MATH113 Mathematics I
Lecture 11
7.1 Inverse Functions and Their Derivatives
7.2 Natural Logarithms
7.3 Exponential Functions
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neil.course@okan.edu.tr
7.1
Inverse
Functions and
Their
Derivatives
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7.1 Inverse Functions and Their Derivatives
What is the opposite of adding 3 to a number?
What is the inverse of
f : Z → Z, f (x) = x + 3?
6
6
6
5
5
5
4
4
4
3
3
2
2
1
1
1
0
0
0
-1
-1
-1
-2
-2
-2
3
2
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f
7.1 Inverse Functions and Their Derivatives
What is the opposite of adding 3 to a number?
What is the inverse of
f : Z → Z, f (x) = x + 3?
6
6
6
5
5
5
4
4
4
3
2
The inverse of
Note that
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f
is
3
f
2
f −1
3
2
1
1
1
0
0
0
-1
-1
-1
-2
-2
-2
f −1 : Z → Z, f −1 (x) = x − 3.
f −1 ◦ f (x) = x
and
f ◦ f −1 (x) = x.
7.1 Inverse Functions and Their Derivatives
Some functions have an inverse.
g : [0, 3] → [0, 9], g(x) = x2
√
: [0, 9] → [0, 3], g −1 (x) = x.
For example
g −1
√
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42 =
√
16 = 4
and
has the inverse
√ 2
9 = 32 = 9.
7.1 Inverse Functions and Their Derivatives
Some functions have an inverse.
g : [0, 3] → [0, 9], g(x) = x2
√
: [0, 9] → [0, 3], g −1 (x) = x.
For example
g −1
√
42 =
√
16 = 4
and
has the inverse
√ 2
9 = 32 = 9.
But some functions do not have an inverse.
For example, consider
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h : [−3, 3] → [0, 9], h(x) = x2 .
7.1 Inverse Functions and Their Derivatives
Some functions have an inverse.
g : [0, 3] → [0, 9], g(x) = x2
√
: [0, 9] → [0, 3], g −1 (x) = x.
For example
g −1
√
42 =
√
16 = 4
has the inverse
√ 2
9 = 32 = 9.
and
But some functions do not have an inverse.
For example, consider
h : [−3, 3] → [0, 9], h(x) = x2 .
22 = 4
So what do we dene
4 of 81
and
h−1 (4)
to be?
(−2)2 = 4.
2
or
−2?
7.1 Inverse Functions and Their Derivatives
9
8
7
6
h(x) = x2 5
4
3
3
2
2
1
1
0
0
-1
-2
-3
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7.1 Inverse Functions and Their Derivatives
One-to-One Functions
Denition
A function
domain
D
f (x)
is called
x1 ̸= x2
for all
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one-to-one (or injective) (bire-bir) on a
if
x1 , x2 ∈ D.
=⇒
f (x1 ) ̸= f (x2 )
7.1 Inverse Functions and Their Derivatives
Example
x3
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is one-to-one on any domain.
7.1 Inverse Functions and Their Derivatives
Example
x3 is one-to-one on any domain.
√
x is one-to-one on [0, ∞) because
x1 ̸= x2
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=⇒
√
x1 ̸=
√
x2 .
7.1 Inverse Functions and Their Derivatives
Example
x2
is not one-to-one on
R
because
(−1)2 = 12 .
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7.1 Inverse Functions and Their Derivatives
Example
sin x
is not one-to-one on
sin
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[0, π]
because
π
1
5π
= = sin .
6
2
6
7.1 Inverse Functions and Their Derivatives
Example
sin x
is not one-to-one on
sin
However,
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sin x
[0, π]
because
π
1
5π
= = sin .
6
2
6
is one-to-one on
[0, π2 ].
7.1 Inverse Functions and Their Derivatives
Theorem (The Horizontal Line Test for One-to-One Functions)
A function y = f (x) is one-to-one if and only if its graph
intersects each horizontal line at most once.
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7.1 Inverse Functions and Their Derivatives
Inverse Functions
Denition
Suppose that
range
R.
The
f
is a one-to-one function on a domain
inverse function f −1
f −1 (b) = a
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if
is dened by
f (a) = b.
D
with
7.1 Inverse Functions and Their Derivatives
Inverse Functions
Denition
f
Suppose that
range
R.
The
is a one-to-one function on a domain
inverse function f −1
f −1 (b) = a
The domain of
We read
f −1
f −1
R
f (a) = b.
and the range of
as f inverse.
Please note that
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is
if
is dened by
f −1
does not mean
1
f (x) .
f −1
is
D.
D
with
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7.1 Inverse Functions and Their Derivatives
Remark
If
f −1
is the inverse of
f,
f −1 ◦ f (x) = x
then
(for all
x
in the domain of
f)
and
f ◦ f −1 (y) = y
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(for all
y
in the range of
f ).
7.1 Inverse Functions and Their Derivatives
Finding Inverses
Now let's talk about how to nd an inverse of a function.
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7.1 Inverse Functions and Their Derivatives
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7.1 Inverse Functions and Their Derivatives
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7.1 Inverse Functions and Their Derivatives
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7.1 Inverse Functions and Their Derivatives
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7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of
1 Solve for
2 Swap
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x
x
y = 12 x + 1,
in terms of
and
y:
y:
expressed as a function of
x.
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of
1 Solve for
x
y = 12 x + 1,
in terms of
expressed as a function of
y:
1
y = x+1
2
2y = x + 2
2y − 2 = x
x = 2y − 2.
2 Swap
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x
and
y:
x.
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of
1 Solve for
x
y = 12 x + 1,
in terms of
expressed as a function of
y:
1
y = x+1
2
2y = x + 2
2y − 2 = x
x = 2y − 2.
2 Swap
x
and
y:
y = 2x − 2.
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x.
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of
1 Solve for
x
y = 12 x + 1,
in terms of
expressed as a function of
y:
1
y = x+1
2
2y = x + 2
2y − 2 = x
x = 2y − 2.
2 Swap
x
and
y:
y = 2x − 2.
The inverse of
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f (x) = 12 x + 1
is
f −1 (x) = 2x − 2.
x.
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of
1 Solve for
x
y = 12 x + 1,
in terms of
expressed as a function of
y:
1
y = x+1
2
2y = x + 2
2y − 2 = x
x = 2y − 2.
2 Swap
x
and
y:
y = 2x − 2.
The inverse of
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f (x) = 12 x + 1
is
f −1 (x) = 2x − 2.
x.
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of the function
function of
x.
1 Solve for
2 Swap
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x
x
in terms of
and
y:
y:
y = x2 , x ≥ 0,
expressed as a
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of the function
function of
x.
1 Solve for
x
in terms of
y = x2 , x ≥ 0,
y:
y = x2
√
√
y = x2 = |x| = x
√
x = y.
2 Swap
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x
and
y:
expressed as a
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of the function
function of
x.
1 Solve for
x
in terms of
y = x2 , x ≥ 0,
y:
y = x2
√
√
y = x2 = |x| = x
√
x = y.
2 Swap
x
and
y:
y=
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√
x.
expressed as a
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of the function
function of
x.
1 Solve for
x
in terms of
y = x2 , x ≥ 0,
y:
y = x2
√
√
y = x2 = |x| = x
√
x = y.
2 Swap
x
and
y:
y=
The inverse of
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y = x2 , x ≥ 0,
is
√
y=
x.
√
x.
expressed as a
7.1 Inverse Functions and Their Derivatives
Example
Find the inverse of the function
function of
x.
1 Solve for
x
in terms of
y = x2 , x ≥ 0,
y:
y = x2
√
√
y = x2 = |x| = x
√
x = y.
2 Swap
x
and
y:
y=
The inverse of
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y = x2 , x ≥ 0,
is
√
y=
x.
√
x.
expressed as a
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7.1 Inverse Functions and Their Derivatives
Note that
f f −1 (x) = x
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7.1 Inverse Functions and Their Derivatives
Note that
f f −1 (x) = x
d
d
f f −1 (x) =
x=1
dx
dx
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7.1 Inverse Functions and Their Derivatives
Note that
f f −1 (x) = x
d
d
f f −1 (x) =
x=1
dx
dx
d −1
f ′ f −1 (x) ·
f (x) = 1
dx
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7.1 Inverse Functions and Their Derivatives
Note that
f f −1 (x) = x
d
d
f f −1 (x) =
x=1
dx
dx
d −1
f ′ f −1 (x) ·
f (x) = 1
dx
d −1
1
f (x) = ′ −1
.
dx
f (f (x))
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7.1 Inverse Functions and Their Derivatives
Theorem
Suppose that
for an interval I ;
exists on I ; and
f ′ (x) ̸= 0 on I ,
then f −1 is dierentiable at every point in its domain (the range
of f ).
f :I→R
f′
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7.1 Inverse Functions and Their Derivatives
Theorem
Suppose that
for an interval I ;
exists on I ; and
f ′ (x) ̸= 0 on I ,
then f −1 is dierentiable at every point in its domain (the range
of f ).
f :I→R
f′
Moreover
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′
f −1 (x) =
1
f ′ (f −1 (x))
.
7.1 Inverse Functions and Their Derivatives
Remark
This theorem says that
df −1
dx
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=
x=b
1
df
dx x=f −1 (b)
.
′
1
7.1 Inverse Functions and Theirf −1
Derivatives
(x) =
f ′ (f −1 (x))
Example
Let
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f (x) = x2 , x ≥ 0.
Find
′
f −1 (4).
′
1
7.1 Inverse Functions and Theirf −1
Derivatives
(x) =
f ′ (f −1 (x))
Example
Let
f (x) = x2 , x ≥ 0.
Method 1:
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′
f −1 (4).
f (x) = x2 , x ≥ 0
√
x is 2√1 x .
The inverse of
derivative of
Find
is
f −1 (x) =
√
x
and the
′
1
7.1 Inverse Functions and Theirf −1
Derivatives
(x) =
f ′ (f −1 (x))
Example
Let
f (x) = x2 , x ≥ 0.
Find
Method 1:
′
f −1 (4).
f (x) = x2 , x ≥ 0
√
x is 2√1 x .
The inverse of
derivative of
Method 2:
is
f −1 (x) =
Using the theorem (with
√
x
and the
f ′ (x) = 2x),
we calculate
that
′
f −1 (x) =
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1
f ′ (f −1 (x))
=
1
2 (f −1 (x))
=
1
√ .
2 ( x)
′
1
7.1 Inverse Functions and Theirf −1
Derivatives
(x) =
f ′ (f −1 (x))
Example
Let
f (x) = x2 , x ≥ 0.
Find
Method 1:
′
f −1 (4).
f (x) = x2 , x ≥ 0
√
x is 2√1 x .
The inverse of
derivative of
Method 2:
is
f −1 (x) =
Using the theorem (with
√
x
and the
f ′ (x) = 2x),
we calculate
that
′
f −1 (x) =
Either way,
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1
f ′ (f −1 (x))
=
1
2 (f −1 (x))
′
1
1
f −1 (4) = √ = .
4
2 4
=
1
√ .
2 ( x)
′
1
7.1 Inverse Functions and Theirf −1
Derivatives
(x) =
f ′ (f −1 (x))
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df −1
=
7.1 Inverse Functions and Theirdx
Derivatives
df
x=b
1
dx x=f −1 (b)
Example
−1
f (x) = x3 − 2, x > 0. Find the value of dfdx at
x = 6 = f (2) without nding a formula for f −1 (x).
Let
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df −1
=
7.1 Inverse Functions and Theirdx
Derivatives
df
x=b
1
dx x=f −1 (b)
Example
−1
f (x) = x3 − 2, x > 0. Find the value of dfdx at
x = 6 = f (2) without nding a formula for f −1 (x).
Let
Look at the formula in the yellow box. We want to calculate
df −1
dx at
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x = 6.
So we need
df
dx at
x = 2.
df −1
=
7.1 Inverse Functions and Theirdx
Derivatives
df
x=b
1
dx x=f −1 (b)
Example
−1
f (x) = x3 − 2, x > 0. Find the value of dfdx at
x = 6 = f (2) without nding a formula for f −1 (x).
Let
Look at the formula in the yellow box. We want to calculate
df −1
dx at
Since
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x = 6.
So we need
df
dx
df
dx at
x = 2.
= 3x2
x=2
x=2
= 12,
df −1
=
7.1 Inverse Functions and Theirdx
Derivatives
df
x=b
1
dx x=f −1 (b)
Example
−1
f (x) = x3 − 2, x > 0. Find the value of dfdx at
x = 6 = f (2) without nding a formula for f −1 (x).
Let
Look at the formula in the yellow box. We want to calculate
df −1
dx at
x = 6.
Since
So we need
df
dx
df
dx at
x = 2.
= 3x2
x=2
x=2
= 12,
we have that
df −1
dx
23 of 81
=
x=6
1
df
dx x=2
=
df −1
=
7.1 Inverse Functions and Theirdx
Derivatives
df
x=b
1
dx x=f −1 (b)
Example
−1
f (x) = x3 − 2, x > 0. Find the value of dfdx at
x = 6 = f (2) without nding a formula for f −1 (x).
Let
Look at the formula in the yellow box. We want to calculate
df −1
dx at
x = 6.
Since
So we need
df
dx
df
dx at
x = 2.
= 3x2
x=2
x=2
= 12,
we have that
df −1
dx
23 of 81
=
x=6
1
df
dx x=2
=
1
.
12
7.1 Inverse Functions and Their Derivatives
24 of 81
7.2
Natural
Logarithms
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7.2 Natural Logarithms
Definition of the Natural Logarithm Function
Denition
The
natural logarithm function ln : (0, ∞) → R is dened by
ln x =
Z
1
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x
1
dt.
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
1
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1
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
t
If x ≥ 1, then
1
y = ln x
1x
area =
Z
1
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x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
If x ≥ 1, then
1
t
y = ln x
1
1
x
area =
Z
1
27 of 81
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
If x ≥ 1, then
1
t
y = ln x
1
1
x
area =
Z
1
27 of 81
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
t
If x ≥ 1, then
y = ln x
1
1
x
area =
Z
1
27 of 81
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
t
If x ≥ 1, then
y = ln x
1
x
1
area =
Z
1
27 of 81
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
t
If x ≥ 1, then
y = ln x
1
x
1
area =
Z
1
27 of 81
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
t
If x ≥ 1, then
y = ln x
1
x
1
area =
Z
1
27 of 81
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
x
1
y
y=
1
t
If x ≥ 1, then
y = ln x
1
x
1
area =
Z
1
27 of 81
x
1
dt
t
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
y=
1
t
x
1
dt
t
If x ≥ 1, then
y = ln x
1
x
1
area =
Z
1
27 of 81
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
If x < 1, then
y=
1
t
1
y =x1ln x
area =
Z
1
27 of 81
x
1
dt = −
t
Z
1
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
If x < 1, then
y=
1
t
1
x1
y = ln x
area =
Z
1
27 of 81
x
1
dt = −
t
Z
1
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
If x < 1, then
y=
1
t
1
x 1
y = ln x
area =
Z
1
27 of 81
x
1
dt = −
t
Z
1
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
If x < 1, then
y=
1
t
1
x
1
y = ln x
area =
Z
1
27 of 81
x
1
dt = −
t
Z
1
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
If x < 1, then
y=
1
t
1
x
y = ln x
1
area =
Z
1
27 of 81
x
1
dt = −
t
Z
1
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
y
If x < 1, then
y=
1
t
1
x
1
area =
y = ln x
27 of 81
Z
1
x
1
dt = −
t
Z
1
x
1
dt
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
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x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
1
x
1
dt
t
There is an important number between 2 and 3 where the
natural logarithm is equal to 1.
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7.2 Natural Logarithms1
y=
ln x =
t
1
y
y = ln x
1
1
area =
Z
1
e
1
dt = 1
t
e
Denition
The
number e is the number which satises
ln(e) =
Z
1
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Z
e
1
dt = 1.
t
x
1
dt
t
7.2 Natural Logarithms
ln x =
Z
x
1
1
dt
t
The Derivative of y = ln x
Recall the rst part of the Fundamental Theorem of Calculus:
d
dx
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Z
a
x
f (t) dt = f (x).
7.2 Natural Logarithms
ln x =
Z
x
1
1
dt
t
The Derivative of y = ln x
Recall the rst part of the Fundamental Theorem of Calculus:
d
dx
It follows that
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Z
x
f (t) dt = f (x).
a
d
d
ln x =
dx
dx
Z
1
x
1
1
dt = .
t
x
7.2 Natural Logarithms
If
u(x)
is dierentiable; and
u(x) > 0,
then it follows by the Chain Rule that
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7.2 Natural Logarithms
If
u(x)
is dierentiable; and
u(x) > 0,
then it follows by the Chain Rule that
d
1 du
ln u =
.
dx
u dx
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7.2 Natural Logarithms
Proof that
ln bx = ln b + ln x.
First note that
d
b
1
ln(bx) =
=
dx
bx
x
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7.2 Natural Logarithms
Proof that
ln bx = ln b + ln x.
First note that
d
d
b
1
ln(bx) =
= =
ln x.
dx
bx
x
dx
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7.2 Natural Logarithms
Proof that
ln bx = ln b + ln x.
First note that
d
d
b
1
ln(bx) =
= =
ln x.
dx
bx
x
dx
By the second corollary of the Mean Value Theorem, this means
that
ln bx = ln x + C
for some constant
35 of 81
C.
7.2 Natural Logarithms
Proof that
ln bx = ln b + ln x.
First note that
d
d
b
1
ln(bx) =
= =
ln x.
dx
bx
x
dx
By the second corollary of the Mean Value Theorem, this means
that
ln bx = ln x + C
for some constant
C.
Putting in
x = 1,
ln(b · 1) = ln 1 + C
35 of 81
we have
7.2 Natural Logarithms
Proof that
ln bx = ln b + ln x.
First note that
d
d
b
1
ln(bx) =
= =
ln x.
dx
bx
x
dx
By the second corollary of the Mean Value Theorem, this means
that
ln bx = ln x + C
for some constant
C.
Putting in
x = 1,
we have
ln b = ln(b · 1) = ln 1 + C = 0 + C = C.
35 of 81
7.2 Natural Logarithms
Proof that
ln bx = ln b + ln x.
First note that
d
d
b
1
ln(bx) =
= =
ln x.
dx
bx
x
dx
By the second corollary of the Mean Value Theorem, this means
that
ln bx = ln x + C
for some constant
C.
Putting in
x = 1,
we have
ln b = ln(b · 1) = ln 1 + C = 0 + C = C.
Therefore
ln bx = ln x + ln b.
35 of 81
7.2 Natural Logarithms
Proof that
ln xr = r ln x
(assuming
Using the Chain Rule with
r ∈ Q.)
u(x) = xr ,
we nd that
d
d
1 du
1
1
d
ln xr =
ln u =
= r rxr−1 = r · =
r ln x.
dx
dx
u dx
x
x
dx
36 of 81
7.2 Natural Logarithms
Proof that
ln xr = r ln x
(assuming
Using the Chain Rule with
r ∈ Q.)
u(x) = xr ,
we nd that
d
d
1 du
1
1
d
ln xr =
ln u =
= r rxr−1 = r · =
r ln x.
dx
dx
u dx
x
x
dx
So we must have
ln xr = r ln x + C
for some constant
36 of 81
C.
7.2 Natural Logarithms
Proof that
ln xr = r ln x
(assuming
Using the Chain Rule with
r ∈ Q.)
u(x) = xr ,
we nd that
d
d
1 du
1
1
d
ln xr =
ln u =
= r rxr−1 = r · =
r ln x.
dx
dx
u dx
x
x
dx
So we must have
ln xr = r ln x + C
for some constant
are nished.
36 of 81
C.
Putting in
x = 1,
we nd
C=0
and we
7.2 Natural Logarithms
The Graph and Range of ln x
y
Note rst that
y = ln x
1
x
1
d
1
ln x = > 0
dx
x
for all
x > 0.
So
ln x
increasing function.
37 of 81
is an
7.2 Natural Logarithms
The Graph and Range of ln x
y
Note rst that
d
1
ln x = > 0
dx
x
y = ln x
1
x
1
for all
x > 0.
So
ln x
is an
increasing function.
Moreover
d2
d 1
1
ln x =
= − 2 < 0.
2
dx
dx x
x
So
y = ln x
down.
37 of 81
is concave
7.2 Natural Logarithms
y
ln 2 =
1
Z
1
2
1
dt
t
y=
0.5
1
t
x
1
38 of 81
2
7.2 Natural Logarithms
y
ln 2 =
1
0.5
area =
Z
1
1
2
2
1
dt
t
y=
1
t
x
1
Note that
38 of 81
2
1
ln 2 > .
2
7.2 Natural Logarithms
y
ln 2 =
1
0.5
area =
Z
1
1
2
2
1
dt
t
y=
1
t
x
1
Note that
It follows that
1
ln 2 > .
2
1
n
ln 2 = n ln 2 > n
= .
2
2
n
38 of 81
2
7.2 Natural Logarithms
y
y = ln x
1
x
1
So
n
= ∞.
n→∞ 2
lim ln 2n ≥ lim
n→∞
Since
ln x
is an increasing function, we also have that
lim ln x = ∞.
x→∞
39 of 81
7.2 Natural Logarithms
y
y = ln x
1
x
1
Using
t=
1
x , we also have that
lim ln x = lim ln
x→0+
40 of 81
t→∞
1
= lim ln t−1 = lim − ln t = −∞.
t→∞
t→∞
t
7.2 Natural Logarithms
y
y = ln x
1
x
1
The domain of
The range of
41 of 81
ln x
ln x
is
is
(0, ∞).
R.
7.2 Natural Logarithms
The Integral
If
u(x)
then
is dierentiable; and
1
u
du
u(x) ̸= 0,
Z
42 of 81
R
1
du = ln |u| + C.
u
7.2 Natural Logarithms
The Integral
If
u(x)
then
is dierentiable; and
u = f (x),
then
du =
Z
42 of 81
1
u
du
u(x) ̸= 0,
Z
If
R
1
du = ln |u| + C.
u
du
dx
dx = f ′ (x) dx.
Thus
f ′ (x)
dx = ln |f (x)| + C.
f (x)
7.2 Natural Logarithms
Example
Calculate
Z
π
2
− π2
Z
du
= ln |u| + C
u
4 cos θ
dθ.
3 + 2 sin θ
The idea is to rewrite this integral in the form
43 of 81
Z
du
.
u
Z
7.2 Natural Logarithms
Example
Calculate
Z
π
2
− π2
du
= ln |u| + C
u
4 cos θ
dθ.
3 + 2 sin θ
The idea is to rewrite this integral in the form
Let
43 of 81
u = 3 + 2 sin θ.
Then
du = 2 cos θ dθ.
Z
du
.
u
7.2 Natural Logarithms
Example
Calculate
Z
π
2
− π2
Z
du
= ln |u| + C
u
4 cos θ
dθ.
3 + 2 sin θ
The idea is to rewrite this integral in the form
Let
43 of 81
Z
du
.
u
u = 3 + 2 sin θ. Then du = 2 cos θ dθ. Moreover
π
π
u = 3 + 2 sin = 3 + 2 = 5
θ=
2
2 π =⇒
π
θ=−
u = 3 + 2 sin −
= 3 − 2 = 1.
2
2
Z
7.2 Natural Logarithms
Example
Calculate
Z
π
2
− π2
du
= ln |u| + C
u
4 cos θ
dθ.
3 + 2 sin θ
The idea is to rewrite this integral in the form
Let
π
2
− π2
43 of 81
du
.
u
u = 3 + 2 sin θ. Then du = 2 cos θ dθ. Moreover
π
π
u = 3 + 2 sin = 3 + 2 = 5
θ=
2
2 π =⇒
π
θ=−
u = 3 + 2 sin −
= 3 − 2 = 1.
2
2
Therefore
Z
Z
2 · 2 cos θ
dθ =
3 + 2 sin θ
Z
1
5
2
du
u
Z
7.2 Natural Logarithms
Example
Calculate
Z
π
2
− π2
du
= ln |u| + C
u
4 cos θ
dθ.
3 + 2 sin θ
The idea is to rewrite this integral in the form
Let
π
2
− π2
43 of 81
du
.
u
u = 3 + 2 sin θ. Then du = 2 cos θ dθ. Moreover
π
π
u = 3 + 2 sin = 3 + 2 = 5
θ=
2
2 π =⇒
π
θ=−
u = 3 + 2 sin −
= 3 − 2 = 1.
2
2
Therefore
Z
Z
2 · 2 cos θ
dθ =
3 + 2 sin θ
Z
1
5
h
i5
2
du = 2 ln |u| = 2 ln |5| − 2 ln |1| = 2 ln 5.
u
1
Break
We will continue at 3pm
44 of 81
7.2 Natural Logarithms
The Integrals of tan x, cot x, sec x and
cosec x
We can also use
to integrate
45 of 81
Z
1
du = ln |u| + C
u
tan x, cot x, sec x
and
cosec x.
Z
7.2 Natural Logarithms
du
= ln |u| + C
u
The Integral of tan x
If
u = cos x,
Z
du = − sin x dx and
Z
sin x
tan x dx =
dx =
cos x
=
then
=
=
=
46 of 81
.
Z
7.2 Natural Logarithms
du
= ln |u| + C
u
The Integral of tan x
If
u = cos x,
Z
du = − sin x dx and
Z
Z
sin x
−du
tan x dx =
dx =
cos x
u
=
then
=
=
=
46 of 81
.
Z
7.2 Natural Logarithms
du
= ln |u| + C
u
The Integral of tan x
If
u = cos x,
Z
du = − sin x dx and
Z
Z
sin x
−du
tan x dx =
dx =
cos x
u
= − ln |u| + C = − ln |cos x| + C
then
=
=
=
46 of 81
.
7.2 Natural Logarithms
Z
du
= ln |u| + C
u
The Integral of tan x
If
u = cos x,
Z
du = − sin x dx and
Z
Z
sin x
−du
tan x dx =
dx =
cos x
u
= − ln |u| + C = − ln |cos x| + C
then
= ln |cos x|−1 + C
1
= ln
+C
|cos x|
= ln |sec x| + C.
46 of 81
7.2 Natural Logarithms
Z
du
= ln |u| + C
u
The Integral of cot x
If
u = sin x, then
Z
Z
cos x
cot x dx =
dx =
sin x
(you ll in the details)
47 of 81
= ... =
.
7.2 Natural Logarithms
Z
du
= ln |u| + C
u
The Integral of cot x
If
u = sin x, then
Z
Z
Z
cos x
du
cot x dx =
dx =
= ... =
sin x
u
(you ll in the details)
47 of 81
.
7.2 Natural Logarithms
Z
du
= ln |u| + C
u
The Integral of cot x
If
u = sin x, then
Z
Z
Z
cos x
du
cot x dx =
dx =
= . . . = − ln |cosec x| + C.
sin x
u
(you ll in the details)
47 of 81
Z
du
7.2 Natural
Logarithms
= ln |u|
+ C (tan x)′ = sec2 x (sec x)′ = sec x tan x
u
The Integral of sec x
This time, we need to do an extra step. We will multiply and
divide by
48 of 81
u = (sec x + tan x).
Z
du
7.2 Natural
Logarithms
= ln |u|
+ C (tan x)′ = sec2 x (sec x)′ = sec x tan x
u
The Integral of sec x
This time, we need to do an extra step. We will multiply and
u = (sec x + tan x). We calculate that
Z
Z
sec x + tan x
dx
sec x dx = sec x
sec x + tan x
divide by
=
=
=
48 of 81
.
Z
du
7.2 Natural
Logarithms
= ln |u|
+ C (tan x)′ = sec2 x (sec x)′ = sec x tan x
u
The Integral of sec x
This time, we need to do an extra step. We will multiply and
u = (sec x + tan x). We calculate that
Z
Z
sec x + tan x
dx
sec x dx = sec x
sec x + tan x
Z
sec2 x + sec x tan x
=
dx
sec x + tan x
divide by
=
=
48 of 81
.
Z
du
7.2 Natural
Logarithms
= ln |u|
+ C (tan x)′ = sec2 x (sec x)′ = sec x tan x
u
The Integral of sec x
This time, we need to do an extra step. We will multiply and
u = (sec x + tan x). We calculate that
Z
Z
sec x + tan x
dx
sec x dx = sec x
sec x + tan x
Z
sec2 x + sec x tan x
=
dx
sec x + tan x
Z
du
=
u
=
divide by
48 of 81
.
Z
du
7.2 Natural
Logarithms
= ln |u|
+ C (tan x)′ = sec2 x (sec x)′ = sec x tan x
u
The Integral of sec x
This time, we need to do an extra step. We will multiply and
u = (sec x + tan x). We calculate that
Z
Z
sec x + tan x
dx
sec x dx = sec x
sec x + tan x
Z
sec2 x + sec x tan x
=
dx
sec x + tan x
Z
du
=
u
= ln |u| + C = ln |sec x + tan x| + C.
divide by
48 of 81
du
7.2 Natural
= ln |u|Logarithms
+ C (cot x)′ = − cosec2 x
Z
u
(cosec x)′ = − cosec x cot x
The Integral of cosec x
This is similar, except that we will multiply and divide by
u = (cosec x + cot x) instead. We calculate that
Z
Z
cosec x + cot x
dx
cosec x dx = cosec x
cosec x + cot x
=
=
=
49 of 81
.
du
7.2 Natural
= ln |u|Logarithms
+ C (cot x)′ = − cosec2 x
Z
u
(cosec x)′ = − cosec x cot x
The Integral of cosec x
This is similar, except that we will multiply and divide by
u = (cosec x + cot x) instead. We calculate that
Z
Z
cosec x + cot x
dx
cosec x dx = cosec x
cosec x + cot x
Z
cosec2 x + cosec x cot x
=
dx
cosec x + cot x
=
=
49 of 81
.
du
7.2 Natural
= ln |u|Logarithms
+ C (cot x)′ = − cosec2 x
Z
u
(cosec x)′ = − cosec x cot x
The Integral of cosec x
This is similar, except that we will multiply and divide by
u = (cosec x + cot x) instead. We calculate that
Z
Z
cosec x + cot x
dx
cosec x dx = cosec x
cosec x + cot x
Z
cosec2 x + cosec x cot x
=
dx
cosec x + cot x
Z
−du
=
u
=
49 of 81
.
du
7.2 Natural
= ln |u|Logarithms
+ C (cot x)′ = − cosec2 x
Z
u
(cosec x)′ = − cosec x cot x
The Integral of cosec x
This is similar, except that we will multiply and divide by
u = (cosec x + cot x) instead. We calculate that
Z
Z
cosec x + cot x
dx
cosec x dx = cosec x
cosec x + cot x
Z
cosec2 x + cosec x cot x
=
dx
cosec x + cot x
Z
−du
=
u
= − ln |u| + C = − ln |cosec x + cot x| + C.
49 of 81
50 of 81
51 of 81
7.2 Natural Logarithms
Logarithmic Differentiation
Sometimes when we are trying to nd a derivative, it becomes
easier if we take
52 of 81
ln
of both sides of an equation.
7.2 Natural Logarithms
Logarithmic Differentiation
Sometimes when we are trying to nd a derivative, it becomes
easier if we take
ln
of both sides of an equation.
Example
Find
52 of 81
dy
dx
1
if
(x2 + 1)(x + 3) 2
y=
x−1
for
x > 1.
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
y=
=
=
53 of 81
(x2 + 1)(x + 3) 2
x−1
.
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
ln y = ln
=
=
53 of 81
(x2 + 1)(x + 3) 2
x−1
.
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
(x2 + 1)(x + 3) 2
x−1
1
2
= ln (x + 1)(x + 3) 2 − ln(x − 1)
ln y = ln
=
53 of 81
.
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
(x2 + 1)(x + 3) 2
x−1
1
2
= ln (x + 1)(x + 3) 2 − ln(x − 1)
ln y = ln
= ln(x2 + 1) +
53 of 81
1
ln(x + 3) − ln(x − 1).
2
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
(x2 + 1)(x + 3) 2
x−1
1
2
= ln (x + 1)(x + 3) 2 − ln(x − 1)
ln y = ln
= ln(x2 + 1) +
1
ln(x + 3) − ln(x − 1).
2
Now we can dierentiate both sides.
d
d
ln y =
dx
dx
53 of 81
ln(x2 + 1) +
1
ln(x + 3) − ln(x − 1)
2
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
(x2 + 1)(x + 3) 2
x−1
1
2
= ln (x + 1)(x + 3) 2 − ln(x − 1)
ln y = ln
= ln(x2 + 1) +
1
ln(x + 3) − ln(x − 1).
2
Now we can dierentiate both sides.
d
d
1
ln y =
ln(x2 + 1) + ln(x + 3) − ln(x − 1)
dx
dx
2
1 dy
2x
1
1
1
= 2
+ ·
−
y dx
x +1 2 x+3 x−1
53 of 81
7.2 Natural Logarithms
We are going to take our equation and apply
ln
to both sides.
1
(x2 + 1)(x + 3) 2
x−1
1
2
= ln (x + 1)(x + 3) 2 − ln(x − 1)
ln y = ln
= ln(x2 + 1) +
1
ln(x + 3) − ln(x − 1).
2
Now we can dierentiate both sides.
d
d
1
ln y =
ln(x2 + 1) + ln(x + 3) − ln(x − 1)
dx
dx
2
1 dy
2x
1
1
1
= 2
+ ·
−
y dx
x +1 2 x+3 x−1
dy
2x
1
1
=y
+
−
.
dx
x2 + 1 2x + 6 x − 1
53 of 81
7.2 Natural Logarithms
Thus
dy
=y
dx
=
54 of 81
2x
1
1
+
−
x2 + 1 2x + 6 x − 1
!
1
(x2 + 1)(x + 3) 2
2x
1
1
+
−
.
x−1
x2 + 1 2x + 6 x − 1
7.3
Exponential
Functions
55 of 81
7.3 Exponential Functions
y
e
y = ln x
1
x
1
56 of 81
ln is a one-to-one
(−∞, ∞).
e
function with domain
(0, ∞)
and range
7.3 Exponential Functions
=
y
y = ln−1 x
x
y
e
y = ln x
1
x
1
e
So there exists and inverse function
and range
56 of 81
(0, ∞).
ln−1
with domain
(−∞, ∞)
7.3 Exponential Functions
Denition
The inverse of the function
function and is written
ln x
is called the
exp x = ex .
57 of 81
exponential
7.3 Exponential Functions
Denition
The inverse of the function
function and is written
ln x
is called the
exponential
exp x = ex .
Remark
We have
eln x = x
(for all
x > 0)
and
ln (ex ) = x
57 of 81
(for all
x).
58 of 81
7.3 Exponential Functions
59 of 81
7.3 Exponential Functions
59 of 81
7.3 Exponential Functions
The Derivative and Integral of ex
ln (ex ) =
60 of 81
x
7.3 Exponential Functions
The Derivative and Integral of ex
d
d
ln (ex ) =
x
dx
dx
60 of 81
7.3 Exponential Functions
The Derivative and Integral of ex
d
d
ln (ex ) =
x
dx
dx
1 d x
e =1
ex dx
60 of 81
7.3 Exponential Functions
The Derivative and Integral of ex
d
d
ln (ex ) =
x
dx
dx
1 d x
e =1
ex dx
d x
e = ex .
dx
by the Chain Rule.
60 of 81
7.3 Exponential Functions
The Derivative and Integral of ex
d
d
ln (ex ) =
x
dx
dx
1 d x
e =1
ex dx
d x
e = ex .
dx
by the Chain Rule.
Theorem
d x
e = ex .
dx
60 of 81
61 of 81
7.3 Exponential Functions
Since
d x
e = ex ,
dx
we also have
Theorem
Z
62 of 81
ex dx = ex + C.
63 of 81
7.3 Exponential Functions
y
y
=
x
y = ex
y = ln x
1
x
1
ex has domain (−∞, ∞)
lim ex = ∞; and
x→∞
lim ex = 0
x→−∞
64 of 81
and range
(0, ∞);
7.3 Exponential Functions
Laws of Exponents
Theorem
1
ea eb = ea+b
(proof in textbook)
65 of 81
7.3 Exponential Functions
Laws of Exponents
Theorem
1
2
ea eb = ea+b
1
e−x = x
e
(proof in textbook)
65 of 81
7.3 Exponential Functions
Laws of Exponents
Theorem
1
2
ea eb = ea+b
1
e−x = x
e
(proof in textbook)
65 of 81
3
ea
= ea−b
eb
7.3 Exponential Functions
Laws of Exponents
Theorem
1
2
ea eb = ea+b
1
e−x = x
e
(proof in textbook)
65 of 81
3
4
ea
= ea−b
eb
(ea )b = eab .
7.3 Exponential Functions
The General Exponential Function ax
Denition
For any numbers
base a is
66 of 81
a>0
and
x ∈ R,
the
ax = ex ln a .
exponential function with
7.3 Exponential Functions
The General Exponential Function ax
Denition
For any numbers
base a is
a>0
and
x ∈ R,
the
ax = ex ln a .
Denition
For any
x>0
and for any
n ∈ R,
xn = en ln x .
66 of 81
exponential function with
7.3 Exponential Functions
Now we can prove
Theorem
For any x > 0 and any n ∈ R,
d n
x = nxn−1 .
dx
67 of 81
7.3 Exponential Functions
Now we can prove
Theorem
For any x > 0 and any n ∈ R,
d n
x = nxn−1 .
dx
If x ≤ 0, then the formula holds whenever
exist.
67 of 81
d n
dx x
, xn and xn−1 all
7.3 Exponential Functions
Proof.
1 If
x > 0,
then we calculate
d n
d n ln x
d
n
x =
e
= en ln x ·
(n ln x) = xn · = nxn−1 .
dx
dx
dx
x
68 of 81
7.3 Exponential Functions
Proof.
2 If
x < 0,
and if
y ′ , y = xn
and
xn−1
all exists, then
ln |y| = ln |x|n = n ln |x| .
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7.3 Exponential Functions
Proof.
2 If
x < 0,
and if
y ′ , y = xn
and
xn−1
all exists, then
ln |y| = ln |x|n = n ln |x| .
Using Implicit Dierentiation, we have
ln |y| =
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n ln |x|
7.3 Exponential Functions
Proof.
2 If
x < 0,
and if
y ′ , y = xn
and
xn−1
all exists, then
ln |y| = ln |x|n = n ln |x| .
Using Implicit Dierentiation, we have
d
d
ln |y| =
n ln |x|
dx
dx
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7.3 Exponential Functions
Proof.
2 If
x < 0,
and if
y ′ , y = xn
and
xn−1
all exists, then
ln |y| = ln |x|n = n ln |x| .
Using Implicit Dierentiation, we have
d
ln |y| =
dx
y′
=
y
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d
n ln |x|
dx
n
x
7.3 Exponential Functions
Proof.
2 If
x < 0,
and if
y ′ , y = xn
and
xn−1
all exists, then
ln |y| = ln |x|n = n ln |x| .
Using Implicit Dierentiation, we have
d
ln |y| =
dx
y′
=
y
d
n ln |x|
dx
n
x
y
xn
y′ = n = n
= nxn−1 .
x
x
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7.3 Exponential Functions
Proof.
3 For
x=0
and
n ≥ 1,
we just use the denition of the
derivative to show that
(0 + h)n − 0n
= lim hn−1 = 0.
h→0
h→0
h
(xn )′ (0) = lim
(Note
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0n−1
does not exist if
n < 1.)
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7.3 Exponential Functions
The number e Expressed as a Limit
Theorem
1
e = lim (1 + x) x .
x→0
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7.3 Exponential Functions
The number e Expressed as a Limit
Theorem
1
e = lim (1 + x) x .
x→0
Proof.
Let
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f (x) = ln x.
Then
f ′ (x) =
1
x and
f ′ (1) = 1.
7.3 Exponential Functions
The number e Expressed as a Limit
Theorem
1
e = lim (1 + x) x .
x→0
Proof.
Let
f (x) = ln x.
Then
f ′ (x) =
1
x and
f ′ (1) = 1.
Therefore
f (1 + h) − f (1)
=
h→0
h
1 = f ′ (1) = lim
=
=
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=
=
.
7.3 Exponential Functions
The number e Expressed as a Limit
Theorem
1
e = lim (1 + x) x .
x→0
Proof.
Let
f (x) = ln x.
Then
f ′ (x) =
1
x and
f ′ (1) = 1.
Therefore
f (1 + h) − f (1)
f (1 + x) − f (1)
= lim
x→0
h→0
h
x
1 = f ′ (1) = lim
=
=
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=
=
.
7.3 Exponential Functions
The number e Expressed as a Limit
Theorem
1
e = lim (1 + x) x .
x→0
Proof.
Let
f (x) = ln x.
Then
f ′ (x) =
1
x and
f ′ (1) = 1.
Therefore
f (1 + h) − f (1)
f (1 + x) − f (1)
= lim
x→0
h→0
h
x
ln(1 + x) − 0
1
= lim
= lim ln(1 + x)
x→0
x→0 x
x
1 = f ′ (1) = lim
1
= lim ln(1 + x) x =
x→0
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.
7.3 Exponential Functions
The number e Expressed as a Limit
Theorem
1
e = lim (1 + x) x .
x→0
Proof.
Let
f (x) = ln x.
Then
f ′ (x) =
1
x and
f ′ (1) = 1.
Therefore
f (1 + h) − f (1)
f (1 + x) − f (1)
= lim
x→0
h→0
h
x
ln(1 + x) − 0
1
= lim
= lim ln(1 + x)
x→0
x→0 x
x
1 = f ′ (1) = lim
1
= lim ln(1 + x) x = ln
x→0
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1
lim (1 + x) x .
x→0
7.3 Exponential Functions
Proof continued.
1 = ln
1
lim (1 + x) x
x→0
Taking the exponential of both sides gives
1
e = lim (1 + x) x
x→0
as required.
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7.3 Exponential Functions
Proof continued.
1 = ln
1
lim (1 + x) x
x→0
Taking the exponential of both sides gives
1
e = lim (1 + x) x
x→0
as required.
Remark
By taking very small
x
in
1
(1 + x) x ,
we can calculate
e ≈ 2.718281828459045
to 15 decimal places.
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7.3 Exponential Functions
The derivative of ax
d x
d x ln a
d
a =
e
= ex ln a ·
(x ln a) = ax ln a.
dx
dx
dx
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7.3 Exponential Functions
The derivative of ax
d x
d x ln a
d
a =
e
= ex ln a ·
(x ln a) = ax ln a.
dx
dx
dx
e
is a special number because
d x
e = ex ln e = ex · 1 = ex .
dx
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7.3 Exponential Functions
By the Chain Rule, if
u
is dierentiable then
d u
du
a = au ln a
dx
dx
for
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a > 0.
7.3 Exponential Functions
By the Chain Rule, if
u
is dierentiable then
d u
du
a = au ln a
dx
dx
for
a > 0.
Equivalently
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Z
au du =
au
+ C.
ln a
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7.3 Exponential Functions
Logarithms with Base a
If
a>0
and
a ̸= 1,
then
ax
it has an inverse function.
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is a one-to-one function. Therefore
7.3 Exponential Functions
Logarithms with Base a
If
a>0
and
a ̸= 1,
then
ax
is a one-to-one function. Therefore
it has an inverse function. Since
must be dierentiable.
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(ax )′ ̸= 0,
the inverse function
7.3 Exponential Functions
Logarithms with Base a
If
a>0
and
a ̸= 1,
then
ax
is a one-to-one function. Therefore
it has an inverse function. Since
must be dierentiable.
(ax )′ ̸= 0,
the inverse function
Denition
a > 0 and a ̸= 1, then the inverse of ax is
of x with base a and is denoted by loga x.
If
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called the
logarithm
7.3 Exponential Functions
=
y
y = 2x
x
y
y = log2 x
2
1
x
1 2
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7.3 Exponential Functions
Remark
We have
aloga x = x
(for all
x > 0)
and
loga (ax ) = x
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(for all
x).
7.3 Exponential Functions
Remark
Note that
y = loga x
ay = x
ln ay = ln x
y ln a = ln x
ln x
y=
.
ln a
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7.3 Exponential Functions
Remark
Note that
y = loga x
ay = x
ln ay = ln x
y ln a = ln x
ln x
y=
.
ln a
Therefore
loga x =
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ln x
.
ln a
7.3 Exponential Functions
Derivatives and Integrals Involving loga x
Note that
d
d
loga u =
dx
dx
ln u
ln a
=
1 d
1 1 du
ln u =
· ·
.
ln a dx
ln a u dx
d
1 1 du
loga u =
· ·
dx
ln a u dx
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80 of 81
Next Time
7.5 Indeterminate Forms and L’Hôpital’s Rule
7.6 Inverse Trigonometric Functions
7.7 Hyperbolic Functions
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