to compute the derivatives of the inverse trigonometric functions, we will need to simplify...
TRANSCRIPT
Quiz 7.7
3 2lim 9 ?x
xe x x
0
3 2 29 3 2 6 2 6lim lim lim lim 0
x x x xx x x x
x x x x x
e e e e
sin is not
does not
one-to-
have a
on
n inverse.
ey x
y
*** Our restricted domains will be the largest
interval (containing quadr
sin
ant
with restricted domai
I) in which is
strictly mon
n ,2 2
otonic.
f
x
,12
, 12
1,
2
1,2
1
12
2
x
ysiny x 1
1
y
x2
2
11
y
x
2
2
sin with
restricted domain
y x1siny x
Restricted : /2 2
1
s
/
i
: 1
n x
D x x
R y y
1sin
: / 1 1
: /2 2
D x x
R y y
x
1sin is the unique in , such that sin2 2
y x y x
1,2
1,2
11
y
x
2
2
1siny x
. . sin 12
i e
,12
, 12
1
1
y
x2
2
sin with
restricted domain
y x
x
sin x
0 2 3 4 6 6 4 3 2
1 3
2
2
2
1
2 0
1
2
2
2
3
21
x
1sin x
1 3
2
2
2
1
2 0
1
2
2
2
3
21
2
3
4
6
0
6
4
3
2
,12
, 12
1,
2
1,2
1
12
2
x
ysiny x 1
1
y
x2
2
11
y
x
2
2
sin with
restricted domain
y x1siny x
1sin sin ?4
1 5sin
5Expla siin why n .
4 4
15 5sin sin sin sin
4 44 4
4
sin
Restricted : /
2
2
:
/ 1 1
x
D x x
R y y
1 2
2sin
4
1cos is the unique in 0, such that cosy x y x
To compute the derivatives of the inverse trigonometric functions, we will need to simplify composite expressions such as cos(sin−1 x) and tan(sec−1 x). This can be done in two ways:
1. by referring to the approprate ri
2. by using trig identi
ght .
ties.
, 1
0,1 1,
1,0
1
12
2
x
ycosy x 1
1
y
x2
11
y
xcos with
restricted domain
y x1cosy x
cosine of "the whose sine is "xSimplify cos(sin−1 x) and tan(sin−1 x).
-1 2c 1os sin cosx x
-1
2tan sin
1tanx
x
x
2 2 1b x
1
1
2
1
1
sin sin
1sin
cos sin
1
1
f x x f x g x x
dx
d xx x
THEOREM 1 Derivatives of Arcsine and Arccosine
1
Derivative of an inverse ''
g xf g x
1
1
2
1
1
cos cos
1cos
sin c
1
1os
f x x f x g x x
dx
dx x x
21 x
Derivatives of Arcsine and Arccosine
1 1
2 2
1 1sin , cos
1 1
d dx x
dx dxx x
"the whose cosine is "x
THM 2
Psi
1sin x
1
2
1sin
1
dx
dx x
f (x) = arcsin(x2)
1 ' ?
2f
1 2
4
2 1 1sin '
4 4
2 11 116
15
15
1
151
156
d xx f
dx x
1tan is the unique in , such that tan2 2
y x y x
1cot is the unique in 0, such that coty x y x
1sec is the unique in [0, ) ( , ] such that sec2 2
y x y x
secy x
x
y y
1secy x
x
1csc is the unique in [ ,0) (0, ] such that csc2 2
y x y x
2
2
1
1
2
2
x
111csc x cscf
y
cscy x
x
y
1cscy xx
THEOREM 2 Derivatives of Inverse Trigonometric Functions
1 12 2
1 1
2 2
1 1tan , cot
1 11 1
sec , csc1 1
d dx x
dx x dx xd d
x xdx dxx x x x
122
3
9 6
1 3tan
1 1 23 1
3
dx
dx x x x
Chain Rule
Day 2
THEOREM 2 Derivatives of Inverse Trigonometric Functions
1 12 2
1 1
2 2
1 1tan , cot
1 11 1
sec , csc1 1
d dx x
dx x dx xd d
x xdx dxx x x x
1
20
2
1csc 1
1 1 1
1 1
1
2 31
x
x
x xx
x
x x
ede
dx e e
e
e e
Chain Rule
The formulas for the derivatives of the inverse trigonometric functions yield the following integration formulas.
Integral Formulas
1
2
12
1
2
sin1
tan1
sec1
dxx C
xdx
x Cxdx
x Cx x
In this list, we omit the integral formulas corresponding to the derivatives of cos−1 x, cot−1 x, and csc−1 x
1
2Write in terms of cos .
1
dxx
x
2
1
1cos1
c s o
dxx C
x C
x
????Why
We can use these formulas to express the inverse trigonometric functions as definite integrals. For example, because sin−1 0 = 0, we have:
1
20
sin for 1 11
x dtx x
t
1
2
1
2
1sin
1
sin1
dx
dx xdx
x Cx
Area model, in terms of .x
0C
2nd Fundamental THM of Calculus
12
tan1
dxx C
x
11
0tan
40
4x
1
20
?1
dx
x
221
22
1
2
1sec 2 s
12 sec
11
ec 2
2
duu
u u
2 2u x du dx
1
2sec
1
dxx C
x x
Using Substitution1
21/ 2
?4 1
dx
x x
12 ?
2u u
4 4
3 3u x du dx
222 16 4
9 16 9 1 3 19 3
x xx
0 0
2 21 1
01
1
31443 1 1
1 1sin 0
4 4 2 8
du du
u u
u
Using Substitution
1
2sin
1
dxx C
x
0
23/ 4
?9 16
dx
x
2, so I can use my new derivative rules
1
dx
x
Quiz 7.8 1sin 7 ?d
xdx
THEOREM 1 Derivatives of Arcsine and Arccosine
1 1
2 2
1 1sin , cos
1 1
d dx x
dx dxx x
1
2sin 7
7
1 7
dx
dx x
Chain Rule:
Quiz 7.84
?1
dx
x x
2 2u x du xdx
2
1
1 2
2 11
sec21
sec2
d
x
u
u u
u C
C
22 2
2
2 1
xdx
x x
2 0u No , but think
about our .u