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Unit 1
Derivatives of and
Integrals Yielding
Transcendental Functions

Unit 1
...
1
...

dx
1

y  2Arcsin x 3
dy
1
 2
dx
1
1 x 3

 

2

1  23
 x
3

Example 1
...
2

1

...
1
...

x 

 Arccsc x3 
Dx 

x 

1
1
2
3
x
 3x  Arccsc x 
3
3 2
2 x
x  x  1

2
x

 

Theorem



1
1 u

2

du  Arcsin u  C

1
 1  u 2 du  Arctan u  C

u

1
u 1
2

du  Arcsec u  C

Theorem



1

u
du  Arcsin  C
2
2
a
a u

1
1
u
 a2  u 2 du  a Arctan a  C
1

1
u
 u u 2  a2 du  a Arcsec a  C

Example 1
...
4
Evaluate the following
...

16  sin 2 x
cos x dx

2
2
4   sin x 


du

42  u 2
u
 Arcsin    C
4
 sin x 
 Arcsin 
C
 4 

Let

u  sin x
du  cos x dx

2
...
 2
x  2x  3
dx
 2
 x  2x   3
dx
 2
 x  2x  1  1  3
dx

2
 x  1  2


dx

 x  1

2



 2

2




dx

 x  1

2



 2

2

du
u 
2

 2

2

1
 u 

Arctan 
C
2
 2
1
 x 1

Arctan 
C
2
 2 

Let

u  x 1
du  dx

Required Exercises

Answer Exercises in TC7
...
7 (33-55) on page 505-507
Exercises 5
...


End of Unit 1
...
2
Derivatives of and Integrals
Yielding Logarithmic
Functions

First Fundamental Theorem of Calculus
Let the function f be continuous on  a, b and

let x be any number in  a, b
...

dx

Example 1
...
1
Use the First Fundamental Theorem of Calculus
dy
to find
...
y  

x

2

t  4 dt
2

2
...

dx
y  ln x
x1
y   dt
1 t
dy 1
 by the First FTC
...
2
...




1
...
y  ln x  5x  2
4

2

y  ln  x  5x  2 
4

2

1

2

1
4
2
y  ln  x  5x  2 
2

ln ab  b ln a

1
4
2
y  ln  x  5x  2 
2
dy 1
1
  4
  4 x3  10 x 
2
dx 2 x  5x  2
3
...
2
...

1
1
f ' x  

log5  tan  3x   ln 2  tan  3x   ln5





sec2  3x   3

Example 1
...
4
Find the derivative of the function given by





y  log  5x  3  2 x  7 
...
2
...

csc2  2 x 
1
...
 2
dx   2
dx   2
dx
x 9
x 9
x 9
x
For  2
dx,
x 9
2
x
Let
u  x 9
dx
 x2  9
du  2 xdx
1 du
 
du
 xdx
2 u
2
1
 ln u  C
2
1
 ln x2  9  C
2

Therefore,
x 1
x
1
 x2  9 dx   x2  9 dx   x2  9 dx
1
1
x
2
 ln x  9  Arctan  C
2
3
3

3
...
2
...




sec  Arcsin x 
1  x2

dx

  sec u du

Let

u  Arcsin x
dx
du 
1  x2

 ln sec u  tan u  C
 ln sec  Arcsin x   tan  Arcsin x   C

Required Exercises

Answer Exercises in TC7
...
2 (5-36) on page 449-450
Exercises 5
...


Reading Assignment

There is a READING ASSIGNMENT
found in the fb group
...
2

Unit 1
...
3
...


2x  1 x  1
5

y 

2

x 1

1

3

4

2

2x  1 x  1
5

ln y  ln

2

x 1

1

3

4

2
4

ln y  ln 2 x  1  ln x  1  ln x  1
5

2

3

1

2

1
ln y  5ln 2 x  1  4ln x  1  ln x3  1
2
1 dy
1
1
1 1
  5
2  4 2
 2 x   3  3x 2
y dx
2x  1
x 1
2 x 1
2

1 dy
1
1
1 1
  5
2  4 2
 2 x   3  3x 2
y dx
2x  1
x 1
2 x 1
2

1 dy
10
8x
3x
 
 2
 3
y dx 2 x  1 x  1 2 x  2
 10
dy
8x
3x 2 
 y
 2
 3

dx
 2x  1 x  1 2x  2 

dy  2 x  1  x  1  10
8x
3x 2 

 2
 3


3
dx
x 1
 2x  1 x  1 2x  2 
5

2

4

Logarithmic Differentiation
For y  f  x  ,
i
...

ii
...

iii
...

iv
...


Example 1
...
2
Differentiate the function given by y  x
y  x

cos  2 x 


...

Exercises 5
...
5 (14-20) on page 476
Check your answers on A-155 and A-156
...
4
Derivatives of and
Integrals Yielding
Exponential Functions

Exponential Function

Let a  0 and a  1
...

x

ya
x  log a y
1 dy
1

y ln a dx
dy
 y ln a
dx
dy
x
 a ln a
dx
x

Theorem

If u is a differentiable function of x then
Dx  a

u

a

u

ln a  Dxu

Dx  eu   eu  Dxu

Example 1
...
1

Find the derivative of the function defined

 
...
4
...

dx
x
y
x y
e e e
dy
dy 
x y 
e e
 e 1  
dx
 dx 
x
y dy
x y
x  y dy
e e
e e
dx
dx
dy y x  y
e  e   ex  y  ex

dx
dy e x  y  e x
 y x y
dx e  e
x

y

Theorem

 e du  e
u

u

C
u

a
 a du  ln a  C
u

Example 1
...
3

Evaluate the integrals
...


eArcsec x

x

x2  1

dx

  eu du
 eu  C
 eArcsec x  C

Let

u  Arcsec x
1
du 
dx
x x2  1

2
...
4
...

x

2x csc  2x  dx

1

 csc u du
ln 2
1

 ln csc u  cot u  C
ln 2
ln csc  2x   cot  2x 

C
ln 2

Let

u  2x
du  2x ln 2 dx
du
 2x dx
ln 2

Required Exercises

Answer Exercises in TC7
...
4 (5-53) on page 467-468
Exercises 5
...


End of Unit 1
...
5
Derivatives of and
Integrals Yielding
Hyperbolic Functions

Hyperbolic Functions

e e
sinh x 
2
x

x

e e
cosh x 
2
x

x

sinh x
tanh x 
cosh x

1
csch x 
sinh x

cosh x
coth x 
sinh x

1
sech x 
cosh x

Circular and Hyperbolic Functions

Properties of Hyperbolic Functions
y  sinh x
Domain 
Range 

y

5
4
3
2
1

-5

-4

-3

-2

-1

1
-1
-2
-3
-4
-5

2

3

4

5

x

y  cosh x
Domain 

y

5

4

Range  1,  

3

2

1

-3

-2

-1

1
-1

2

3

x

y  tanh x
Domain 
Range   1,1
y

1
...
0
0
...
5
-1
...
5

2

3

x

Hyperbolic Identities
x

  x

e e
sinh   x  
2
x
x
e e

2
x
x
e e

2
  sinh x

P  x    cosh x,sinh x 
cosh x  sinh x  1
2

2

2

2

cosh x sinh x
1


2
2
2
cosh x cosh x cosh x
2
2
1  tanh x  sech x

sinh   x    sinh x
cosh   x   cosh x
cosh x  sinh x  1
2

2

1  tanh 2 x  sech 2 x
coth x  1  csch x
2

2

y  sinh x
e x  e x
y
2
dy 1 x  x
   e  e  1 
dx 2
e x  e x

2
 cosh x

y  cosh x
e x  e x
y
2
dy 1 x  x
   e  e  1 
dx 2
e x  e x

2
 sinh x

Theorem
If u is a differentiable function of x then
Dx  sinh u   cosh u  Dxu
Dx  cosh u   sinh u  Dxu
Dx  tanh u   sech 2 u  Dxu
Dx  csch u    csch u coth u  Dxu
Dx  sech u    sech u tanh u  Dxu
Dx  coth u    csch u  Dxu
2

Example 1
...
1
Find the derivative of the function
...
y  sinh  e

Arccot x



dy
1
Arccot x
Arccot x
 cosh  e
  e  1  x2
dx
2
...
f  x   2 cosh x
x

3

f  x   2  cosh x 

3

x

f '  x   2x  3cosh 2 x  sinh x  cosh 3 x  2 x ln 2
4
...
5
...

sinh  log x 
1
...




coth

2

x

x

dx

 2 coth u du
2

 2  csch u  1 du
2

 2   coth u  u   C
 2coth x  2 x  C

Let u  x
du 

1

dx

2 x
1
2du 
dx
x

3
...




e

4dx
x

e



x 2
2

 2 
   x  x  dx
e e 
  sech x dx
2

 tanh x  C

Required Exercises

Answer Exercises in TC7
...
9 (1-32) on page 524
Check your answers on A-158
...
5

Unit 1
...


3

-2

-1

1
-1

-2

-3

2

3

x

Inverse Hyperbolic Sine Function

The inverse hyperbolic sine function is
defined as
y  Argsinh x if and only if x  sinh y

y  Argsinh x
Domain 
Range 

y

3

2

1

-3

-2

-1

1
-1

-2

-3

2

3

x

Hyperbolic Cosine Function
y

f  x   cosh x

4

3

Domain 
Range  1,  

2

1

f is not a 1-1
function
...


y  Argcosh x
Domain  1,  
Range  0,  

y

4

3

2

1

-2

-1

1

-1

2

3

4

x

Other Inverse Hyperbolic Functions

y  Arg tanh x if and only if x  tanh y
where  1  x  1 and y 
...


Other Inverse Hyperbolic Functions

y  Argsech x if and only if x  sech y
where 0  x  1 and y  0
...


y  Argsinh x
sinh y  x
dy
cosh y   1
dx
dy
1

dx cosh y
dy
1

dx
1  x2

cosh 2 y  sinh 2 y  1
cosh 2 y  1  sinh 2 y
cosh y  1  sinh 2 y

Theorem
If u is a differentiable function of x then
Dx  Argsinh u  

Dx  Argcosh u  

1
1 u

2

1
u2  1

Dxu

Dxu

1
Dxu
Dx  Arg tanh u  
2
1 u

Dx  Argcsch u   

Dx  Argsech u   

1
u 1 u
1
u 1 u

2

1
Dx  Argcoth u  
Dxu
2
1 u

2

Dxu

Dxu

Example 1
...

dx
1
...
y 
Argsech x
1
1
Argsech x 
 Arg tanh x 
2
1 x
dy
x 1  x2

2
dx
 Argsech x 

Required Exercises

Answer Exercises in TC7
...
9 (40-48) on page 535
Check your answers on A-158
...
6



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