Calculus AB Chapter 5 Note-Taking Guide Name _______________________ Logarithms, Exponential, and Transcendental Functions Per ____ Date ________________ 5.1 Remember the power rule for integration:

1

1

nn xx dx C

n

+

= ++∫ which works for all cases except one: 11 dx x dx

x−=∫ ∫

In our PTF problems we have seen this pop up. 5.1 Definition of a Natural Logarithm:

1

1ln , 0x

x dt xt

= >∫

The domain of the natural logarithm function is the set of all positive real numbers (0, )x∈ ∞ The range of the natural logarithm function is all real numbers ( , )y∈ −∞ ∞

Slope field for ( ) 1f xx

=

So 1 lndx x cx

= +∫

( ) ln occurs when the initial condition is (0,1)F x x=

5.1 Properties of Natural Logs If a and b are positive numbers and n is rational, then the following properties are true:

1.) ( )ln 1 0=

2.) ( )ln ln lna b a b= +

3.) ln lnna n a=

4.) ln ln lna a bb

= −

5.1 Expanding Logarithmic Expressions:

a.) 6ln5x

b.) ln 3 2x +

c.) ( )22

3 2

3ln

1

x

x x

+ +

5.1 The number e 2.71828182846...e ≈ • All logarithms have a base. The base for a natural log is e. log lne x x= . • That also means that a base e and natural logarithm are inverses of each other:

ln xe x= and ln xe x=

• Expansion of our integration definition: 11

1 ln ln ln1 ln 1e edt t e et

= = − = =∫

Evaluate Natural logarithms: a.) ln 2 b.) ln 32 c.) ln 0.1

5.1 Derivative of the Natural Logarithm:

1.) [ ] 1ln , 0d x xdx x

= > 2.) [ ] 1 'ln , 0d du uu udx u dx u

= = >

5.1 Using the Natural Logarithms in Differentiation:

a.) ( )ln 2d xdx

= b.) ( )2ln 1d xdx

+ =

u = u = du = du =

c.) [ ]lnd x xdx

= d.) ( )3lnd xdx

=

*product rule *chain rule

5.1 Using Logarithmic Properties to Differentiate

e.) Differentiate ln 1y x= + f.) Differentiate ( )22

3

1ln

2 1

x x

x

+

−

5.1 Using Logarithms to help differentiating non-logarithmic functions We can natural log both sides of an equation to allow us to separate complicated functions.

( ) ( )2 2

2 2

2 2ln ln

1 1

x xy y

x x

− −= ⇒⇒ =

+ + now we use our properties . . .

( ) ( )1/22 2ln ln 2 ln 1y x x= − − +

( ) ( )21ln 2ln 2 ln 12

y x x= − − + then we will differentiate . . .

( )2

' 1 222 2 1

y xy x x

= − − + common denominator

2

2' 1 22 1

12xy

y x x + +

= − −

2x

( )2

221

xxx−−+

( )( )2 2

2

' 2 2 22 1

y x x xy x x

+ − +=

− + ( )( )2

2

' 2 22 1

y x xy x x

y y + +⇒⇒ =

− +

( )( )( )

2 2

22

2 2 2'2 11

x x xyx xx

− + +⇒⇒ =

− ++

( ) 22'

xy

−⇒⇒ =

( ) ( )

2

1/22

2 221

x xxx

+ +−+

( )2 1x +

( )( )( )

2

3/22

2 2 2'

1

x x xy

x

− + +⇒⇒ =

+

5.1 Derivative involving logarithm and absolute value:

The only change is the domain is no longer limited to 0x > but changes to 0x ≠ 1 'ln , 0d du uu u

dx u dx u = = ≠

Example: find f’(x) if ( ) ln cosf x x= u = du =

5.1 Finding Relative Extrema: ( )2ln 2 3y x x= + +

5.2 Logarithmic Rule for Integration

1.) 1 lndx x Cx

= +∫ 2.) 1 lndu u Cu

= +∫ 3.) ' lnu u C

u= +∫

5.2

a.) 2 dxx

=∫ b.) 1

4 1dx

x −∫ u =

u’ =

5.2 Find the area of the region bounded by the function ( ) 2 1

xf xx

=+

, the x-axis, and x = 3.

5.2 Recognizing Quotient Forms of the Log Rule

a.) 2

33 1x dxx x

++∫ b.)

2sectan

xdxx∫ c.) 2

12

x dxx x

++∫ d.)

13 2

dxx +∫

5.2 Using Long Division before Integrating: 2

21

1x x dx

x+ ++∫

5.2 U Substitution (more complicated) and Log rule

( )22

1x

x +∫

5.2 Example: Solve the differential equation

1ln

dydx x x

= (hint: find the integral)

5.2 Breaking down a trigonometric function to integrate: a.) tan xdx∫ b.) sec xdx∫

5.2 /4 2

01 tan xdx

π+∫

5.2 Find the average value of ( ) tanf x x= on the interval 0,

4π

5.3 Inverse Functions f-1(x): • Inverse functions will “undo” each other. • Reflective Property: When taking each coordinate of a function, if you switch the x and y

values, those coordinates will represent values of the inverse function. • Reflective Property: When a function is reflected over the line y = x, you will get the graph of

the inverse function. • The domain of a function will be the range of the inverse and vice versa. • To find the inverse: When looking at the equation of a function, if you switch the x and y

values, then solve for y. Define the domain of the inverse as the range of the original function. • The composition of a function and its inverse will result in x:

( )( )1f f x x− = or ( )( )1f f x x− =

• Note: ( )1 1( )

f xf x

− ≠ this is used to note that its an inverse not a reciprocal function.

• Existence of an Inverse: A function will have an inverse if it is 1-to-1 (passes vertical and horizontal line test)

• Existence of an Inverse: If a function is monotonic (increasing over the entire domain, or decreasing over the entire domain) the function is 1-to-1. To determine if the function is monotonic, calculate the derivative. If the derivative is positive or negative over the entire domain, then it is monotonic and an inverse exists!

5.3 Determine or Verify Inverse Functions Determine if the following are inverse functions:

3( ) 2 1f x x= − and ( ) 31

2xg x +

=

5.3 Which of the functions has an inverse function? (hint: graph and check if it is 1-to-1, verify using the derivative)

a.) ( ) 3 1f x x x= + − b.) ( ) 3 1f x x x= − +

5.3 Find the inverse function: ( ) 2 3f x x= −

5.3 Continuity and Differentiability of Inverse Function: Let f be a function whose domain is an interval I. If f has an inverse function, then the following statements are true.

1.) If is f continuous on its domain, then f -1 is continuous on its domain. 2.) If is f increasing on its domain, then f -1 is increasing on its domain. 3.) If is f decreasing on its domain, then f -1 is decreasing on its domain. 4.) If is f differentiable on an interval containing c then f -1(c) ≠ 0 is differentiable at f(c).

5.3 The derivative of an inverse function:

Let f be a function that is differentiable on an interval I. If f has an inverse function f, then g is differentiable at any x for which ( )( )' 0f g x ≠ . Moreover,

( ) ( )( ) ( )( )1' , ' 0'

g x f g xf g x

= ≠

5.3 Let ( ) 31 1

4f x x x= + −

a.) What is the value of ( )1f x− when 3x = ? b.) What is the value of ( ) ( )1 'f x− when 3x = ?

5.3 The example above has shown us that the graphs of inverse functions have ________________

slopes. So dydx

=

5.3 Example: ( ) 2 , 0f x x x= ≥ and ( )1f x x− = . Show that the slopes of the graphs are reciprocals at

each of the following points. a.) ( )2,4 and ( )4, 2 b.) ( )3,9 and ( )9,3

5.4 Definition of the natural exponential function: The inverse function of the natural logarithm ( ) lnf x x= is called the natural exponential function and is denoted by

( )1 xf x e− = That is,

xy e= if and only if lnx y=

5.4 Solving Exponential and Logarithmic Functions: a.) Solve 17 xe += b.) Solve ( )ln 2 3 5x − =

5.4 Operations of Exponential Functions: Let a and b be any real numbers

1.) a b a be e e +=

2.) a

a bb

e ee

−=

Properties of the Natural Exponential Function( ) xf x e=

1.) The domain of is ( ),−∞ ∞ and the range is ( )0,∞ . 2.) It is continuous, increasing, and 1-to-1 on its entire

domain. 3.) The graph is concave up on its entire domain. 4.) lim 0x

xe

→−∞= and lim x

xe

→∞= ∞

5.4 Derivatives of the Natural Exponential Function Let u be differentiable function of x.

1.) x xd e edx

=

2.) 'u ud e e udx

=

5.4 a.) 2 1xd e

dx− b.) 3/ xd e

dx−

5.4 Locating Relative Extrema Find the relative extrema of ( ) xf x xe=

5.4 Integration Rules for Exponential Functions Let u be a differentiable function of x.

1.) x xe dx e C= +∫ 2.) u ue du e C= +∫

5.4 Examples: a.) 3 1xe dx+∫ b.)

2

5 xxe dx−∫

c.) 1/

2

xe dxx∫ d.) cossin xxe dx∫

5.4 Areas bounded by exponential functions

a.) 1

0

xe dx−∫ b.) 1

0 1

x

xe dx

e+∫ c.) ( )0

1cosx xe e dx

− ∫

5.5A Definition of exponential function with base a If a is a positive real number (a ≠ 1) and x is any real number, then the exponential function with base a is denoted by xa and is defined by ( )ln a xxa e= If a = 1, then 1 1xy = = which is a constant function. Properties of exponential function with base a

1.) 0 1a = 2.) x y x ya a a +=

3.) x

x yy

a aa

−= 4.) ( )yx x ya a=

5.5A Definition of logarithm function with base a If a is a positive real number (a ≠ 1) and x is any real number, then the logarithmic function with base

a is denoted by loga x and is defined by 1log ln

lna x xa

=

Properties of exponential function with base a

1.) log 1 0a = 2.) log log loga a axy x y= +

3.) log log loga a ax x yy= − 4.) log logn

a ax n x=

5.5A Properties of Inverse Functions: Exponential and Logarithmic Functions

1.) xy a= if and only if logax y= *Common logarithm: 10log logx x=

2.) loga xa x= for x > 0 3.) log x

a a x= for all x 5.5A Solve for x:

a.) 1381

x = b.) 2log 4x = −

5.5A Derivatives for bases other than e Let a be a positive real number (a ≠ 1) and let u be a differentiable function of x.

1.) ( )lnx xd a a adx

= 2.) ( )lnu ud dua a adx dx

=

3.) [ ] ( )1log

lnad xdx a x

= 4.) [ ] ( )1log

lnad duudx a u dx

=

Find the derivative of each:

a.) 2xy = b.) 32 xy = c.) 10log cosy x=

5.5A Integrating an Exponential Function to Another Base 1

lnx xa dx a C

a = + ∫

Example: 2xdx =∫

5.5A Now we will start seeing a mix of different types of derivatives so it is important to know which rule to use when. Derive each:

a.) ed edx

b.) xd edx

c.) ed xdx

d.) xy x=

5.5B A limit involving e Suppose you deposit P dollars in an account that earns an annual interest rate r (in decimal form). We can calculate the amount of money in the account if we know how often interest is compounded

(calculated and deposited in the account) in one year (n) using the formula: $ 1ntrP

n = +

As we compound interest more and more often we end up with a limit as n→∞ which gives us a new formula:

1 1lim 1 limx x

x x

x ex x→∞ →∞

+ + = =

Now we can use our math skills to rewrite our original equation so if we continuously compound the interest we will get a new formula: $ rtPe=

/ 1 1$ lim 1 lim 1 lim 1 lim 1 lim 1/ / /

rtnnt nt nt ntr

n n n n n

r r r rP P P P Pn n n r n r n r→∞ →∞ →∞ →∞ →∞

= + = + = + = + = +

If /x n r= then

1l1$ lim 1/

im 1

rtn rtr

x

n

rt

nP P P

n re

x→∞ →∞

= + = =

+

5.5B A deposit of $2500 is made in an account that pays an annual interest rate of %5. Find the balance in the account at the end of 5 years if the interest is compounded:

a.) Quarterly b.) Monthly c.) Continuously

5.5B A bacterial culture is growing according to the logistic growth function

0.41.25 , 0

1 0.25 ty te−= ≥

+

Where y is the weight of the culture in grams and t is the time in hours. Find the weight of the culture after

a.) 0 hours b.) 1 hour c.) 10 hours

c.) What is the limit as t approaches infinity?

5.6 Inverse Trigonometric Functions Since no trig function is 1-to-1, it doesn’t have a true inverse unless their domains are restricted so they are 1-to-1.

5.6 Definitions of Inverse Trigonometric Functions Function Domain Range

arcsiny x= iff sin y x= 1siny x−=

1 1x− ≤ ≤ 2 2

yπ π−≤ ≤

arccosy x= iff cos y x= 1cosy x−=

1 1x− ≤ ≤ 0 y π≤ ≤

arctany x= iff tan y x= 1tany x−=

x−∞ < < ∞

2 2yπ π−

< <

arccoty x= iff cot y x= 1coty x−=

x−∞ < < ∞ 0 y π≤ ≤

arcsecy x= iff sec y x= 1secy x−=

1x ≥ 0 y π≤ ≤ ,

2y π≠

arccscy x= iff csc y x= 1cscy x−=

1x ≥ 2 2

yπ π−≤ ≤ , 0y ≠

5.6 Graphs of the inverse trig functions

5.6 Evaluate each function: (Double check the range of the function)

a.) 1arcsin2−

b.) arccos(0) c.) ( )arctan 3 d.) arcsin(0.3)

5.6 Property of Inverse Functions: ( )( )1f f x x− = and ( )( )1f f x x− =

Example: ( )arctan 2 34

x π− =

5.6 Right triangles and inverse trigonometric functions: (Draw the triangles)

1.) Given arcsiny x= where 02

y π< < 2.) Given

5arcsec2

y

=

Find cos y Find tan y

5.6 Derivatives of Inverse Trigonometric Functions:

a.) [ ]2

'arcsin1

d uudx u

=−

d.) [ ]2

'arccsc1

d uudx u u

−=

−

b.) [ ]2

'arccos1

d uudx u

−=

− e.) [ ]

2

'arcsec1

d uudx u u

=−

c.) [ ] 2'arctan

1d uudx u

=+

f.) [ ] 2'arccot

1d uudx u

−=

+

5.6 Examples:

1.) ( )arcsin 2d xdx

2.) ( )arctan 3d xdx

3.) arcsind xdx

4.) ( )2arcsec xd edx

5.6 Example: Simplifying after the derivative 2arcsin 1y x x x= + −

5.7 Integrals involving Inverse Trigonometric Functions: Let u be a differentiable function of x, and let a > 0.

a.) 2 2

arcsindu u Caa u

= +−

∫ b.) 2 21 arctandu u C

a u a a= +

+∫ c.) 2 2

1 arcsecudu C

a au u a= +

−∫

5.7

1.) 24

dxx−

∫

2.) 22 9dx

x+∫

3.) 24 9

dxx x −∫

5.7 Integration by substitution:

2 1x

dxe −

∫

u =

5.7 Rewriting as the sum of two quotients

2 2 2

2 2 4 4 4x xdx

x x x+

= +− − −

∫ ∫ ∫

5.7 Completing the square to rewrite as a difference of two squares 2 2a b−

2 4 7dx

x x− +∫

5.7 Completing the square with a negative coefficient

Find the area of the region bounded by the graph of ( )2

13

f xx x

=−

and 32

x = and 94

x =

5.7 Comparing Integration Problems:

1.) 2 1

dxx x −∫ 2.)

2 1xdxx −

∫ 3.) 2 1

dxx −

∫