CHAPTER
26
Derivatives of Transcendental Functions
26.1
DERIVATIVES OF THE SINE AND COSINE FUNCTIONS
1.
y ¼ sin 3x; y0 ¼ 3 cos 3x
31.
f ðxÞ ¼ sin x; f 0 ðxÞ ¼ cos x; f 00 ðxÞ ¼ sin x
3.
y ¼ 3 cos 2x; y0 ¼ 3ð sin 2xÞ 2 ¼ 6 sin 2x
y ¼ sin x2 þ 1 ; y0 ¼ 2 x cos x2 þ 1
33.
y ¼ sin x; y0 ¼ cos x; y00 ¼ sin x; y000 ¼
cos x
y ¼ 4 sin2 3x ¼ 4ðsin 3xÞ2 ; y0 ¼ 4 2ðsin 3xÞ
ðcos 3xÞð3Þ ¼ 24 sin 3x cos 3x
y ¼ cos 3x2 2 ; y0 ¼ 6x sin 3x2 2
pffiffiffi
y ¼pffiffisin x ¼ sin x1=2 ; y0 ¼ cos x1=2 12 x1=2 ¼
cospffiffi x
2 x
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
y ¼ cos 2x3 4; y0 ¼ sin 2x3 4 1=2 2 3x2 sin pffiffiffiffiffiffiffiffiffi
1
2x3 4
3
pffiffiffiffiffiffiffiffiffi
6x
¼
3
2 2x 4
35.
y ¼ sin x; y0 ¼ cos x; y00 ¼ sin x; and y000 ¼
d4
cos x; yð4Þ ¼ sin x, so dx
4 ðsin xÞ ¼ sin x
pffiffiffi
(a) When x ¼ 4, then f ðxÞ ¼ 2 cos 4 ¼
pffiffiffi pffiffi2 ¼ 1. The line is tangent at the point
2
2
4 ; 1 . The slope of the tangent line at this point
pffiffiffi
is f 0 4 . Since f 0 ðxÞ ¼ 2 sin x, the slope of
pffiffiffi
the tangent line is 2 sin 4 ¼ 1. Using the
5.
7.
9.
11.
13.
37.
2x 4
15.
17.
19.
21.
23.
25.
27.
29.
slope-intercept form for the equation of a line,
y ¼ x2 þ sin2 x; y0 ¼ 2x þ 2 sin x cos x
the equation of the tangent line is
y y0 ¼ m x x0
y 1 ¼ 1 x 4
¼ x þ
4
y ¼ x þ
4
(b)
0
y ¼ sin x cos x; y ¼ sin xð sin xÞ þ cos x cos x ¼
cos2 x sin2 x ¼ cos 2x
2 cos x
;
y¼
sin 2x
sin 2xð2 sin xÞ 2 cos x cos 2x 2
y0 ¼
sin2 2x
2 sin x sin 2x 4 cos x cos 2x
¼
sin2 2x
2 sin xð2 sin x cos xÞ 4 cos x 2 cos2 x 1
¼
ð2 sin x cos xÞ2
2
4 cos x sin x þ 2 cos2 x 1
¼
2
2
4 sin2 x cos x2
4 cos x sin x þ cos x þ cos2 x 1
¼
2
2
42 sin x cos x
4 cos x cos x
cos x
¼
¼
sin2 x
4 sin2 x cos2 x
y ¼ x2 sin x; y0 ¼ 2x sin x þ x2 cos x
pffiffiffi
pffiffiffi
y ¼ x sin x; y0 ¼ 12 x1=2 sin x þ x cos x
y ¼ ðsin 2xÞðcos 3xÞ; y0 ¼ sin 2xð 3 sin 3xÞ þ
2 cos 2x cos 3x ¼ 2 cos 2x cos 3x 3 sin 2x sin 3x
y ¼ sin3 x4 ; y0 ¼
3sin2 x4 cos x4 4x3 ¼
12x3 sin2 x4 cos x4
y ¼ sin2 x þ cos2 x ¼ 1; y0 ¼ 0
39.
(a) The
slope of the tangent line to h at this point
is h0 2 . Since h0 ðxÞ ¼ 4 cosð4xÞ, we see that
h0 2 ¼ 4 cosð2Þ ¼ 4. Thus, the slope of the
normal line to h at this point is 14 and its equation
is y y0 ¼ m x x0
1
y0¼
x
4
2
1
y¼ xþ
4
8
265
266
CHAPTER 26
DERIVATIVES OF TRANSCENDENTAL FUNCTIONS
(b)
41.
43.
45.
As in Example 27.7, V 0 ¼ 45;240 cos 377ð1:571Þ
¼ 3470:332 or about 3470 V.
The current to the capacitor is given by I ¼ q0 ðtÞ ¼
0:25 sinðt 1:45Þ. When t ¼ 7:6 s, the current is I ¼ q0 ð7:6Þ ¼ 0:25 sinð7:6 1:45Þ ¼
0:25 sinð6:15Þ ¼ 0:033 A.
pffiffi
I 0 ðÞ ¼ 2M cosðÞ sinðÞ; I 0 3 ¼ 23 M 0:866M cd/rad
47.
To find the maximum and minimum, we first
find the critical values. I 0 ðtÞ ¼ 2 cos t sin t ¼ 0
sin t
or 2 cos t ¼ sin t and so, cos
t ¼ tan t ¼ 2. This
occurs when t 1:107 and t 4:249. The maximum is at Ið1:107Þ ¼ 2 sinð1:107Þ þ cosð1:107Þ 2:24 A. The minimum is Ið4:249Þ ¼ 2 sinð4:249Þþ
cosð4:249Þ 2:24 A.
49.
(a) This sketch is from the end of 2000, when
t ¼ 100, to end of 2008, when t ¼ 108. Since the
marks at the bottom of the screen are one (year)
apart you should be able to determine how the
graph for the present year should look.
26.2
1.
(b) The given formula is for t years after 1900,
which means that on December 31, 1900, t ¼ 0,
on January, 1, 1901, t 0:023, on December 31,
1901, t ¼ 1. So, December 31, 2001, t ¼ 101 and
January 1, 2002, t ¼ 101:003: That means the July
1995 would be about when t ¼ 94:5. Evaluating
Cð94:5Þ 360:2423 ppm
(c) April 2002 was about when t ¼ 101:25 and
Cð101:25Þ 375:8506 ppm.
(d) C 0 ðtÞ ¼ ð0:010345Þ sinð6:283tÞ þ ð6:283Þ
ð0:010345tþ2:105676Þ cosð6:283tÞþ0:027566t 0:844566 part per million per year t years after
1900.
(e) C0 ð101:25Þ 2:3285 ppm/yr.
(f) This sketch is from the end of 2000, when
t ¼ 100, to the end of 2008, when t ¼ 108. Since
the marks on the x-axis are one (year) apart you
should be able to determine how the graph for
the present year should look.
DERIVATIVES OF THE OTHER TRIGONOMETRIC FUNCTIONS
pffiffiffi
pffiffiffi
pffiffiffi
y ¼ tan2 x; y0 ¼ 2 tan x sec2 x 12 x1=2 ¼
pffiffi 2 pffiffi
tan xpsec
ffiffi x
13.
x
3.
y ¼ sec 5x; y0 ¼ 5 sec 5x tan 5x
5.
y ¼ cscð2x 1Þ; y0 ¼ 2 cscð2x 1Þ cotð2x 1Þ
7.
2
y ¼ sin x tan x; y0 ¼ sin
x þ tan x cos x ¼
x sec
2
2
sin x sec x þ sin x ¼ sin x sec x þ 1
9.
y ¼ tan3 x; y0 ¼ 3 tan2 x sec2 x
y ¼ sec4 x2 ; y0 ¼
4 sec3 x2 sec x2 tan x2 ð2xÞ ¼
8x sec4 ðx2 tan x2
11.
15.
17.
19.
y ¼ tanx cot x; y0 ¼ tan x csc2 x þ cot x sec2 x ¼ sec2 x cot x csc2 x tan x
y ¼ sin2 xcot x; y0 ¼ 2 sin x cos x cot x þ sin2 x 2
x
1
csc2 x ¼ 2 sin x cos x cos
sin x sin x sin2 x ¼
2
2 cos x 1 ¼ cos 2x
y ¼ tan 1x ; y0 ¼ sec2 1x x2 ¼ x12 sec2 1x
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1=2 0 1 y ¼ 1 þ tan x2 ¼ 1 þ tan x2
; y ¼2 1þ
2 2
2 2
x
sec
x
2 1=2
tan x
sec x ð2xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi2
1þtan x
SECTION 26.3
21.
y¼
y0 ¼
¼
tan x
;
1 þ sec x
ð1 þ sec xÞ sec2 x tan xðsec x tan xÞ
35.
ð1 þ sec xÞ2
sec2 x þ sec3 x sec x tan2 x
¼
33.
ð1 þ sec xÞ2
2
2
sec x sec x þ sec x tan x
ð1 þ sec xÞ2
sec x½sec x þ 1
sec x
¼
¼
2
1 þ sec x
ð1 þ sec xÞ
23.
cot x
0
Here y ¼ 1csc
x ; so y ¼
ð1csc xÞ csc2 x cot xðcsc x cot xÞ
ð1csc xÞ2
csc2 xþcsc3 xcsc x cot2 x
ð1csc xÞ2
csc xð1csc xÞ
ð1csc xÞ2
25.
27.
¼
Since y ¼ x sin y, then y0 ¼ xðcos yÞy0 þ sin y, or
y0 xðcos yÞy0 ¼ sin y, or y0 ð1 x cos yÞ ¼ sin y
y
and so y0 ¼ 1xsincos
y.
39.
Here x þ y ¼ sinðx þ yÞ, and differentiating, we
obtain 1 þ y0 ¼ cosðx þ yÞ½1 þ y0 ¼ cosðx þ yÞ þ
y0 cosðx þ yÞ, or y0 y0 cosðxþ yÞ ¼ cosðx þ yÞ1,
1cosðxþyÞ
and so y0 ¼ cosðxþyÞ1
1cosðxþyÞ or cosðxþyÞ1.
n0 ¼ csc cot ¼ csc 8 cot 8 ¼ 6:31
41.
¼
csc x csc2 xcot2 xcsc x
2
ð1csc xÞ
¼
43.
2xðtan 2xÞ
45.
sec x
0
Here y ¼ 1þtan
x, and so y ¼
3
sec
x tan xþtan2 xsec2 x
ð1þtan xÞðsec x tan xÞsec x
¼
¼
2
2
ð1þtan xÞ
ð1þtan xÞ
sec xðtan x1Þ
ð1þtan xÞ2
31.
d
dx cot u
d
¼ dx
cos u sin u
Differentiating I ¼ 16 cot2 ð2Þ, produces
I 0 ¼ 2 16 cotð2Þ csc2 ð2Þ ð2Þ
¼ 64 cotð2Þ csc2 ð2Þ
0 7
When ¼ 7
16 , then I 16 ¼
2 7 csc 2 16 ¼
64 cot 2 7
16
7 2 7 64 cot 8 csc 8 1055:05 mV/rad.
csc x
¼ 1csc
x
2=5
y ¼ tan 2x; y0 ¼ 2 sec2 2x ¼ 2ðsec 2xÞ2 ; y00 ¼
4ðsec 2xÞðsec 2x tan 2xÞ2 ¼ 8 sec2 2x tan 2x
y ¼ x tan x; y0 ¼ x sec2 x þ tan x; y00 ¼ xð2 sec x sec x tan xÞ þ sec2 x þ sec2 x ¼ 2x sec2 x tan x þ
2 sec2 x ¼ 2 sec2 xðx tan x þ 1Þ
37.
y ¼ ðcsc x þ 2 tan xÞ3 ; y0 ¼ 3ðcsc x þ 2 tan xÞ2 csc x cot x þ 2 sec2 x ¼ 3ðcsc x þ 2 tan xÞ2 2 sec2 x csc x cot x
y ¼ ðtan 2xÞ3=5 ; y0 ¼ 35 ðtan 2xÞ2=5 sec2 2x ð2Þ ¼
6
2
5 sec
29.
267
¼
T
W
1
T
d=dt ¼
T 2 w2
1þ w
1
T
d=dt ¼
T2
W2
1 þ W2
¼ arctan
d=dt ¼
T
þ T2
W2
d
sin u dx
cos u cos u2 sin u
¼
sin2 u
du
sin uð sin uÞ du
dx cos u cos u dx
¼
sin2 u
sin2 ucos2 u du
dx
¼ csc2 u du
dx
sin2 u
26.3
DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
1.
2
y ¼ sin1 2x, so y0 ¼ pffiffiffiffiffiffiffiffiffi
.
14x2
3.
y ¼ tan1 2x, so y0 ¼
5.
2x
y ¼ cos1 1 x2 and y0 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi ¼
1
2
2
1 þ x4
7.
2
¼ 4þx
2.
1 1x2
2x ffi
pffiffiffiffiffiffiffiffiffiffi
2x2 x4
2
¼
y ¼ sin1 ð1 2xÞ, so y0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1ð12xÞ
ffiffiffiffiffiffiffi2ffi
¼ p1
xx
9.
13x2
y ¼ cos1 x3 x ; y0 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi ¼
1 x3 x
2
13x
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1x2 þ6x4 x6
2 ffi
pffiffiffiffiffiffiffiffiffiffi
4x4x2
268
11.
CHAPTER 26
DERIVATIVES OF TRANSCENDENTAL FUNCTIONS
1
0
y ¼ sec ð4x þ 2Þ; y ¼
ð2xþ1Þ
ð4xþ2Þ
4
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼
2
ð4xþ2Þ 1
ð4xþ2Þ 1
1 ffi
þ sin1 x
y ¼ x sin1 x; y0 ¼ x pffiffiffiffiffiffiffi
1x2
15.
x ffi
y ¼ x2 cos1 x; y0 ¼ 2x cos1 x pffiffiffiffiffiffiffi
1x2
17.
Since y ¼ tan1
2x
dy
dx
2x
sin
1x2
1
1x2
az ¼ xel sec el
az
d
¼ xel
sec el
del
el
az
¼ xel sec el tan el
el
x ffi
pffiffiffiffiffiffiffiffiffiffiffi
2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
sin
1x2
2x2 x4
1 ffi
¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffi.
sin1 1x2
23.
25.
2x2
y ¼ x cot1 1 þ x2 ; y0 ¼ cot1 1 þ x2 þ
2x2
x 1 2 2x ¼ cot1 1 þ x2 2þ2x
2 þx4
2
1þ 1þx
1 ffi sin1 x
x pffiffiffiffiffiffiffi
1
1x2
Since y ¼ sinx x, then y0 ¼
x2
1
1
1 ffi
1 ffi
¼ xpffiffiffiffiffiffiffi
sinx2 x ¼ 1x pffiffiffiffiffiffiffi
sinx x .
1x2
1x2
26.4
APPLICATIONS
1.
If y ¼ sin x, then y0 ¼ cos x. Since cos x > 0 on
2 ; 2 ; sin x is increasing on 2 ; 2 .
3.
If y ¼ arccos x, then y0 ¼
1 ffi
pffiffiffiffiffiffiffi
1x2
or sin x ¼ 0:84307. Hence, x ¼ 3:7765; x ¼
5:6483; x ¼ 1:0030, or x ¼ 2:1386. Maxima are
; 1:5 ; minima are at 2 ; 1 and
at 6 ; 1:5 ; 5
6
3
2 ; 3 . Inflection points are at ð1:0030; 1:2646Þ;
ð2:1386; 1:2646Þ; ð3:7765; 0:8896Þ and ð5:6483;
0:8896Þ.
when jxj < 1.
0
Hence y < 0 which means that the arccos x is
decreasing for jxj 1.
5.
d
1 ffi du
¼ dx
cos1 u ¼ pffiffiffiffiffiffiffi
.
1u2 dx
33.
1 1x2
dy
dx
In this problem, let P be a point on the track
directly in front of the camera and x the distance the car has traveled past P, and the angle
from P to the camera to x. Then ¼
x
1
1 dx
42 dx
tan1 42
and d
x 2 42
dt ¼
dt ¼ 422 þx2 dt .
1 þ 42
x
x
When ¼ 15 , then ¼ tan1 42
or tan ¼ 42
and x ¼ 42 tan 15 11:25387. Substituting this
value of x and the given information that
dx
d
dt ¼ 320 ft/s in the formula for dt produces
d
42
dt ¼ 422 þð11:25387Þ2 ð320Þ 7:1087 rad/s.
x
y ¼ x tan1 ðx þ 1Þ; y0 ¼ tan1 ðx þ 1Þ þ 1þðxþ1Þ
2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Here we are given y ¼ sin1 1 x2 , and so
1=2
2x
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
y0 ¼ 12 sin1 1 x2
2 ¼
1
31.
4x2
x
ffiqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ffi
pffiffiffiffiffiffiffi
1x2
2
1 du
2
¼ sin
y dx. Since sin y þ cos y ¼ 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi
we have sin y ¼ 1 cos2 y ¼ 1 u2 , and so
1
2x1 2 , then y0 ¼
d
Let y ¼ cos1 u or u ¼ cos y. Hence du
dx ¼ dx cos y ¼
sin y dy
dx so
1þ
2x2ð2x1Þ2 4x4xþ2 1
¼ 4x2 þ4x2 4xþ1
¼
4x2
4x2
2
8x2 4xþ1
21.
29.
2
2x1 y ¼ ðarcsin xÞ1=3 ; y0 ¼ 13 ðarcsin xÞ2=3
1
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
3 1 x ðarcsin xÞ2=3
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
13.
19.
27.
f ðxÞ ¼ 2 sin x þ cos 2x; f 0 ðxÞ ¼ 2 cos x 2 sin 2x.
The critical values occur when f 0 ðxÞ ¼ 0. Hence,
we have 2 cos x 2 sin 2x ¼ 0; 2 cos x 4 sin x
cos x ¼ 0. Factoring, we obtain 2 cos xð1 2 sin xÞ
¼ 0. If cos x ¼ 0 ) x ¼ 2 ; 3
2 and if 1 2 sin x ¼
.
0 ) sin x ¼ 12 ) x ¼ 6 ; 5
6 Differentiating again,
00
we get f ðxÞ ¼ 2 sin x 4 cos 2x ¼ 0, or
2ðsin x þ 2ð1 2 sin2 xÞÞ ¼ 0, or 2ðsin x þ 2 4 sin2 xÞ ¼ 0, and so 2 4 sin2 x sin x 2 ¼ 0.
The quadratic formula yields sin x ¼ 0:59307
7.
hðxÞ ¼ cos x sin x. To find the critical values,
we take the derivative and set it equal to 0. Thus,
SECTION 26.4
h0 ðxÞ ¼ sin x cos x ¼ 0 and so, sin x ¼
cos x, or tan x ¼ 1, which means that x ¼ 3
4 or
7
00
.
The
second
derivative
is
h
ðxÞ
¼
cos
x þ
4
sin x and setting it equal to 0 produces sin x ¼
cos x or tan x ¼ 1. Thus, x ¼ 4 or x ¼ 5
4 . As a
7 pffiffiffi
result, the maxima are 4 ; 3 ; the minimum is
pffiffiffi
3
4 ; 3 , and the inflection points are
4; 0
5 and 4 ; 0 .
13.
3
y ¼ tan
x; at x ¼ 4, we get y ¼ 1, so point is
slope, we take the derivative
4 ; 1 . To find the
y0 ¼ 3 tan x sec2 x and at 4, we get y0 ¼ 3 pffiffiffi
12 ð 2Þ2 ¼ 6. Hence, the equation of the tangent
line is y 1 ¼ 6 x 4 .
15.
The horizontal distance from the light to the
plane is 8000 ft. The velocity is dx
dt ¼ 400 mi/h ¼
5280 ft
1 hr
x
¼
586:7
ft/s;
tan ¼ 6000
.
400 mile
hr
1 mi
3600 s
2 d
1 dx
Taking derivatives we have sec dt ¼ 6000 dt . Also,
4
tan ¼ 8000
6000 ¼ 3, and so ¼ 0:9273 rad. Hence,
y
1
d
dt
π
2π
x
–1
9.
kðxÞ ¼ cos2 x þ sin x. To find the critical values,
we take the derivative and set it equal to 0. k0 ðxÞ ¼
2 cos x sin x þ cos x ¼ cos xð2 sin x þ 1Þ ¼ 0.
This yields cos ¼ 0 which means that there are
critical values at x ¼ 2 and x ¼ 3
2 . We also get
sin x ¼ 12 ; and so there are additional critical values at x ¼ 6 and x ¼ 5
6 . For inflection points, we
take the second derivatives and set them equal to
0: k00 ðxÞ ¼ cos xð2 cos xÞ sin xð2
sin x þ 1Þ
¼
2 cos2 x þ 2 sin2 x sin x ¼ 2 1 2 sin2 x þ
2 sin2 x sin x ¼ 6 sin2 sin x 2 ¼ ð3 sin x 2Þ ð2 sin x þ 1Þ ¼ 0. Hence, sin x ¼ 23 ) x ¼
0:7297 or x ¼ 2:4119, or sin x ¼ 12 ) x ¼ 7
6
or 11
that this function has the fol6 . Thus, we see
5
lowing maxima: 6 ; 1:25 and 6 ; 1:25 ; min
points:
ima: 2 ; 1 and 3
2 ; 1 ; and inflection
;
0
and
(0.7297, 1.2222), (2.4119, 1.2222), 7
6
11 6 ; 0 .
y
2π
11.
y ¼ sin x, at x ¼ 6, we get y ¼ sin 6 ¼ 0:5.
Thus,
we
want to find the line tangent at this point
;
0:5
. To find the slope of the tangent line we
6
take the derivative y0 ¼pcos
ffiffi x and evaluate it when
x ¼ 6 to get cos 6 ¼ 23. Using the point-slope
form for the equation of the
we
pffiffi line,
obtain the
desired equation: y 0:5 ¼ 23 x 6 .
dx=dt
6000 sec2 ðdx=dtÞ cos2 6000
¼
¼
ð586:7Þ cos2 ð0:9273Þ
6000
¼
17.
d
dt
19.
d
dt 30
21.
(a) Q ¼ 1:33ð2Þ5=2 tan 2 ¼ 7:523 tan 2 ; Q ¼
1
2
2 ð7:5236Þ sec 2
(b) Q0 ð30Þ ¼ 3:7618 sec2 15 ¼ 3:7518ð1:0353Þ2
¼ 4:0319 m3 =s
(c) Qð40Þ ¼ 3:7168 sec2 20 ¼ 4:2602
0
(d) Q ¼ k sec2 2; critical values at 2 ¼ 90 or
¼ 180 . Q is increasing so maximum at ¼
180 .
23.
I ¼ sin2 3t; I 0 ¼ 6 sin 3t cos 3t; I 0 ð1:5Þ ¼
6ðsin 4:5Þðcos 4:5Þ ¼ 1:2364 A/s
25.
xðtÞ ¼ 14 cos 2t; vðtÞ ¼ 12 sin 2t; aðtÞ ¼ vðtÞ ¼
cos 2t; aðtÞ ¼ 0
when
2t ¼ 2 þ 2k
or
3
þ
2k.
By
the
first
derivative
test
the
maximums
2
3
occur at 2t þ 3
or t ¼
2 þ 2k or t ¼ 4 þ k
3þ4k
3
;
k
¼
0;
1;
2;
3;
.
.
.
:
Thus,
v
¼
12 sin
4
34 1
1
¼ 2 ð1Þ ¼ 2 m/s ¼ 0:5 m/s.
2
27.
f ðxÞ ¼ cosx þ x 2; f 0 ðxÞ ¼ sinx þ 1. Use first
x¼ 3.
f xn
xnþ1
f xnþ1
n xn
x
–1
¼
0:0352 rad/s.
1
π
269
¼ 1 rpm ¼ 2 rad/min ¼ 0:10472 rad/s .
x
tan ¼ 500
. Taking derivatives we have sec2 2
d
1 dx
dx
2
d
sec 4
dt ¼ 500 dt or dt ¼ sec dt 500 ¼
ð0:10472Þð500Þ ¼ 104:72 m/s.
rpm ¼ 3:1416 rad/s. We know that tan ¼ 8x,
and taking derivatives of both sides we have
1 dx
dx
2
sec2 d
dt ¼ 8 dt . Hence, dt ¼ 8 sec 40 ð3:1416Þ
¼ 42:8284 mi/s.
0
1 3
0:0100
2:98835 6:716105
5
2 2:98835 6:71610
2:9883 3:105109
29.
2
gðxÞ ¼ sin x þ x4 2x; g0 ðxÞ ¼ cos x þ 34 x2 2; x ¼ 0 is one answer. By Newton’s method x ¼
2:71995214. Since gðxÞ is odd, x ¼ 2:71995214
is also an answer.
270
CHAPTER 26
26.5
1.
3.
5.
9.
11.
13.
15.
DERIVATIVES OF LOGARITHMIC FUNCTIONS
Since y ¼ log 5x, its derivative is y0 ¼ 15 x loge
5 ¼ 1x ln110.
y ¼ ln x2 þ 4x ; y0 ¼ x2xþ4
2 þ4x
qffiffiffiffiffiffiffiffiffiffiffi
1=2
0
2
y ¼ ln x þ 4x ; y ¼ 12 ln x2 þ 4x
2xþ4
x2 þ4x
7.
DERIVATIVES OF TRANSCENDENTAL FUNCTIONS
xþ2ffi
¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 1þln x3=2
x 1 ln x 1ln x
y ¼ x 2
¼ x2
x
0
29.
Since y ¼ x2 lnðsin 2 xÞ, then y0 ¼ 2 x lnðsin 2 xÞ þ
2 x2
2
sin 2 x cos 2 x ¼ 2 x lnðsin 2 xÞ þ 2 x cot 2 x, and so
y00 ¼ 2 lnðsin 2xÞþ2x sin12x cos 2x 22x2 csc2 2x 2þ 4x cot 2x ¼ 2 lnðsin 2xÞþ 4x cot 2x4x2 csc2 2x
þ4x cot 2x ¼ 2 lnðsin 2xÞ þ 8x cot 2x 4x2 csc2 2x.
31.
(a) N 0 ðtÞ ¼ 5 lnð0:045tÞ 0:225t
0:045t 0:5 ¼
5 lnð0:045tÞ5:5. Solving 5 lnð0:045tÞ5:5 ¼
0, or lnð0:045tÞ ¼ 1:1, we have 0:045t ¼ e1:1
1
and so, t ¼ 0:045
e1:1 7:397 h. The maximum
number of medflies will be Nð7:397Þ 108:96 109.
(b) Graphing NðtÞ,we see that it is 0 at about 31.67 h,
so the medflies will be eradicated within the 36 h
period.
33.
(a) In order to find the critical values, we determine when P0 ðtÞ ¼ 0:008 lnð0:025tÞ 0:008 ¼ 0.
The maximum value is when lnð0:025tÞ þ 1 ¼ 0
or when t ¼ 40e1 14:72 h
(b) Pð14:72Þ ¼ 0:014 0:008ð14:72Þ ln 0:025ð14:72Þ ¼ 0:014 0:11776 lnð0:368Þ 0:132 ppm.
35.
(a) H ¼xlnxð1xÞlnð1xÞ
dH
1
1
¼ð1lnxx Þ½1lnð1xÞþ
ð1xÞð1Þ
dx
x
1x
dH
¼lnx1½lnð1xÞ1
dx
dH
¼lnx1þlnð1xÞþ1
dx
dH
¼lnxþlnð1xÞ
dx
1=2 0 1 1þx 1=2 1x y ¼ ln 1þx
; y ¼ 2 ln 1x
1þx 1x
ð1xÞþð1þxÞ 1 1þx 1=2 1x 2 ¼ 2 ln 1x
¼
1þx
ð1xÞ2
ð1xÞ2
1=2
1
1þx 1=2
1
ln 1þx
¼ 1x
2
ð1þxÞð1xÞ ln 1x
1x
pffiffiffiffiffiffiffiffi
3
x2 þ4
ffiffiffiffiffiffiffi
ffi ; y0 ¼ 4x
y ¼ ln p4x
3
x2 þ4
p
ffiffiffiffiffiffiffiffiffiffiffiffi
ffi
1=2
2
2x
x þ 4 12x2 4x3 12 x2 þ 4
¼
x2 þ 4
pffiffiffiffiffiffiffiffi
x2 þ 4 12x2 4x4
4x2 x2 þ4 3x2
x2 þ4
1
¼
¼
3=2
x2 þ4
4x3
4x3
x2 þ4
2x
2 þ12
x x2 þ4
17.
y ¼ ðln xÞ2 ; y0 ¼ 2 ln x 1x ¼ 2 lnx x
19.
y ¼ lnðln xÞ; y0 ¼ ln1x 1x ¼ x ln1 x
21.
Since y ¼ ln x, then y0 ¼ 1x ¼ x1 , and y00 ¼
1x2 ¼ 1
x2 .
y ¼ 1x ln x; y0 ¼ 1x 1x þ ln x 1x2 ¼
23.
1
x2
25.
ln x
x2
¼
1ln x
x2 ;
y00 ¼
x2 1x ð1 ln xÞ2x
x4
¼ x 2xxð14 ln xÞ ¼ 1 2xþ3 2 ln x ¼ 2 lnxx3 3 ¼
qffiffiffiffiffiffiffiffi
2
2
3
y ¼ ln sinx3 x ¼ 12 ln sinx3 x ; y0 ¼ 12 sinx 2 x 2 x3 sin x cos x 3x2 sin2 x
x6
ln x2 3
x3
x 3 sin x
¼ 2 x cos
¼ cot x 23x ;
2 sin x x
y00 ¼ csc2 x 32 ð1Þ
x2
¼
ln x2
x2 ln x2
1
Here y ¼ ln 1x and its derivative is y0 ¼ 1=x
1x2 ¼ 1x. By an alternate method: y ¼
ln x1 ¼ ln x, so y0 ¼ 1 1x ¼ 1x.
y ¼ ln tan x; y0 ¼ tan1 x sec2 x ¼ sec x csc x
y¼
pffiffiffiffiffiffiffiffiffi
1=2 0
1=2 1
; y ¼ 12 ln x2
x2 ln x2 ¼ ln x2
1=2
1
2
2x ¼ 1 1=2 ; y00 ¼ x21
2x
ln x2 x 2 ln x
x ln x2
2 1=2
1
1
2 1=2
þ ln x
¼ x2 ln x2 1=2 þ ln x
y¼
2
ln x2 þ4x x2 þ4x
ln x
x ;
27.
3
2x2
csc2 x
(b) lnxþlnð1xÞ¼ 0
lnð1xÞ¼ lnx
1x¼ x
1¼ 2x
x¼ 0:5
(c) Hð0:5Þ
0:6931
SECTION 26.6
26.6
DERIVATIVES OF EXPONENTIAL FUNCTIONS
3.
y ¼ 4x ; y0 ¼ 4x 1 ln 4 ¼ 4x ln 4
pffi
pffiffi
pffiffi
x
y ¼ 5 x ; y0 ¼ 5 x 2p1 ffiffix ln 5 ¼ 25pffiffix ln 5
5.
y ¼ 2sin x ; y0 ¼ 2sin x cos x ln 2 ¼ cos x 2sin x ln 2
7.
y ¼ ex
1.
2
þx
; y0 ¼ ð2x þ 1Þex
4
2
4
11.
y ¼ xe2 ; y0 ¼ x
13.
x
2 x
0
e ex ð2xÞ
x4
; y ¼
x
x2 ex 1þex 2x
x4
¼
xex 22ex
x3
¼
y ¼ etan x ; y0 ¼ etan x sec2 x ¼ sec2 x etan x
17.
y ¼ sin ex ; y0 ¼ cos ex ex 2x ¼ 2xeex cos ex
y ¼ 3x x3 1 ; y0 ¼ 3x 3x2 þ x3 1 3x ln 3 ¼
3x2 3x þ x3 1 3x ln 3
19.
21.
2
2
2
23.
y ¼ ðsin xÞx ; Use logarithmic differentiation. Then,
ln y ¼ x lnðsin xÞ, or 1y y0 ¼ x sin1 x cos x þ
ln ðsin xÞ, and so y0 ¼ y½x cot x þ ln ðsin xÞ or
y0 ¼ ðsin xÞx ½x cot x þ lnðsin xÞ.
25.
e
y ¼ ln sin e3x ; y0 ¼ sin1e3x cos e3x e3x 3 ¼ 3e sincos
e3x ¼
3e3x cot e3x
27.
y ¼ ey þ y þ x; this is equivalent to 0 ¼ ey þ x.
By implicit differentiation we obtain 0 ¼ ey y0 þ 1
0
y
or ey y0 ¼ 1 so y0 ¼ 1
ey or y ¼ e
d
d ex þ ex
¼ 12 ex þ ex ð1Þ ¼
dx ðcosh xÞ ¼ dx
2
29.
41.
ex ex
2
¼ sinh x
2
1
Here f ðxÞ ¼ sinh1 7x, so f 0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi
7¼
2
ð7xÞ þ1
3x
jðxÞ ¼ x sinh1 1x. Using the product rule with
the chain rule we obtain j0 ðxÞ ¼ sinh1 1x þ x 1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2
þ
1
x
y ¼ xcos x . Using logarithmic differentiation, we
obtain ln y ¼ ln xcos x or ln y ¼ cos x ln x, or
1 0
1
0
1
y y ¼ cos x x þ ð sin xÞ ln x, and so y ¼ y x
cos x sin x ln x , or y0 ¼ xcos x cosx x ðsin xÞ
ðln xÞ .
3x
2
7
pffiffiffiffiffiffiffiffiffiffiffi
.
49x2 þ1
15.
2
x
¼ 2e2 ¼ ex
y ¼ esinh 3x ; y0 ¼ esinh 3x cosh 3x 3 ¼
3ðcosh 3xÞesinh 3x
pffiffiffiffiffi
pffiffiffiffiffi
gðxÞ ¼ tanh x3 þ 4; g0 ðxÞ ¼ sech2 x3 þ 4 pffiffiffiffiffiffiffiffiffiffiffiffiffi
1=2
3
2
1
ffi sech2 x3 þ 4
3x2 ¼ 2p3xffiffiffiffiffiffiffi
2 x þ4
x3 þ4
jðxÞ ¼ cosh3 x4 þ sin x ; j0 ðxÞ ¼ 3 cosh2 x4 þ
sin x sinh x4 þ sin x 4x3 þ cos x ¼ 3 4x3 þ
cos x cosh2 x4 þ sin x sinh x4 þ sin x
39.
2
e x e x 33.
37.
¼ e ðx2Þ
x3
þ
cosh x þ sin hx ¼
3
y ¼ 4x ; y0 ¼ 4x 4x3 ln 4 ¼ 4x3 4x ln 4
e x þ e x 31.
35.
þx
9.
x
y ¼ 1þe
x2
ex ðx2Þ2
x3
271
43.
1
x2
¼ sinh1
1
x
1
qffiffiffi
¼
x x12 þ 1
jxj
ffiffiffiffiffiffiffi2ffi. Recall that
sinh1 1x qffiffiffiffiffiffiffiffi ¼ sinh1 1x xpjxj
1þx
2
x 1þx
2
x
pffiffiffiffiffi
x2 ¼ jxj.
The rate at which N is growing is N 0 ðtÞ ¼ 615 e0:02t ð0:02Þ ¼ 12:3e0:02t . After 7 days, N is
growing at the rate of N 0 ð7Þ ¼ 12:3e0:14 ¼ 10:69
people/day.
45.
(a) Using the product rule, we obtain I ¼ q0 ðtÞ ¼
5et cos 2:5t 12:5et sin 2:5t,
(b) q0 ð0:45Þ 8:57 A
47.
(a) N 0 ðtÞ ¼ 106 ð0:1Þ sech2 ð0:1tÞ ¼
10;000 sech2 ð0:1tÞ;
(b) N 0 ð6:93Þ 6401 bacteria/h,
(c) Nð6:93Þ ¼ 699;906 7 105 bacteria
272
CHAPTER 26
26.7
1.
DERIVATIVES OF TRANSCENDENTAL FUNCTIONS
APPLICATIONS
y ¼ x ln x. The domain is ð0; 1Þ. y0 ¼ x 1x þ
1 ln x ¼ 1 þ ln x. Setting y0 ¼ 0 we obtain the critical values 1 þ ln x ¼ 0; ln x ¼ 1; x ¼ e1 ¼
0:3679; y00 ¼ 1x ; since x > 0, y00 is always positive so y is always concave up. This also means
that the critical value yields a minimum.
Sum
mary: maxima: none, minima: e1 ; e1 or
ð0:3679 0:3679Þ; inflection points: none.
7.
x
x
y ¼ sinh x; y0 ¼ cosh x ¼ e þe
2 , which is never
x
x
0. y00 ¼ sinh x ¼ e e
,
which
is
0 when ex ¼ ex
2
or when x ¼ 0. Inflection point at x ¼ 0. Summary:
Maxima: none, minima: none, inflection point (0, 0)
y
4
3
2
y
1
2
–4 –3 –2 –1
–1
1
2
3
4
x
–2
1
–3
1
2
–4
x
3
9.
3.
y ¼ xex ; y0 ¼ xex þ ex ¼ ex ðx þ 1Þ. The critical
values occur when y0 ¼ 0. Since ex cannot be 0,
the only one is when x þ 1 ¼ 0 or x ¼ 1.
y00 ¼ ex ð1Þ þ ðx þ 1Þex ¼ 2ex þ xex ¼ ex ð2 þ xÞ.
Inflection point is when x ¼ 2; y00 ð1Þ > 0 so at
x ¼ 1 there is a minimum. Summary:
Maxima;
1
none;
minimum:
1;
e
;
inflection
point:
2; 2e2
y
2
1
–5
–4
–3
–2
–1
1
x
y ¼ ex cos x; 2 x 2; y0 ¼ ex sin x cos xex ¼ ex ðsin x þ cos xÞ. Critical values occur
when sin x þ cos x ¼ 0 or sin x ¼ cos x or tan x ¼
3
7
00
1. This is at 5
4 ; 4 ; 4 , and
4. y ¼
x
x
e ðcos x sin xÞ þ e ðsin x þ cos xÞ ¼ ex
ð cos x þ sin x þ sin x þ cos xÞ ¼ 2ex sin x. Inflection points occur when sin x ¼ 0 or x ¼
00
2; ; 0; ; 2. At x ¼ 5
> 0 so
4 ; y
00
y is minima. At x ¼ 4 ; y < 0 so y is max00
> 0 so y is minima. At
ima. At x ¼ 3
4 ; y
7
00
x ¼ 4 ; y < 0 so y is maxima. Summary:
7
; 4 ; 0:002896 , minima at
maxima 4 ; 1:5509
5
3
and
4 ; 35:8885
4 ; 0:06702 ; Inflection
points at ð2; 535:49Þ; ð; 23:1407Þ; ð0; 1Þ;
ð; 0:0432Þ and ð2; 0:001867Þ.
–1
5.
y
Here y ¼ ln x21þ1, and so y0 ¼
x2 þ1
1
2x
x2 þ1
2
¼
4
2x
x2 þ1.
–2π
Since the denominator cannot be 0, the only critical
x2 þ1 ð2Þþ2 xð2 xÞ
00
2
¼
value is when x ¼ 0. y ¼
2 x 2þ4 x
x2 þ1
2
2
–4
–12
π
2π
x
–20
–28
x2 þ1
2
–π
–36
2
¼ 2 x 22 . Inflectionpoints when x2 x2 þ1
1 ¼ 0 or at x ¼ 1. At x ¼ 0; y00 < 0 so y is a maximum. Summary: maxima; (0, 0); minima:
none;
inflection points: 1; ln 12 ; 1; ln 12 .
y
11.
y ¼ x3 ln x; y0 ¼ 3x2 ln x þ x2 ; at x ¼ 1; y0 ¼ 3
12 ln 1 þ 12 ¼ 12 ¼ 1 so the slope is 1.
y 0 ¼ 1ðx 1Þ or y ¼ x 1
13.
y ¼ x2 ex ; y0 ¼ 2xex þ x2 ex ; at x ¼ 1; y0 ¼
2e þ e ¼ 3e. The equation of the line is y e ¼
3eðx 1Þ
15.
(a) sðtÞ ¼ e3t ; vðtÞ ¼ s0 ðtÞ ¼ 3e3t ,
(b) v 0 ðtÞ ¼ 9e3t which is never undefined or 0
so velocity is never maximum.
1
–4 –3 –2 –1
1
–1
–2
–3
2
3
4
x
CHAPTER 26 REVIEW
17.
19.
sðtÞ ¼ sin et ;
t
t
(a) vðtÞ ¼ s0 ðtÞ ¼ cos et et ; vðtÞ
¼e cos e ;
0
t
t
t 2t
aðtÞ ¼ v ðtÞ ¼ e cos e sin e e or aðtÞ ¼
et cos et et sin et
(b) vðtÞ ¼ 0 when cos et ¼ 0 or et ¼ 2; this
gives t ¼ ln 2 0:4516 s; a ln 2 ¼ 2 0 2
2
2
sin 2 ¼ 2 ; 2 2:4674 units/ss.
2
273
2
21.
i ¼ 1 et =2L ; i ¼ Lt et =2L ; The critical value
is when t ¼ 0. When t < 0; i0 < 0 and when
t > 0; i0 > 0 so this yields a maximum.
23.
PðtÞ ¼ 100e0:015t ; P0 ðtÞ ¼ 1:5e0:015t ; P0 ð50Þ
¼ 1:5e0:01550 ¼ 0:7085 W/day
25.
f ðxÞ ¼ x ln x 2; f 0 ðxÞ ¼ 1 1x. The roots are
0.1586; 3.1462.
27.
hðxÞ ¼ ex cos x; h0 ðxÞ ¼ ex cos x ex sin x ¼
ex ðcos x sin xÞ. Newton’s method yields
1.570796327 and several other answers. Reexamining hðxÞ we can see that ex is never 0 and
cos x ¼ 0 when x ¼ 2 þ k where k is an
integer.
2=3
ðxÞ ¼ xex ; this will have a maximum when
2=3
2=3
0 ðxÞ ¼ 0. 0 ðxÞ ¼ x ex 23 x1=3 þ ex ¼
2=3 2
ex 1 23 x2=3 ¼ 0 when 23 x3 ¼ 1 or x2=3 ¼ 32 or
3=2
x ¼ 32
1:8371 m. We can use the first derivative test to verify that this is a maximum.
CHAPTER 26 REVIEW
1.
y ¼ sin 2x þ cos 3x; y0 ¼ cos 2x 2 sin 3x 3 ¼
2 cos 2x 3 sin 3x
3.
y ¼ tan2 3x; y0 ¼ 2 tan 3x sec2 3x 3 ¼
6 tan 3x sec2 3x
5.
y ¼ x2 sin x. Using the product rule we obtain
y0 ¼ x2 cos x þ 2x sin x ¼ xðx cos x þ 2 sin xÞ
7.
1
3
ffi
y ¼ sin1 ð3x 2Þ; y0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
12x9x2 3
9.
11.
13.
25.
1ð3x2Þ
2
0
y ¼ arctan 3x ; y ¼
1
1þ9x4
6x
6x ¼ 1þ9x
4
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1
ffi
Here y ¼ arcsin 1 x2 , and so y0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
1ð1xÞ
1=2
1
2
xffiffiffiffiffiffiffiffi
p
ffiffiffiffiffiffiffi2ffi .
ð2xÞ ¼ pffiffiffi
¼ jxjpx
2 1 x
x2 1x2
1x
y ¼ x2 ex ; y0 ¼ 2xex þ x2 e2 ¼ ex x2 þ 2x . The
critical values occur when x2 þ 2x ¼ 0 or x ¼ 0
and x ¼ 2. For the inflection points,
we find
y00 ¼ 2ex þ 2xex þ x2 ex þ 2xex ¼ ex x2 þ 4x þ 2 .
Inflection points occur when x2 þ 4x þp
2 ffiffi¼
ffi 0. The
quadratic formula yields x ¼ 2 2. When
x ¼ 2, then y00 < 0, and so y is a maximum.
When x ¼ 0, then y00 > 0, and so y is a minimum.
Summary: Maxima: ð2; 0:5413Þ; minima: (0, 0);
pffiffiffi
pffiffiffi
inflection points: ð2 2; 0:3835Þ and ð2 þ 2;
0:1910Þ.
y
1 ffi
1 ffi
If x > 0 then y0 ¼ pffiffiffiffiffiffiffi
; x < 0; y0 ¼ pffiffiffiffiffiffiffi
1x2
1x2
y ¼ log4 3x2 5 ; y0 ¼ 3x215 ð6xÞ ln14 ¼ 3x6x
2 5 6
4
1
ln 4
2
15.
2 ln x3 þ4
y ¼ ln2 x3 þ 4 ; y0 ¼ x3 þ4 3x2 ¼
6x2 ln x3 þ4
x3 þ4
17.
–10
2
3
2
2
3
2
¼ ð4x3Þ
3x
x3
2
ð4x3Þ2 8x3 ð4x3Þ
ð4x3Þ4
¼ 3ð4x3Þ8x
xð4x3Þ ¼
4x9
xð4x3Þ
2
2
19.
y ¼ e4x ; y0 ¼ 8xe4x
21.
y ¼ esin x ; y0 ¼ 2x cos x2 esin x
23.
y ¼ ex sin 5x; y0 ¼ ex cos 5x 5 þ ð1Þex
sin 5x ¼ ex ð5 cos 5x sin 5xÞ
2
2
–6
–4
–2
2
27.
ð4x3Þ ð4x3Þ 3x x 2ð4x3Þ4
x
0
y ¼ ln ð4x3Þ
2 ; y ¼
x3
ð4x3Þ4
–8
x
y ¼ ex=2 sin 2x on ½0; 2; y0 ¼ 12 ex=2 sin 2x þ
2 cos 2xex=2 ¼ ex=2 2 cos 2x 12 sin 2x . Since
1
e2x never equals 0, the critical values occur when
2 cos 2x 12 sin 2x ¼ 0. Solving we get 2 cos 2x ¼
1
2 sin 2x or tan 2x ¼ 4; so 2x ¼ 1:325817; 4:45741;
7:6090, and 10.75060, and as a result x ¼ :6629;
2:2337; 2:38045; 5:3753. Taking the second deri
vative, we obtain y00 ¼ 12 ex=2 2 cos 2x 1
x=2
ð4 sin 2x cos 2xÞ ¼ ex=2 2 sin 2x þ e
cos 2x þ 14 sin 2x 4 sin 2x cos 2xÞ ¼ ex=2 15
2 cos 2x 15
4 sin 2x . Solving 2 cos 2x þ 4 sin 2x
274
CHAPTER 26
DERIVATIVES OF TRANSCENDENTAL FUNCTIONS
sin 2x
¼ 0, we obtain 15
4 sin 2x ¼ 2 cos 2x, or cos 2x ¼
2
8
, or tan 2x ¼ 14
which means that 2x ¼
15
31.
1
0
1
1
y ¼ arctan x; y0 ¼ 1þx
2 ; y ð1Þ ¼ 1þ1 ¼ 2
1
(a) The slope of the tangent is 2 and its equation
is y ¼ 4 ¼ 12 ðx 1Þ,
(b) The slope of the normal is 2 and its equation
is y ¼ 4 ¼ 2ðx 1Þ
33.
y ¼ ln 3x has derivative y0 ¼ 1x and, when x ¼ 1,
the derivative is y0 ð1Þ ¼ 1.
(a) The slope of the tangent at the point (1, 0) is
1 and its equation is y ¼ x 1.
(b) The slope of the normal line at the point (1, 0) is
1 and its equation is y ¼ x þ 1 or y ¼ 1 x.
35.
y ¼ lnðsin xÞ has the derivative y0 ¼ sin1 x cos x ¼
tan x and, when x ¼ 4, the derivative is y0 4 ¼
tan 4 ¼ 1.
4
2:6516; 5:7932; 8:9348, and 12:0764, and as a
result, that x ¼ 1:3258; 2:8966; 4:4675; 6:0382.
Summary: maxima: (0.6629, 0.6964), (3.8045,
0.1448); minima: ð2:2337; 0:3175Þ; ð5:3753;
0:0660Þ; inflection points: (1.3258, 0.2425),
ð2:8966; 0:1106Þ, (4.4674, 0.0504), and
ð6:0382; 0:0230Þ
y
0.5
π
2π
x
(a) The slope of the tangent is 1 and its equation is
pffiffiffi
y þ ln 2 ¼ x 4.
–0.5
29.
With y ¼ x3 ln x, we get y0 ¼ 3x2 ln x þ x2 ¼
x2 ð3 ln þ1Þ. Solving 3 ln x þ 1 ¼ 0, we obtain
ln x ¼ 13 or e1=3 0:7165. For the second
derivative, we have y00 ¼ 6x ln x þ 3x þ 2x ¼
6x ln x þ 5x ¼ xð6 ln xþ5Þ. Solving 6 ln xþ5 ¼ 0,
5
we find ln x ¼ 56 or x ¼ e6 0:4346. At
x ¼ 0:7165; y00 is positive so y has a minimum
at this point. Summary: Maxima: none; minima; ð0:7165; 0:1226Þ; inflection point (0.4346,
0.0684)
(b) The slope
pffiffiffi of the normal is 1 and its equation
is y þ ln 2 ¼ 4 x.
37.
T ¼ 75e0:05109t þ 29
(a) Tð10Þ ¼ 75e0:5109 þ 29 ¼ 64:9967 C,
(b) T 0 ¼ ð0:05109Þ75e0:05109t and T 0 ð10Þ ¼
ð0:05109Þ75e0:5109 ¼ 2:2989 C
39.
Using the same set-up as Example 27.20 we have
8
tanð þ Þ ¼ 48
tan ¼
x and tan ¼ x. Writing
tan ðþÞ , we obtain tan ¼ tan ðþÞ
48
8
40
tanðþÞtan x x
¼ 1þtanðþÞ
¼ x384 ¼ x2 40x
þ384.
tan ¼
48
8
1þ 2
1þ x x
x
0
Hence, ¼ arctan x240x
þ384 and ¼
2
1
x2 þ384 4040xð2xÞ
x2 þ384
¼
2
2
2
x2 þ384
x2 þ384 þð40xÞ2
1 þ x240x
þ384
y
40x2 þ1536080x2
2
x2 þ384
¼ 1536040x2
x2 þ384
2
þð40xÞ2
. The derivative is
0 when 15360 40x2 ¼ 0 or 40x2 ¼ 15360, or
x2 ¼ 384, which is x ¼ 19:5959. By the first derivative test this is a maximum. Answer 19.5959 ft.
x
1
CHAPTER 26 TEST
1.
d
f 0 ðxÞ ¼ cos 7x dx
ð7xÞ ¼ 7 cos 7x
3.
d
h0 ðxÞ ¼ e2x dx
ð2xÞ ¼ 2e2x
x2 x2 d
1
d
1
k0 ðxÞ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
e dx 2 2ffi dx e ¼ pffiffiffiffiffiffiffiffiffiffi
2
1e2x
5.
1 ex
2
x2
ffiffiffiffiffiffiffiffiffiffi2
x ¼ p2xe
1e2x
7.
d
Using the product rule: g0 ðxÞ ¼ 7x2 þ 3x dx
2 d 2
tan 5x dx 7x þ 3x . Then, by using the chain
rule we obtain g0 ðxÞ ¼ 7x2 þ 3x ð2 tan 5xÞ 2 2 sec 5x 5þ tan 5x ð14x þ 3Þ ¼ ðtan 5xÞ 10 7x2
þ3x sec2 5x þ ð14x þ 3Þðtan 5xÞ .
9.
The current, i, at a particular time is found by
taking the derivative of the charge. So, iðtÞ ¼
q0 ðtÞ ¼ 3et cos 2:5 t 7:5et sin 2:5 t. At t ¼
0:65 sec; ið0:65Þ 3:82 A.