Thomas Calculus 13th [Solutions]

536 Chapter 7 Transcendental Functions
11.
e
e
2 2
1 dx e 1/2 1
2 1/2
(ln ) ,
ln e
x x x x
dx 1
u du where u ln x, du 1 dx;
x
1/2
2
2 u 2 2 1 2 2 2
1
2
x e u 1, x e u 2
12.
4 4 4ln 4
(1 ln t)( t ln t) dt ( t ln t)(1 ln t) dt u du , where u t ln t, du ( t) 1 ln t 1 dt 1 ln t dt;
2 2 2ln 2
t
t 2 u 2ln 2, t 4 u 4ln 4
4 ln 4
1 2 1 2 2 1 2 2 (2ln 2) 2
2
u
(4ln 4) (2ln 2) (8ln 2) (2ln 2) (16 1) 30(ln 2)
2 2ln 2 2 2 2
y y 1 y y 1
13. 3 2 ln 3 ln 2 y(ln 3) ( y 1) ln 2 (ln 3 ln 2) y ln 2 ln 3 y ln 2 y ln 2
2 3
ln
2
14.
y y 2 y y 2
4 3 ln 4 ln 3 y ln 4 ( y 2) ln 3 2ln 3 (ln 3 ln 4) y
(ln12) y 2ln 3
y
ln 9
ln12
15.
2 2 2 2 2
2 2 2 2
9e y x e y x ln e y ln x 2 y(ln e) ln x y 1 ln x ln x ln x ln | x| ln 3
9 9 9 2 9 9 3
16.
y
y
3 3ln x ln 3 ln(3ln x) y ln 3 ln(3ln x)
y
ln(3ln x) ln 3 ln(ln x)
ln 3 ln 3
17.
18.
ln( 1) ( ln ) ln
ln( y 1) x ln y e y e x y e x e y y 1 ye x y ye x 1 y 1 e x 1 y 1
x
1 e
ln(10ln ) ln 5 ln /2 /2
ln(10ln y) ln 5x e y e x 10ln y 5x ln y x e y e x y e
x
2
19 . (a)
(b)
(c)
ln x
ln 2
ln x
ln 3
2
2
lim log2
x
ln 3 ln 3
log lim lim
same rate
x 3 x
x x
ln 2 ln 2
lim x lim x lim 2x
1
1 2x
lim 1 1 same rate
x x x x x x
x
x
100
x
lim lim xe lim e
x xe x
100x
x
100
faster
(d) lim x 1
x tan x
(e)
(f )
x
x
x
2
faster
1 1 1
2
1 x
1 sin x
lim csc x lim lim lim 1 1
1
1 2
x x
x
x
x x x
1
x
e e
2x
lim sinh x lim lim 1 e 1
x
x
x e x 2e
x
2 2
x
1
x
2
same rate
same rate
20. (a) lim x
3 lim x
2
3
0 slower
x
x 2 x
(b) lim ln 2x
lim ln 2 ln x lim ln 2 1 1 same rate
2
x ln x x
2(ln x) x
2ln x 2 2
3 2 2
(c) lim 10x 2x lim 30x 4x lim 60x
4 lim 60 0 slower
x x x x
x e x e x e x e
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