5.2 The Natural Logarithmic Function

[Pages:8]5.2 The Natural Logarithmic Function

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Recall d

. Dierentiate the following

Warmup:

sin(x) = cos(x)

functions.

dx

3

3

3

1

5

sin(x ), x sin(x), x sin(x + 3x )

Definition: ln(x)

Define the natural logarithmic function by

Z

x

1

ln(x) =

dt.

1t

y

1

x

t

Recall some facts about definite integrals:

For

a

function

f (t)

that

is

integrable

on

[a,

b]

( a

), b

we

have

the

following.

1. The definite integral

Z

b

evaluates to the signed area f (t) dt

a

between and the axis, between and .

f (t)

t

ab

Signed area means that it takes a negative value if it falls

below the -axis. t

2. For any in , c [a, b]

Z

c

f (t) dt = 0.

c

3. Reversing the order of integration gives

Z

Z

a

b

f (t) dt = f (t) dt.

b

a

ln( ) Some examples of x

Z

x

1

ln(x) = dt. 1t

y

y

ln(2) = 0.6731...

ln(2.718...) = 1

1

2

t

y

ln(0.5) = -0.6931...

1

y

2.718... t

ln(1) = 0

0.5 1

t

1

t

Back to some facts about definite integrals:

y

Z

x

1

ln(x) =

dt

1t

1

x

t

1. The definite integral

Z

b

evaluates to the signed area. . . f (t) dt

a

Conclusion: For

,

.

x > 1 ln(x) > 0

R

2. For any in , c

Conclusion:

c [a, b] f (t) dt = 0.

ln(1) = 0

c

3. Reversing the order of integration gives

Z

Z

a

b

f (t) dt = f (t) dt.

b

a

Conclusion: For

,

.

0 < x < 1 ln(x) < 0

ln( ) Some facts about x

y

Z

x

1

ln(x) =

dt

1t

1

x

t

1.

for

,

at

,

for

ln(x) < 0 0 < x < 1 ln(x) = 0 x = 1 ln(x) > 0

x

>

, 1

and

ln(x)

is

undefined

for

x

. 0

2. By the fundamental theorem of calculus,

Z

x

d

d1 1

ln(x) =

dt = .

dx

dx 1 t

x

So, for example,

is monotonically increasing since

ln(x)

for

.

1/x > 0 x > 0

3. We have the following algebraic properties:

and

p

ln(ab) = ln(a) + ln(b)

ln(a ) = p ln(a)

These both follow from the fact that d

.

ln(x) = 1/x

dx

You try:

1. Use the algebraic rules

p

ln(ab) = ln(a) + ln(b) ln(a ) = p ln(a) ln(a/b) = ln(a) ln(b)

to expand the expressions

ln( ( ))

()

(write in terms of a bunch of f x 's where f x is as simple as possible)

2

3

2

3

2

3

(x + 2) and (x + 2) sin(x)

ln (x + 2) , ln

, ln

,

5x + 5

5x + 5

and to contract the expressions (write in terms of one ln(. . . ))

and ln(x) + ln(2), 3 ln(x) ln(2), 5 (3 ln(x) ln(2)) .

2. Dierentiate the following functions. Simplify where you can.

3

px + 1

ln(x + 2), ln(sin(x)), ln

.

x2

(Hint:

Lots

of chain rule!!

d

ln(f (x)) =

dx

f

1 (x)

f

0) (x)

Derivatives with absolute values

Example: Calculate d | |.

ln x

Recall

dx

(

|| x x 0

x=

.

x x 0

dx

ln( ) Graphing x

y

1

ex

Define the number by e ln(e) = 1

(such a number exists by the intermediate value theorem).

e = 2.718 . . .

Section 5.3: The natural exponential function

Note that since

is always increasing, it is one-to-one, and

ln(x)

therefore invertible! Define

as the inverse function of ,

exp(x)

ln(x)

e.g.

if and only if

()

exp(x) = y

x = ln(y).

Some facts about

:

exp(x)

1. Since

, we have

. Similarly,

ln(1) = 0

exp(0) = 1

ln(e) = 1

implies

.

exp(1) = e

2. Domain: (range of ) ln(x) (

1, 1)

Range: (domain of ) 1 ln(x) (0, )

3. Graph:

y

exp(x)

x ln(x)

Section 5.3: The natural exponential function

Some facts about

:

exp(x)

1.

.

exp(0) = 1, exp(1) = e

2. Domain: 1 1 ; Range: 1 .

( ,)

(0, )

3. Graph:

y

exp(x)

x ln(x)

4. We have x

, and so () gives

ln(e ) = x ln(e) = x 1 = x

x

implies

ln(e ) = x

So

ln( ) x

for

and

e = x x > 0,

x

exp(x) = e

x

for all .

ln(e ) = x

x

Exercise: Use logarithmic dierentiation to calculate d x. e

dx

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