On this section,
22 = 4,
32 = 9,
52 = 25,
62 = 36,
72 = 49,
1.
Properties of exponentials
use this table; do not use a calculator.
23 = 8,
24 = 16,
25 = 32,
26 = 64,
3
4
5
3 = 27,
3 = 81,
3 = 243,
36 = 729,
3
4
5
5 = 125,
5 = 625,
5 = 3125,
56 = 15625,
3
4
5
6 = 216,
6 = 1296,
6 = 7776,
73 = 343,
74 = 2401,
75 = 16807.
(73 )4
77
2. 75 7−3
3
7
3.
3
4. 35 25
5. 6251/2
6. 7−4
7. 100
8. 4log4 15
9. log8 86
10. log6 7776
√
11. log7 343
12. log5 (25 · 125)
13. log6
216
1296
14. log3 (2434 )
1
2401
16. log10 100
15. log7
ln 16
ln 2
18. log8 128
17.
1
27 = 128,
37 = 2187,
Calculus review
Evaluate the following derivatives.
d
(7x5 )
dx
d 8
20.
(x − 3x4 )
dx
3
d √
x+ 5
21.
dx
x
19.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
d
(12e−4x )
dx
d x
(2 )
dx
2
d
(4e−x )
dx
d
ln x
dx
d
ln(8x)
dx
d
ln(x2 + 1)
dx
d
ln(ex + 1)
dx
d
sin(x)
dx
d
(3 cos x)
dx
d
tan(5x)
dx
d
cot(2x − 3)
dx
d
sec(x2 )
dx
d
csc(ex )
dx
d
3 arcsin x
dx
d
arctan(x/3)
dx
d
arcsec(x + 3)
dx
d 2 3x
(x e )
dx
d 4
(x cos(2x))
dx
d sin x
dx x
41. Let f (x) = x7 g(x), where g(1) = 5 and g 0 (1) = 3.
Find f 0 (1).
42. Let f (x) = 3e−2x g(x), where g(0) = 2 and
g 0 (0) = 4. Find f 0 (0).
43. Suppose that y = y(x) is defined by the relation
dy
y 5 + y = x. Use implicit differentiation to find
.
dx
44. Suppose that y = y(x) is defined by the relation
dy
y +x2 y 3 = x4 . Use implicit differentiation to find
.
dx
Evaluate the following integrals.
Z 32
√
5
45.
x dx
1
Z
46.
Z
47.
Z
2x3 +
3
dx
x5
3e7x dx
−1
48.
−e
9
dx
x
Z
49.
3 sin(5x) dx
Z
50.
8 cos(2x) dx
Use u-substitution to evaluate the following integrals.
Z π
51.
cos4 x sin x dx
0
Z
52.
Z
53.
esin 3x cos 3x dx
e7x
dx
e7x + 5
Z
54.
tan x dx
Z
55.
cot x dx
Z
56.
2
sec2 x + sec x tan x
dx
sec x + tan x
Z
csc2 x + csc x cot x
dx
csc x + cot x
sec x dx =
Z
57.
Z
csc x dx =
Use integration by parts to evaluate the following integrals.
Z
58.
x sin(3x) dx
Z
59.
Z
60.
Z
61.
x2 ln x dx
ex sin x dx
cos2 x dx
Use trigonometric substitution to evaluate the following integrals.
Z
1
dx
62.
4 + x2
Z p
63.
9 − x2 dx
Z
64.
√
1
dx
x2 − 1
Z
65.
Use
66.
67.
68.
1
dx
(x + 3)2 + 16
partial fractions to evaluate the following integrals.
Z
1
dx
9 − x2
Z
4x2
dx
(x + 1)(x2 + 1)
Z
x+5
dx
(x + 1)(x + 2)2
3
Answers
1.
(73 )4
= 16807.
77
2. 75 7−3 = 49.
3
7
343
3.
=
.
3
27
4. 35 25 = 1296.
5. 6251/2 = 25.
1
.
6. 7−4 =
2401
7. 100 = 1.
8. 4log4 15 = 15.
9. log8 86 = 6.
10. log6 7776 = 5.
√
3
11. log7 343 = .
2
12. log5 (25 · 125) = 5.
13. log6
216
= −1.
1296
14. log3 (2434 ) = 20.
1
= −4.
2401
16. log10 100 = 2.
15. log7
17.
ln 16
= 4.
ln 2
18. log8 128 =
7
.
3
d
(7x5 ) = 35x4 .
dx
d 8
20.
(x − 3x4 ) = 8x7 − 12x3 .
dx
d √
3
1
15
21.
x + 5 = √ − 6.
dx
x
2 x x
19.
22.
23.
24.
25.
26.
d
(12e−4x ) = −48e−4x .
dx
d x
(2 ) = 2x ln 2.
dx
2
2
d
(4e−x ) = −8xe−x .
dx
d
1
ln x = .
dx
x
d
1
ln(8x) = .
dx
x
4
27.
28.
29.
30.
31.
32.
33.
34.
35.
d
2x
ln(x2 + 1) = 2
.
dx
x +1
d
ex
ln(ex + 1) = x
.
dx
e +1
d
sin(x) = cos x.
dx
d
(3 cos x) = −3 sin x.
dx
d
tan(5x) = 5 sec2 (5x).
dx
d
cot(2x − 3) = −2 csc2 (2x − 3).
dx
d
sec(x2 ) = 2x sec(x2 ) tan(x2 ).
dx
d
csc(ex ) = −ex csc(ex ) cot(ex ).
dx
3
d
3 arcsin x = √
.
dx
1 − x2
d
3
arctan(x/3) =
.
dx
9 + x2
d
1
p
.
37.
arcsec(x + 3) =
dx
|x + 3| (x + 3)2 − 1
36.
d 2 3x
(x e ) = 2xe3x + 3x2 e3x .
dx
d 4
39.
(x cos(2x)) = 4x3 cos(2x) − 2x4 sin(2x).
dx
x cos x − sin x
d sin x
=
.
40.
dx x
x2
38.
41. f 0 (x) = 7x6 g(x) + x7 g 0 (x), so f 0 (1) = 7g(1) + g 0 (1) = 38.
42. f 0 (x) = −6e−2x g(x) + 3e−2x g 0 (x), so f 0 (0) = −6g(0) + 3g 0 (0) = 0.
43. If y 5 + y = x, then
dy
1
= 4
.
dx
5y + 1
44. If y + x2 y 3 = x4 , then
Z
32
45.
1
Z
46.
Z
47.
Z
3
1
3
dx = x4 − 4 + C.
x5
2
4x
3 7x
e + C.
7
−1
9
dx = 9 ln |x| = 9 ln 1 − 9 ln e = −9.
x
−e
3e7x dx =
−1
−e
49.
32
5
315
5 6/5 5
.
x dx = x = 326/5 − =
6
6
6
6
1
2x3 +
48.
Z
√
5
dy
4x3 − 2xy 3
=
.
dx
3x2 y 2 + 1
3
3 sin(5x) dx = − cos(5x) + C.
5
5
Z
50.
8 cos(2x) dx = 4 sin(2x) + C.
Z
51.
0
Z
52.
π
π
1
1
1
2
cos4 x sin x dx = − cos5 x = − cos5 π + cos5 0 = .
5
5
5
5
0
esin 3x cos 3x dx =
1 sin 3x
e
+ C.
3
1
e7x
dx = ln(e7x + 5) + C.
+5
7
Z
Z
Z
sin x
54.
tan x dx =
dx. Using the substitution u = cos x, we see that
cos x
ln | sec x| + C.
Z
Z
Z
cos x
dx. Using the substitution u = sin x, we see that
55.
cot x dx =
sin x
Z
56.
sec x dx = ln | sec x + tan x| + C.
Z
53.
e7x
sin x
dx = − ln | cos x| + C =
cos x
cos x
dx = ln | sin x| + C.
sin x
Z
csc x dx = − ln | csc x + cot x| + C.
57.
Z
58.
Z
59.
Z
60.
Z
61.
1
1
x sin(3x) dx = − x cos(3x) + sin(3x) + C.
3
9
1 3
1
x ln x − x3 + C.
3
9
x2 ln x dx =
ex sin x dx =
cos2 x dx =
1 x
1
e sin x − ex cos x + C.
2
2
1
1
sin x cos x + x + C.
2
2
Z
1
1
x
dx = arctan + C.
4 + x2
2
2
Z p
p
x
1
9
9 − x2 dx = x 9 − x2 + arcsin + C.
63.
2
2
3
Z
1
1 p 2
1 p 2
√
64.
dx = x x − 1 − ln
x − 1 + x + C.
2
2
x2 − 1
Z
1
x+3
1
65.
dx = arctan
+ C.
(x + 3)2 + 16
4
4
Z
Z
1
1/6
1/6
1
1
66.
dx
=
+
dx = − ln |3 − x| + ln |3 + x| + C.
9 − x2
3−x 3+x
6
6
Z
Z
4x2
2
2x − 2
67.
dx =
+
dx = 2 ln |x + 1| + ln(x2 + 1) − 2 arctan x + C.
(x + 1)(x2 + 1)
x + 1 x2 + 1
Z
Z
x+5
4
4
3
3
68.
dx
=
−
−
dx = 4 ln |x + 1| − 4 ln |x + 2| +
+ C.
(x + 1)(x + 2)2
x + 1 x + 2 (x + 2)2
x+2
62.
6