Let f .x/ D csc x ! cot x. Then
f 0 .x/ D csc2 x ! csc x cot x
and
f0
Hence the tangent line is
31. y D e x cos x, x D 0
SOLUTION
4
D2!
p
p
2 " 1 D 2 ! 2:
!# " !
"
p "!
#"
# " !p
Cf
D 2! 2 x!
C
2!1
4
4
4
4
!
"
p "
p
# !p
D 2! 2 xC 2!1C
2!2 :
4
y Df0
!# "!
!# "
x!
Let f .x/ D e x cos x. Then
f 0 .x/ D e x .! sin x/ C e x cos x D e x .cos x ! sin x/;
and f 0 .0/ D e 0 .cos 0 ! sin 0/ D 1. Thus, the equation of the tangent line is
32. y D e x cos2 x,
SOLUTION
xD
y D f 0 .0/.x ! 0/ C f .0/ D x C 1:
!
4
Let f .x/ D e x cos2 x. Then
f 0 .x/ D e x
d
.cos x " cos x/ C e x cos2 x D e x .cos x.! sin x/ C cos x.! sin x// C e x cos2 x
dx
D e x .cos2 x ! 2 sin x cos x/;
and
f0
The tangent line at x D
!
4
!# "
4
D e !=4
%
&
1
1
! 1 D ! e !=4 :
2
2
is
y Df0
!# " !
4
x!
!# "
!
#"
1
# " 1 !=4
Cf
D ! e !=4 x !
C e
:
4
4
2
4
2
!
3
is
S E C T I O N 3.6
33. y D e t .1 ! cos t/,
SOLUTION
tD
Trigonometric Functions
261
!
2
Let f .t/ D e t .1 ! cos t/. Then
f 0 .t/ D e t sin t C e t .1 ! cos t/ D e t .1 C sin t ! cos t/;
and f 0 . !2 / D 2e !=2 . The tangent line at x D !2 is
!# "!
!# "
!
#"
#"
y Df0
t!
Cf
D 2e !=2 t !
C e !=2 :
2
2
2
2
34. y D e " sec $, $ D !4
SOLUTION
Let f .$/ D e " sec $. Then
f 0 .$/ D e " sec $ tan $ C e " sec $ D e " sec $.tan $ C 1/;
and
f0
!# "
4
D e !=4 sec
"
p
#!
#
tan C 1 D 2 2e !=4 :
4
4
Thus, the equation of the tangent line is
!# "!
!# "
!
p
#"
# " p !=4
y Df0
x!
Cf
D 2 2e !=4 x !
C 2e
:
4
4
4
4
In Exercises 35–37, use Theorem 1 to verify the formula.
d
cot x D ! csc2 x
dx
cos x
SOLUTION cot x D
. Using the quotient rule and the derivative formulas, we compute:
sin x
35.
d
d cos x
sin x.! sin x/ ! cos x.cos x/
!.sin2 x C cos2 x/
!1
cot x D
D
D
D
D ! csc2 x:
2
2
dx
dx sin x
sin x
sin x
sin2 x
36.
d
sec x D sec x tan x
dx
SOLUTION
Since sec x D
1
, we can apply the quotient rule and the known derivatives to get:
cos x
d
d 1
cos x.0/ ! 1.! sin x/
sin x
sin x 1
sec x D
D
D
D
D tan x sec x:
dx
dx cos x
cos x cos x
cos2 x
cos2 x
37.
d
csc x D ! csc x cot x
dx
SOLUTION
Since csc x D
1
, we can apply the quotient rule and the two known derivatives to get:
sin x
d
d 1
sin x.0/ ! 1.cos x/
! cos x
cos x 1
csc x D
D
D
D!
D ! cot x csc x:
dx
dx sin x
sin x sin x
sin2 x
sin2 x
38. Show that both y D sin x and y D cos x satisfy y 00 D !y.
SOLUTION Let y D
y 00 D ! cos x D !y.
sin x. Then y 0 D cos x and y 00 D ! sin x D !y. Similarly, if we let y D cos x, then y 0 D ! sin x and
In Exercises 39–42, calculate the higher derivative.
39. f 00 .$/,
SOLUTION
f .$/ D $ sin $
Let f .$/ D $ sin $. Then
f 0 .$/ D $ cos $ C sin $
40.
d2
dt 2
f 00 .$/ D $.! sin $/ C cos $ C cos $ D !$ sin $ C 2 cos $:
cos2 t
SOLUTION
d
d
cos2 t D
.cos t " cos t/ D cos t.! sin t/ C cos t.! sin t/ D !2 sin t cos t
dt
dt
d2
d
cos2 t D
.!2 sin t cos t/ D !2.sin t.! sin t/ C cos t.cos t// D !2.cos2 t ! sin2 t/:
dt
dt 2
262
CHAPTER 3
41. y 00 , y 000 ,
SOLUTION
DIFFERENTIATION
y D tan x
Let y D tan x. Then y 0 D sec2 x and by the Chain Rule,
y 00 D D
42. y 00 , y 000 ,
SOLUTION
d
sec2 x D 2.sec x/.sec x tan x/ D 2 sec2 x tan x
dx
y 000 D 2 sec2 x.sec2 x/ C .2 sec2 x tan x/ tan x D 2 sec4 C4 sec4 x tan2 x
y D e t sin t
Let y D e t sin t. Then
y 0 D e t cos t C e t sin t D e t .cos t C sin t/
y 00 D e t .! sin t C cos t/ C e t .cos t C sin t/ D 2e t cos t
y 000 D 2e t .! sin t/ C 2e t cos t D 2e t .cos t ! sin t/:
43. Calculate the first five derivatives of f .x/ D cos x. Then determine f .8/ and f .37/ .
SOLUTION
"
"
Let f .x/ D cos x.
Then f 0 .x/ D ! sin x, f 00 .x/ D ! cos x, f 000 .x/ D sin x, f .4/ .x/ D cos x, and f .5/ .x/ D ! sin x.
Accordingly, the successive derivatives of f cycle among
f! sin x; ! cos x; sin x; cos xg
"
in that order. Since 8 is a multiple of 4, we have f .8/ .x/ D cos x.
Since 36 is a multiple of 4, we have f .36/ .x/ D cos x. Therefore, f .37/ .x/ D ! sin x.
44. Find y .157/ , where y D sin x.
SOLUTION
Let f .x/ D sin x. Then the successive derivatives of f cycle among
fcos x; ! sin x; ! cos x; sin xg
in that order. Since 156 is a multiple of 4, we have f .156/ .x/ D sin x. Therefore, f .157/ .x/ D cos x.
45. Find the values of x between 0 and 2# where the tangent line to the graph of y D sin x cos x is horizontal.
SOLUTION
Let y D sin x cos x. Then
y 0 D .sin x/.! sin x/ C .cos x/.cos x/ D cos2 x ! sin2 x:
When y 0 D 0, we have sin x D ˙ cos x. In the interval Œ0; 2#%, this occurs when x D
! 3! 5! 7!
4, 4 , 4 , 4 .
46.
Plot the graph f .$/ D sec $ C csc $ over Œ0; 2#% and determine the number of solutions to f 0 .$/ D 0 in this interval
graphically. Then compute f 0 .$/ and find the solutions.
SOLUTION The graph of f .$/ D sec $ C csc $ over Œ0; 2#% is given below. From the graph, it appears that there are two
locations where the tangent line would be horizontal; that is, there appear to be two solutions to f 0 .$/ D 0. Now f 0 .$/ D
sec $ tan $ ! csc $ cot $. Setting sec $ tan $ ! csc $ cot $ D 0 and then multiplying by sin $ tan $ and rearranging terms yields
tan3 $ D 1. Thus, on the interval Œ0; 2#%, there are two solution of f 0 .$/ D 0: $ D !4 and $ D 5!
4 .
y
5
x
1
2
3
4
5
6
−5
47.
Let g.t/ D t ! sin t.
(a) Plot the graph of g with a graphing utility for 0 # t # 4#.
(b) Show that the slope of the tangent line is nonnegative. Verify this on your graph.
(c) For which values of t in the given range is the tangent line horizontal?
Let g.t/ D t ! sin t.
(a) Here is a graph of g over the interval Œ0; 4#%.
SOLUTION
S E C T I O N 3.6
Trigonometric Functions
263
y
12
10
8
6
4
2
2
4
6
8
10 12
x
(b) Since g 0 .t/ D 1 ! cos t $ 0 for all t, the slope of the tangent line to g is always nonnegative.
(c) In the interval Œ0; 4#%, the tangent line is horizontal when t D 0; 2#; 4#.
48.
Let f .x/ D .sin x/=x for x ¤ 0 and f .0/ D 1.
(a) Plot f .x/ on Œ!3#; 3#%.
(b) Show that f 0 .c/ D 0 if c D tan c. Use the numerical root finder on a computer algebra system to find a good approximation to
the smallest positive value c0 such that f 0 .c0 / D 0.
(c) Verify that the horizontal line y D f .c0 / is tangent to the graph of y D f .x/ at x D c0 by plotting them on the same set of
axes.
SOLUTION
(a) Here is the graph of f .x/ over Œ!3#; 3#%.
y
1
0.8
0.4
−5
5
(b) Let f .x/ D
x
10
−10
sin x
. Then
x
x cos x ! sin x
:
x2
f 0 .x/ D
To have f 0 .c/ D 0, it follows that c cos c ! sin c D 0, or
tan c D c:
Using a computer algebra system, we find that the smallest positive value c0 such that f 0 .c0 / D 0 is c0 D 4:493409.
(c) The horizontal line y D f .c0 / D !0:217234 and the function y D f .x/ are both plotted below. The horizontal line is clearly
tangent to the graph of f .x/.
y
1
0.8
0.4
−5
5
10
−10
x
49.
Show that no tangent line to the graph of f .x/ D tan x has zero slope. What is the least slope of a tangent line? Justify
by sketching the graph of .tan x/0 .
SOLUTION
f 0 .x/ D 0
Let f .x/ D tan x. Then f 0 .x/ D sec2 x D
1
.
cos2 x
Note that f 0 .x/ D
1
cos2 x
has numerator 1; the equation
therefore has no solution. Because the maximum value of cos2 x is 1, the minimum value of f 0 .x/ D
Hence, the least slope for a tangent line to tan x is 1. Here is a graph of f 0 .
y
14
12
10
8
6
4
2
−4
−2
2
4
x
1
cos2 x
is 1.
264
CHAPTER 3
DIFFERENTIATION
50. The height at time t (in seconds) of a mass, oscillating at the end of a spring, is s.t/ D 300 C 40 sin t cm. Find the velocity and
acceleration at t D !3 s.
SOLUTION
Let s.t/ D 300 C 40 sin t be the height. Then the velocity is
v.t/ D s 0 .t/ D 40 cos t
and the acceleration is
a.t/ D v 0 .t/ D !40 sin t:
p
# $
# $
At t D !3 , the velocity is v !3 D 20 cm/sec and the acceleration is a !3 D !20 3 cm/sec2 .
51. The horizontal range R of a projectile launched from ground level at an angle $ and initial velocity v0 m/s is R D
.v02 =9:8/ sin $ cos $. Calculate dR=d$ . If $ D 7#=24, will the range increase or decrease if the angle is increased slightly?
Base your answer on the sign of the derivative.
SOLUTION
Let R.$/ D .v02 =9:8/ sin $ cos $.
dR
D R0 .$/ D .v02 =9:8/.! sin2 $ C cos2 $/:
d$
If $ D 7#=24, !4 < $ < !2 , so j sin $j > j cos $j, and dR=d$ < 0 (numerically, dR=d$ D !0:0264101v02 ). At this point,
increasing the angle will decrease the range.
52. Show that if !2 < $ < #, then the distance along the x-axis between $ and the point where the tangent line intersects the
x-axis is equal to jtan $j (Figure 1).
y
y = sin x
x
2
| tan x |
FIGURE 1
Let f .x/ D sin x. Since f 0 .x/ D cos x, this means that the tangent line at .$; sin $/ is y D cos $.x ! $/ C sin $.
When y D 0, x D $ ! tan $. The distance from x to $ is then
SOLUTION
j$ ! .$ ! tan $/j D j tan $j:
Further Insights and Challenges
53. Use the limit definition of the derivative and the addition law for the cosine function to prove that .cos x/0 D ! sin x.
SOLUTION
Let f .x/ D cos x. Then
cos.x C h/ ! cos x
cos x cos h ! sin x sin h ! cos x
D lim
h
h
h!0
%
&
sin h
cos h ! 1
D lim .! sin x/
C .cos x/
D .! sin x/ " 1 C .cos x/ " 0 D ! sin x:
h
h
h!0
f 0 .x/ D lim
h!0
54. Use the addition formula for the tangent
tan.x C h/ D
tan x C tan h
1 C tan x tan h
to compute .tan x/0 directly as a limit of the difference quotients. You will also need to show that lim
h!0
SOLUTION
First note that
sin h
1
tan h
D lim
" lim
D 1.1/ D 1:
h!0 h
h!0 cos h
h!0 h
lim
Now, using the addition formula for tangent,
tan.x C h/ ! tan x
D
h
tan xCtan h
1Ctan x tan h
h
! tan x
tan h
D 1.
h
S E C T I O N 3.6
D
Therefore,
Trigonometric Functions
265
tan h.1 ! tan2 x/
tan h
sec2 x
D
"
:
h.1 C tan x tan h/
h
1 C tan x tan h
d
tan h
sec2 x
tan x D lim
"
dx
1 C tan x tan h
h!0 h
tan h
sec2 x
" lim
h!0 h
h!0 1 C tan x tan h
D lim
D 1.sec2 x/ D sec2 x:
55. Verify the following identity and use it to give another proof of the formula .sin x/0 D cos x.
!
"
! "
sin.x C h/ ! sin x D 2 cos x C 21 h sin 12 h
Hint: Use the addition formula to prove that sin.a C b/ ! sin.a ! b/ D 2 cos a sin b.
SOLUTION
Recall that
sin.a C b/ D sin a cos b C cos a sin b
and
sin.a ! b/ D sin a cos b ! cos a sin b:
Subtracting the second identity from the first yields
sin.a C b/ ! sin.a ! b/ D 2 cos a sin b:
If we now set a D x C
h
2
and b D
h
2,
then the previous equation becomes
%
& % &
h
h
sin.x C h/ ! sin x D 2 cos x C
sin
:
2
2
Finally, we use the limit definition of the derivative of sin x to obtain
!
"
! "
2 cos x C h2 sin h2
d
sin.x C h/ ! sin x
sin x D lim
D lim
dx
h
h
h!0
h!0
! "
%
&
sin h2
h
D lim cos x C
" lim ! " D cos x " 1 D cos x:
h
2
h!0
h!0
2
In other words,
d
.sin x/ D cos x.
dx
56.
Show that a nonzero polynomial function y D f .x/ cannot satisfy the equation y 00 D !y. Use this to prove that
neither sin x nor cos x is a polynomial. Can you think of another way to reach this conclusion by considering limits as x ! 1?
SOLUTION
"
"
Let p be a nonzero polynomial of degree n and assume that p satisfies the differential equation y 00 C y D 0. Then p 00 C p D 0
for all x. There are exactly three cases.
(a) If n D 0, then p is a constant polynomial and thus p 00 D 0. Hence 0 D p 00 C p D p or p % 0 (i.e., p is equal to 0 for all
x or p is identically 0). This is a contradiction, since p is a nonzero polynomial.
(b) If n D 1, then p is a linear polynomial and thus p 00 D 0. Once again, we have 0 D p 00 C p D p or p % 0, a contradiction
since p is a nonzero polynomial.
(c) If n $ 2, then p is at least a quadratic polynomial and thus p 00 is a polynomial of degree n ! 2 $ 0. Thus q D p 00 C p
is a polynomial of degree n $ 2. By assumption, however, p 00 C p D 0. Thus q % 0, a polynomial of degree 0. This is a
contradiction, since the degree of q is n $ 2.
CONCLUSION: In all cases, we have reached a contradiction. Therefore the assumption that p satisfies the differential
equation y 00 C y D 0 is false. Accordingly, a nonzero polynomial cannot satisfy the stated differential equation.
Let y D sin x. Then y 0 D cos x and y 00 D ! sin x. Therefore, y 00 D !y. Now, let y D cos x. Then y 0 D ! sin x and
y 00 D ! cos x. Therefore, y 00 D !y. Because sin x and cos x are nonzero functions that satisfy y 00 D !y, it follows that
neither sin x nor cos x is a polynomial.
266
DIFFERENTIATION
CHAPTER 3
57. Let f .x/ D x sin x and g.x/ D x cos x.
(a) Show that f 0 .x/ D g.x/ C sin x and g 0 .x/ D !f .x/ C cos x.
(b) Verify that f 00 .x/ D !f .x/ C 2 cos x and
g 00 .x/ D !g.x/ ! 2 sin x.
(c) By further experimentation, try to find formulas for all higher derivatives of f and g. Hint: The kth derivative depends on
whether k D 4n, 4n C 1, 4n C 2, or 4n C 3.
Let f .x/ D x sin x and g.x/ D x cos x.
(a) We examine first derivatives: f 0 .x/ D x cos x C .sin x/ " 1 D g.x/ C sin x and g 0 .x/ D .x/.! sin x/ C .cos x/ " 1 D
!f .x/ C cos x; i.e., f 0 .x/ D g.x/ C sin x and g 0 .x/ D !f .x/ C cos x.
(b) Now look at second derivatives: f 00 .x/ D g 0 .x/ C cos x D !f .x/ C 2 cos x and g 00 .x/ D !f 0 .x/ ! sin x D !g.x/ ! 2 sin x;
i.e., f 00 .x/ D !f .x/ C 2 cos x and g 00 .x/ D !g.x/ ! 2 sin x.
" The third derivatives are f 000 .x/ D !f 0 .x/ ! 2 sin x D !g.x/ ! 3 sin x and g 000 .x/ D !g 0 .x/ ! 2 cos x D f .x/ !
(c)
3 cos x; i.e., f 000 .x/ D !g.x/ ! 3 sin x and g 000 .x/ D f .x/ ! 3 cos x.
" The fourth derivatives are f .4/ .x/ D !g 0 .x/ ! 3 cos x D f .x/ ! 4 cos x and g .4/ .x/ D f 0 .x/ C 3 sin x D g.x/ C 4 sin x;
i.e., f .4/ D f .x/ ! 4 cos x and g .4/ .x/ D g.x/ C 4 sin x.
" We can now see the pattern for the derivatives, which are summarized in the following table. Here n D 0; 1; 2; : : :
SOLUTION
k
f
.k/ .x/
g .k/ .x/
4n
4n C 1
4n C 2
4n C 3
f .x/ ! k cos x
g.x/ C k sin x
!f .x/ C k cos x
!g.x/ ! k sin x
g.x/ C k sin x
!f .x/ C k cos x
!g.x/ ! k sin x
f .x/ ! k cos x
58.
Figure 2 shows the geometry behind the derivative formula .sin $/0 D cos $. Segments BA and BD are parallel to the
x- and y-axes. Let " sin $ D sin.$ C h/ ! sin $. Verify the following statements.
(a) " sin $ D BC
(b) †BDA D $ Hint: OA ? AD.
(c) BD D .cos $/AD
Now explain the following intuitive argument: If h is small, then BC & BD and AD & h, so " sin $ & .cos $/h and .sin $/0 D
cos $.
y
D
C
A
B
h
θ
x
O
1
FIGURE 2
SOLUTION
(a) We note that sin.$ C h/ is the y-coordinate of the point C and sin $ is the y-coordinate of the point A, and therefore also of
the point B. Now, " sin $ D sin.$ C h/ ! sin $ can be interpreted as the difference between the y-coordinates of the points B and
C ; that is, " sin $ D BC .
(b) From the figure, we note that †OAB D $, so †BAD D # ! $ and †BDA D $.
(c) Using part (b), it follows that
cos $ D
BD
AD
or
BD D .cos $/AD:
For h “small,” BC & BD and AD is roughly the length of the arc subtended from A to C ; that is, AD & 1.h/ D h. Thus, using
(a) and (c),
" sin $ D BC & BD D .cos $/AD & .cos $/h:
In the limit as h ! 0,
" sin $
! .sin $/0 ;
h
so .sin $/0 D cos $.