Thomas Calculus 13th [Solutions]

Section 3.9 Inverse Trigonometric Functions 183
53. f(x) = sec x
1
df
f ( x) sec x tan x
1 1 1
dx df
1 1
sec(sec b) tan(sec b) 2
x b
b b 1
dx x f
1 ( b)
Since the slope of
sec
1
x is always positive, we choose the right sign by writing
d 1
sec x 1 .
dx
2
x x 1
54. 1 1 1 1
dx dx
cot u tan u d (cot u) d tan u 0
2 dx dx 2 2 2
1 u 1 u
du
du
55. The functions f and g have the same derivative (for x 0), namely 1 . The functions therefore differ by a
x( x 1)
constant. To identify the constant we can set x equal to 0 in the equation f(x) = g(x) + C, obtaining
1 1
1
sin ( 1) 2 tan (0) C 0 C C . For x 0, we have 1 1
2 2
sin x
1 2 tan x
x
2
.
56. The functions f and g have the same derivative for x > 0, namely 1 . The functions therefore differ by a
2
1 x
constant for x > 0. To identify the constant we can set x equal to 1 in the equation f(x) = g(x) + C, obtaining
sin 1 1 tan 1
1 C C C
2
4 4
0.
1 1
For x > 0, we have sin 1 tan 1 .
2
x 1
x
57. (a)
(c)
1 1
sec 1.5 cos 1 0.84107 (b)
1.5
1 1
cot 2 tan 2 0.46365
2
1 1
csc ( 1.5) sin 1 0.72973
1.5
58. (a)
(c)
1 1
sec ( 3) cos 1 1.91063 (b)
3
1 1
cot ( 2) tan ( 2) 2.67795
2
1 1
csc 1.7 sin 1 0.62887
1.7
59. (a) Domain: all real numbers except those having the form 2
k where k is an integer. Range:
y
2 2
(b) Domain: < x < ; Range: < y <
The graph of y tan
1
(tan x ) is periodic, the graph of
x < .
1
y tan(tan x)
x for
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