AP Calculus AB Chapter 5 Review Flashcards | StudBud

AP Calculus AB Chapter 5 Review

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Derivations and integrations of basic logarithmic, exponential, and trigonometric functions. #calculus #AP #math #trigonometry #derivatives #integrals

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Questions 35

Answers

1
d/dx[cscx]
-cscxcotx
sec2x
-csc2x
secxtanx
2
d/dx[arccotu]
-u'/(|u|√(u2-1))
-u'/(1+u2)
u'/√(1-u2)
-u'/√(1-u2)
3
∫cscxcotxdx
sinx+C
tanx+C
-csc2x+C
-cscx+C
4
d/dx[e^x]
ex
lnx(ex)
(1/lnx)ex
5
d/dx[sinx]
sin2x
sec2x
-cotx
cosx
6
d/dx[secx]
-csc2x
sec2x
secxtanx
tanx
7
∫cotxdx
-ln|cosx|+C
ln|sinx|+C
-ln|cscx+cotx|+C
-cosx+C
8
g(x) is the inverse of f(x). g'(x)=?
1/-f'(x)
-f'(x)
f'(g(x))
1/f'(g(x))
9
∫cosxdx
sinx+C
-cosx+C
-ln|cosx|+C
ln|sinx|+C
10
d/dx[log_au]
u'/u
u'/(lna)u
auu'
(1/lna)ax
11
∫a^xdx
(lna)ax+C
x'/(lna)x+C
(1/lna)logau+C
(x')(1/lna)ax+C
12
will you look over your log rules? (5.5)
no
yes
Maybe later
I'll say I will but I won't
13
d/dx[arcsecu]
u'/(1+u2)
-u'/(1+u2)
u'/(|u|√(u2-1))
u'/√(1-u2)
14
d/dx[tanx]
sec2x
secxtanx
-cosx
sinx
15
∫tanxdx
-ln|cosx|+C
ln|sinx|+C
ln|secx+tanx|+C
-ln|cscx+tanx|+C
16
∫ 1/(u√(u²-a²))
(1/a)arcsec(|u|/a)+C
(1/a)arctan(u/a)+C
arcsin(u/a)+C
17
∫cscxdx
ln|sinx|+C
-ln|cscx+cotx|+C
-ln|cosx|+C
ln|secx+tanx|+C
18
d/dx[lnu]
u'/(lna)u
u'/(1+u2)
u'/√(1-u2)
u'/u
19
d/dx[a^x]
(lna)ax
u'(eu)
x'(logax)
u'/(lna)u
20
d/dx[arcsinu]
-u'/√(1-u2)
u'/√(1-u2)
u'/(|u|√(u2-1))
u'/(1+u2)
21
d/dx[arctanu]
-u'/(1+u2)
u'/(1+u2)
u'/√(1-u2)
u'/(|u|√(u2-1))
22
∫ 1/(a²+u²)
arcsin(u/a)+C
(1/a)arctan(u/a)+C
(1/a)arcsec(|u|/a)+C
23
∫ 1/√(a²-u²)
(1/a)arctan(u/a)+C
(1/a)arcsec(|u|/a)+C
arcsin(u/a)+C
24
∫secxtanxdx
-cscx+C
-cosx+C
secx+C
sec2x+C
25
∫sinxdx
ln|secx+tanx|+C
ln|cosx|+C
sinx+C
-cosx+C
26
d/dx[arccscu]
-u'/√(1-u2)
u'/(1+u2)
u'/√(1-u2)
-u'/(|u|√(u2-1))
27
secx is the reciprocal of:
cscx
tanx
sinx
cosx
28
d/dx[cotx]
-csc2x
secxtanx
-cscxcotx
sec2x
29
cotx is the reciprocal of:
secx
sinx
tanx
cosx
30
cscx is the reciprocal of:
tanx
cscx
sinx
cosx
31
d/dx[cosx]
secxtanx
sec2x
-cosx
-sinx
32
∫csc²xdx
tanx+C
-sinx+C
-cotx+C
-ln|cscx+cotx|+C
33
d/dx[arccosu]
-u'/(1+u2)
-u'/√(1-u2)
-u'/(|u|√(u2-1))
u'/√(1-u2)
34
∫sec²xdx
tanx+C
-ln|cosx|+C
secxtanx+C
-cotx+C
35
∫secxdx
ln|sinx|+C
-cosx+C
ln|secx+tanx|+C
-ln|cscx+cotx|+C