0
Quiz 7.7
lim e
x
x 
x
3
 x  9  ?
2
x  x 9
3x  2 x
6x  2
6
 lim
 lim
 lim x  lim x  0
x
x
x 
x 
x 
x  e
e
e
e
3
2
2
y
y  sin x
1


2

1
y
2
y  sin x is not one-to-one
 y does not have an inverse.
1
x


2
 
 ,1
2 

1
y
2
 
1, 
 2

2
x
1
1

 


  , 1
 2

 2
 y  sin x with

1,



2  y  sin 1 x
restricted domain 
  
sin  x  with restricted domain   , 
 2 2
*** Our restricted domains will be the largest
interval (containing quadrant I) in which f is
strictly monotonic.
x
  
y  sin x is the unique  in   ,  such that sin y  x
 2 2
1
i.e. sin
si n x



Restricted D :  x /   x  
2
2

R :  y / 1  y  1

2
1
y
1


x

2
2
1
 

  , 1
 2
 y  sin x with
restricted domain
sin 1  x 
y
 
1, 
 2

D :  x / 1  x  1



R : y /   y  
2
2

 
 ,1
2 
2
1

1

 2

 1,  
2  y  sin 1 x

x
y
y  sin x
1

x


2

2

1
2


3

4


6
2

2

1
2
2
2

1
2
1
x
1

3
2

1



3

2
x


2
 
 ,1
2 

1
 
1, 
 2

2
x
1
2
1

 


  , 1
 2

 2
 y  sin x with

1,



2  y  sin 1 x

restricted domain
sin x
sin x
1

3

2

y
y

4


6




6
4
3
2
0
1
2
2
2
3
2
1
0
1
2
2
2
3
2
1




6
4
3
2
0
0
x
    
sin  sin     ?
  4  4
1
  5
Explain why sin  sin 
  4
1
 5
sin 
 4
  5
.
 
 4

 
 5
1 
  sin     sin  sin 

 4
  4


  
4

sin x



Restricted D :  x /   x  
2
2

R :  y /  1  y  1
 2

sin 
  
4
 2 
1
y  cos 1 x is the unique  in  0,   such that cos y  x
y
y
y  cos x
1


2

1
y
2
1
x
 1,  
 0,1

1
2

 , 1
y  cos x with
restricted domain

x
1, 0 
1
1
y  cos 1 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 right .
2. by using trig identities.
x
b2  x2  1
Simplify cos(sin−1 x) and tan(sin−1 x).
cosine of "the  whose sine is x "
 cos  sin -1 x   cos   1  x 2
 tan  sin x   tan  
-1
x
1  x2
Derivatives of Arcsine and Arccosine
Derivative of an inverse  g '  x  
f  x   sin x  f 1  x   g  x   sin 1 x
"the  whose cosine is x "
1
  Psi
f ' g  x 
 sin 1 x
d
1
1
1
 sin x 

1
dx
cos  sin x 
1  x2
f  x   cos x  f 1  x   g  x   cos 1 x
d
1
1
1
 cos x 

1
dx
 sin  cos x 
1  x2
THEOREM 1
1  x2
THM 2
Derivatives of Arcsine and Arccosine
d
1
1
sin x 
,
2
dx
1 x
d
1
1
cos x  
dx
1  x2
f (x) = arcsin(x2)
1
f '   ?
2
d
1
1
sin x 
2
dx
1 x
d
2x
1
1
1
2
sin  x  
 f '  
4
dx
1
2
1 x
1
16
1
4
4 15



15
15
15
16
  
y  tan x is the unique  in   ,  such that tan y  x
 2 2
1
y  cot 1 x is the unique  in  0,   such that cot y  x


y  sec x is the unique  in [0, )  ( , ] such that sec y  x
2
2
y
y
1
y  sec x
x
y  sec 1 x
x

1

y  csc x is the unique  in [  ,0)  (0, ] such that csc y  x
2
2
y
y



1
2
1



x
2
2
fy csc
csc x
1
1


2
xx
y  csc11 xx
Day 2
THEOREM 2
Derivatives of Inverse Trigonometric Functions
d
1
d
1
1
1
tan x  2 ,
cot x   2
dx
x 1
dx
x 1
d
1
d
1
sec x 
,
csc 1 x  
dx
dx
x x2 1
x
1
x2 1
1 3 
d
3
1
tan  3 x  1 
 2
2
dx
 3x  1  1 9 x  6 x  2
Chain Rule
THEOREM 2
Derivatives of Inverse Trigonometric Functions
d
1
d
1
1
1
tan x  2 ,
cot x   2
dx
x 1
dx
x 1
d
1
d
1
sec x 
,
csc 1 x  
dx
dx
x x2 1
x
1
x2 1
1 e 
d
1
x
csc  e  1  
2
dx
x
x
x 0
 e  1  e  1  1
x

e
 e  1
x
x
 e  1  1
x
2

1
2 3
Chain Rule
The formulas for the derivatives of the inverse trigonometric
functions yield the following integration formulas.
Integral Formulas

dx
Write
1
 sin x  C
1  x2
dx
1
 x 2  1  tan x  C
dx
1

sec
xC
 x x2 1
 

dx
1  x2
dx
1  x2
in terms of cos 1 x.
   cos 1 x  C 
  cos 1 x  C
In this list, we omit the integral formulas corresponding to the
derivatives of cos−1 x, cot−1 x, and csc−1 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:
d
1
1
sin x 
2
dx
1 x
dx

 sin 1 x  C
1  x2
x
sin x  
1
Area model, in terms of x.
0
C 0
dt
1 t
2
for  1  x  1
2nd Fundamental THM of Calculus
1
dx

?
0 x 2  1
dx
1

tan
x

C
 x2  1
1


  tan x    0 
0
4
4
1
Using Substitution
1

dx
?
x
x 4x 1
u  2 x  du  2dx
2
1/ 2
2
 1 
u  2
u ?
 2

1
2
dx
1
x 1
2
1
du
2
u u 1
2
 sec x  C
 sec u 
2
1
1
 sec 2  sec
2
1
2
2
dx
Using Substitution
0

3/ 4
dx
9  16 x
2
?
4
4
u  x  du  dx
3
3
1 x

2
, so I can use my new derivative rules
dx
1 x
1
2
 sin x  C
2
2



16
x
4
x
  
 9  16 x 2  9 1 

3
1


 
9
3  






3
0
0
du
1
du
4

 
2
2
4
1 3 1  u
1 1  u
0
1
1     
1
 sin u    0      
1
4
4   2  8
Quiz 7.8
THEOREM 1
d
1
sin  7 x   ?
dx
Derivatives of Arcsine and Arccosine
d
1
1
sin x 
,
2
dx
1 x
d
1
1
cos x  
2
dx
1 x
d
7
1
sin  7 x  
2
dx
1 7x
Chain Rule:
Quiz 7.8
No
, but think
about our u.
x
dx
x 1
4
2u  0
?
u  x  du  2 xdx
2

du
2u u  1
2
1 1
 sec u  C
2
1 1 2
 sec x  C
2

2 xdx
2x
2
x 
2 2
1