Chapter 5

Logarithmic, Exponential, and Other Transcendental Functions

For x 0 and 0 a 1, y = loga x if and only if x = a y.

The function given by f (x) = loga x is called the

logarithmic function with base a.

Every logarithmic equation has an equivalent exponential form: y = loga x is equivalent to x = a y

A logarithmic function is the inverse function of an exponential function.

Exponential function: y = ax

Logarithmic function: y = logax is equivalent to x = ay

A logarithm is an exponent!

The function defined by f(x) = loge x = ln x

is called the natural logarithm function.

y = ln x

(x 0, e 2.718281)

y

x5

–5

y = ln x is equivalent to e y = x

In Calculus, we work almost exclusively with natural logarithms!

Definition of the Natural Logarithmic Function

Theorem 5.1 Properties of the Natural Logarithmic Function

Natural Log1

1ln for 0

x

x dt xt

lnf x x

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

1. ln 1f x x 2. lnf x x

Natural Log1

1ln for 0

x

x dt xt

: 0,D

Passes through (1,0) and (e,1).

lnf x x : ,R

V.A. at 0x

1-to-1

increasing

concave down

4

2

-2

-4

-5 5

You can’t take the log of zero or a negative.

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

1. ln 1f x x 2. lnf x x(Same graph 1 unit right)

: 0D x

0x x

Theorem 5.2 Logarithmic Properties

. ln1a

Properties of Natural Log:

. lnb e

. lnc ab

. lnn

d a

. lna

eb

Expand:

23. ln 1z z

2

14. ln

e

235. ln 1 ln 1 ln 1

2x x x

Write as a single log:

. ln1a

Properties of Natural Log:

. lnb e

. lnc ab

. lnn

d a

. lna

eb

0

1

ln lna b

lnn a

ln lna b

Expand:

23. ln 1z z

2

14. ln

e

235. ln 1 ln 1 ln 1

2x x x

Write as a single log:

2ln ln 1z z

ln 2ln 1z z

2ln1 lne

0 2lne

0 2 1 2

23 1

ln2 1 1

x

x x

322

2

1ln

1

x

x

Definition of e

Theorem 5.3 Derivative of the Natural Logarithmic Function

Derivative of Logarithmic Functions

The derivative is

'( )(ln ( ) )

(.

)

d f xf x

dx f x

2Find the derivative of ( ) ln 1 .f x x x

2

22

2

( 1)(ln 1)

12 1

1

dx xd dxx x

dx x xx

x x

Example:

Solution:

Notice that the derivative of expressions such as ln|f(x)| has no logarithm in the answer.

Example

3ln xy xln3

xxy

313'

Example

3ln 2 xy 32 xu

xdu 2

3

2'

2

x

xy

Example

xxy ln

Product Rule

1ln1

' xx

xy

xy ln1'

Example

2

3

1ln xy x1ln2

3

xxy

22

3

1

1

2

3'

Example

1

1ln

x

xy 1ln1ln

2

1 xx

1

1

1

1

2

1'

xxy

1

2

2

1'

2xy

1

1'

2

xy

Example

xy secln

xx

xxy tan

sec

tansec'

Example

xxy tansecln

xx

xxxy

tansec

sectansec'

2

x

xx

xxxy sec

tansec

sectansec'

lnd

udx

Theorem:

6. ln 2d

xdx

27. ln 1d

xdx

2

3

18. ln

2 1

x xd

dx x

lnd

udx

Theorem:

6. ln 2d

xdx

1

22x

1 du

u dx

27. ln 1d

xdx

1

x

21ln 1

2

dx

dx

2

1 12

2 1x

x

2 1

x

x

2

3

18. ln

2 1

x xd

dx x

2 31

ln ln 1 ln 2 12

dx x x

dx

22 3

1 1 1 11 2 6

1 2 2 1x x

x x x

2

2 3

1 2 3

1 2 1

x x

x x x

5 3 2

2 3

7 5 1

1 2 1

x x x

x x x

52

3

x

xy

)52ln()3ln(ln xxy

52

2

3

11

xxdx

dy

y

52

2

3

1

xxy

dx

dy

252

11

52

2

3

1

52

3

xxxx

x

dx

dy

Theorem 5.4 Derivative Involving Absolute Value

1. ln cosd

xdx

2

32

12. Find .

1

x dyy

x dx

1. ln cosd

xdx

1

sincos

xx

tanx

2

32

12. Find .

1

x dyy

x dx

Try Logarithmic Differentiation.

2

32

1ln ln

1

xy

x

2 21ln ln 1 ln 1

3y x x

2 2

1 1 1 12 2

3 1 1

dyx x

y dx x x

2 2

1 1 2 2

3 1 1

dy x x

y dx x x

2 2

2 2

2 1 2 11

3 1 1

x x x xdyy

dx x x

132

2 2 2

1 1 4

3 1 1 1

dy x x

dx x x x

22. ln 6 Find .dy

xy x y xy xdx

22. ln 6 Find .dy

xy x y xy xdx

211 2 1 6

dy dy dyx y x y x x y

xy dx dx dx

21 12 6

dy dy dyx xy x y

y dx x dx dx

21 16 2

dy dy dyx x y xy

y dx dx dx x

21 16 2

dyx x y xy

dx y x

2

16 2

1

y xydy xdx x x

y

xy

xy2 2 2

3 2

6 2dy xy xy y x y

dx x x y x y

ln 4y x x x 4. Show that is a solution to the statement .0

dyx y x

dx

ln 4y x x x 4. Show that is a solution to the statement .0

dyx y x

dx

1ln 4

dyx x

dx x

1 ln 4dy

xdx

ln 3dy

xdx

0dy

x y xdx

ln 4 ln 3 0x x x x x x

ln 4 ln 3 0x x x x x x x

0 0

Find the equation of the line tangent to: at (1, 3)xxy ln

xy

11' At (1, 3) the

slope of the tangent is 2

bmxy

b )1(23

12 xy

Find the equation of the tangent line to the graph of the function

2ln6)( xxxf

at the point (1, 6).