Lesson 14: Derivatives of Logarithmic and Exponential Functions

Sections 3.1–3.3
Derivatives of Exponential and
Logarithmic Functions

V63.0121.002.2010Su, Calculus I

New York University

June 1, 2010

Announcements

Today: Homework 2 due
Tomorrow: Section 3.4, review
Thursday: Midterm in class
. . . . . .

Announcements

Today: Homework 2 due
Tomorrow: Section 3.4,
review
Thursday: Midterm in class

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 2 / 54

Objectives for Sections 3.1 and 3.2

Know the definition of an
exponential function
Know the properties of
exponential functions
Understand and apply the
laws of logarithms,
including the change of
base formula.

. . . . . .


Objectives for Section 3.3

Know the derivatives of the
exponential functions (with
any base)
Know the derivatives of the
logarithmic functions (with
any base)
Use the technique of
logarithmic differentiation
to find derivatives of
functions involving roducts,
quotients, and/or
exponentials.

. . . . . .


Outline
Definition of exponential functions
Properties of exponential Functions
The number e and the natural exponential function
Compound Interest
The number e
A limit
Derivatives of Exponential Functions
Exponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithms
Other exponentials
Other logarithms
Logarithmic Differentiation
The power rule for irrational powers
. . . . . .


Derivation of exponential functions

Definition
If a is a real number and n is a positive whole number, then

an = a · a · · · · · a
n factors

. . . . . .


Derivation of exponential functions

Definition
If a is a real number and n is a positive whole number, then

an = a · a · · · · · a
n factors

Examples

23 = 2 · 2 · 2 = 8
34 = 3 · 3 · 3 · 3 = 81
(−1)5 = (−1)(−1)(−1)(−1)(−1) = −1

. . . . . .


Fact
If a is a real number, then
ax+y = ax ay
ax
ax−y = y
a
(ax )y = axy
(ab)x = ax bx
whenever all exponents are positive whole numbers.

. . . . . .


Fact
If a is a real number, then
ax+y = ax ay
ax
ax−y = y
a
(ax )y = axy
(ab)x = ax bx
whenever all exponents are positive whole numbers.

Proof.
Check for yourself:

ax+y = a · a · · · · · a = a · a · · · · · a · a · a · · · · · a = ax ay
x + y factors x factors y factors

. . . . . .


Let's be conventional

The desire that these properties remain true gives us conventions
for ax when x is not a positive whole number.

. . . . . .



For example:
!
an = an+0 = an a0

. . . . . .



For example:
!
an = an+0 = an a0

Definition
If a ̸= 0, we define a0 = 1.

. . . . . .



For example:
!
an = an+0 = an a0

Definition
If a ̸= 0, we define a0 = 1.

Notice 00 remains undefined (as a limit form, it’s indeterminate).

. . . . . .


Conventions for negative exponents

If n ≥ 0, we want
an · a−n = an+(−n) = a0 = 1
!

. . . . . .



If n ≥ 0, we want
an · a−n = an+(−n) = a0 = 1
!

Definition
1
If n is a positive integer, we define a−n = .
an

. . . . . .



If n ≥ 0, we want
an · a−n = an+(−n) = a0 = 1
!

Definition
1
If n is a positive integer, we define a−n = .
an

Fact
1
The convention that a−n = “works” for negative n as well.
an
am
If m and n are any integers, then am−n = n .
a

. . . . . .


Conventions for fractional exponents

If q is a positive integer, we want
!
(a1/q )q = a1 = a

. . . . . .



!
(a1/q )q = a1 = a

Definition
√
If q is a positive integer, we define a1/q = q
a. We must have a ≥ 0 if q
is even.

. . . . . .



!
(a1/q )q = a1 = a

Definition
√
If q is a positive integer, we define a1/q = q a. We must have a ≥ 0 if q
is even.
√q
( √ )p
Notice that ap = q a . So we can unambiguously say

ap/q = (ap )1/q = (a1/q )p

. . . . . .


Conventions for irrational powers

So ax is well-defined if x is rational.
What about irrational powers?

. . . . . .




Definition
Let a > 0. Then
ax = lim ar
r→x
r rational

. . . . . .




Definition
Let a > 0. Then
ax = lim ar
r→x
r rational

In other words, to approximate ax for irrational x, take r close to x but
rational and compute ar .

. . . . . .


Graphs of various exponential functions
y
.

. x
.
. . . . . .

y
.

. = 1x
y

. x
.
. . . . . .

y
.
. = 2x
y

. = 1x
y

. x
.
. . . . . .

y
.
. = 3x. = 2x
y y

. = 1x
y

. x
.
. . . . . .

y
.
. = 10x= 3x. = 2x
y y
. y

. = 1x
y

. x
.
. . . . . .

y
.
. = 10x= 3x. = 2x
y y
. y . = 1.5x
y

. = 1x
y

. x
.
. . . . . .

y
.
. = (1/2)x
y . = 10x= 3x. = 2x
y y
. y . = 1.5x
y

. = 1x
y

. x
.
. . . . . .

x
y
.
. = (1/2)x (1/3)
y y
. = . = 10x= 3x. = 2x
y y
. y . = 1.5x
y

. = 1x
y

. x
.
. . . . . .

y
.
y . = x
. = (1/2)x (1/3)
y . = (1/10)x. = 10x= 3x. = 2x
y y y
. y . = 1.5x
y

. = 1x
y

. x
.
. . . . . .

y
.
y yx
.. = ((1/2)x (1/3)x
y = 2/. )=
3 . = (1/10)x. = 10x= 3x. = 2x
y y y
. y . = 1.5x
y

. = 1x
y

. x
.
. . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .


.
Theorem
If a > 0 and a ̸= 1, then f(x) = ax is a continuous function with domain R and
range (0, ∞). In particular, ax > 0 for all x. If a, b > 0 and x, y ∈ R, then
ax+y = ax ay
ax
ax−y = y
a
(ax )y = axy
(ab)x = ax bx

Proof.

This is true for positive integer exponents by natural definition
Our conventional definitions make these true for rational exponents
Our limit definition make these for irrational exponents, too

.

. . . . . .


.
Theorem
ax+y = ax ay
ax
ax−y = y negative exponents mean reciprocals.
a
(ax )y = axy
(ab)x = ax bx

Proof.


.

. . . . . .


.
Theorem
ax+y = ax ay
ax
ax−y = y negative exponents mean reciprocals.
a
(ax )y = axy fractional exponents mean roots
(ab)x = ax bx

Proof.


.

. . . . . .


Simplifying exponential expressions
Example
Simplify: 82/3

. . . . . .


Example
Simplify: 82/3

Solution
√
3 √
82/3 = 82 =
3
64 = 4

. . . . . .


Example
Simplify: 82/3

Solution
√3 √
82/3 = 82 = 64 = 4
3

( √ )2
8 = 22 = 4.
3
Or,

. . . . . .


Example
Simplify: 82/3

Solution
√3 √
82/3 = 82 = 64 = 4
3

( √ )2
8 = 22 = 4.
3
Or,

Example
√
8
Simplify:
21/2

. . . . . .


Example
Simplify: 82/3

Solution
√3 √
82/3 = 82 = 64 = 4
3

( √ )2
8 = 22 = 4.
3
Or,

Example
√
8
Simplify:
21/2

Answer
2
. . . . . .


Limits of exponential functions

Fact (Limits of exponential y
.
functions) . = (= 2()1/32/3)x
y . 1/ =x( )x
y .
y y y = x . 3x y
. = (. /10)10x= 2x. =
1 . =
y y

If a > 1, then lim ax = ∞
x→∞
and lim ax = 0
x→−∞
If 0 < a < 1, then
lim ax = 0 and y
. =
x→∞
lim a = ∞ x . x
.
x→−∞

. . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .


Compounded Interest

Question
Suppose you save $100 at 10% annual interest, with interest
compounded once a year. How much do you have
After one year?
After two years?
after t years?

. . . . . .


Compounded Interest

Question
After one year?
After two years?
after t years?

Answer

$100 + 10% = $110

. . . . . .


Compounded Interest

Question
After one year?
After two years?
after t years?

Answer

$100 + 10% = $110
$110 + 10% = $110 + $11 = $121

. . . . . .


Compounded Interest

Question
After one year?
After two years?
after t years?

Answer

$100 + 10% = $110
$110 + 10% = $110 + $11 = $121
$100(1.1)t .

. . . . . .


Compounded Interest: quarterly

Question
compounded four times a year. How much do you have
After one year?
After two years?
after t years?

. . . . . .



Question
After one year?
After two years?
after t years?

Answer

$100(1.025)4 = $110.38,

. . . . . .



Question
After one year?
After two years?
after t years?

Answer

$100(1.025)4 = $110.38, not $100(1.1)4 !

. . . . . .



Question
After one year?
After two years?
after t years?

Answer

$100(1.025)4 = $110.38, not $100(1.1)4 !
$100(1.025)8 = $121.84

. . . . . .



Question
After one year?
After two years?
after t years?

Answer

$100(1.025)4 = $110.38, not $100(1.1)4 !
$100(1.025)8 = $121.84
$100(1.025)4t .

. . . . . .


Compounded Interest: monthly

Question
compounded twelve times a year. How much do you have after t
years?

. . . . . .


Compounded Interest: monthly

Question
compounded twelve times a year. How much do you have after t
years?

Answer
$100(1 + 10%/12)12t

. . . . . .


Compounded Interest: general

Question
Suppose you save P at interest rate r, with interest compounded n
times a year. How much do you have after t years?

. . . . . .


Compounded Interest: general

Question
Suppose you save P at interest rate r, with interest compounded n
times a year. How much do you have after t years?

Answer

( r )nt
B(t) = P 1 +
n

. . . . . .


Compounded Interest: continuous

Question
Suppose you save P at interest rate r, with interest compounded every
instant. How much do you have after t years?

. . . . . .


Compounded Interest: continuous

Question
Suppose you save P at interest rate r, with interest compounded every
instant. How much do you have after t years?

Answer

( ( )
r )nt 1 rnt
B(t) = lim P 1 + = lim P 1 +
n→∞ n n→∞ n
[ ( )n ]rt
1
=P lim 1 +
n→∞ n
independent of P, r, or t

. . . . . .


The magic number

Definition
( )
1 n
e = lim 1 +
n→∞ n

. . . . . .


The magic number

Definition
( )
1 n
e = lim 1 +
n→∞ n

So now continuously-compounded interest can be expressed as

B(t) = Pert .

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106 2.71828

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
We can experimentally n
verify that this number 1 2
exists and is 2 2.25
3 2.37037
e ≈ 2.718281828459045 . . .
10 2.59374
100 2.70481
1000 2.71692
106 2.71828

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
3 2.37037
e ≈ 2.718281828459045 . . .
10 2.59374
e is irrational 100 2.70481
1000 2.71692
106 2.71828

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
3 2.37037
e ≈ 2.718281828459045 . . .
10 2.59374
e is irrational 100 2.70481
1000 2.71692
e is transcendental
106 2.71828

. . . . . .


Meet the Mathematician: Leonhard Euler

Born in Switzerland, lived
in Prussia (Germany) and
Russia
Eyesight trouble all his life,
blind from 1766 onward
Hundreds of contributions
to calculus, number theory,
graph theory, fluid
mechanics, optics, and
astronomy

Leonhard Paul Euler
Swiss, 1707–1783
. . . . . .


A limit
.
Question
eh − 1
What is lim ?
h→0 h

. . . . . . .


A limit
.
Question
eh − 1
What is lim ?
h→0 h

Answer

If h is small enough, e ≈ (1 + h)1/h . So

eh − 1
≈1
h

. . . . . . .


A limit
.
Question
eh − 1
What is lim ?
h→0 h

Answer


eh − 1
≈1
h

eh − 1
In fact, lim = 1.
h→0 h
2h − 1
This can be used to characterize e: lim = 0.693 · · · < 1 and
h→0 h
3h − 1
lim = 1.099 · · · > 1
h→0 h
. . . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .


Logarithms

Definition

The base a logarithm loga x is the inverse of the function ax

y = loga x ⇐⇒ x = ay

The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .

. . . . . .


Logarithms

Definition



y = ln x ⇐⇒ x = ey .

Facts

(i) loga (x · x′ ) = loga x + loga x′

. . . . . .


Logarithms

Definition



y = ln x ⇐⇒ x = ey .

Facts

(x)
(ii) loga ′ = loga x − loga x′
x

. . . . . .


Logarithms

Definition



y = ln x ⇐⇒ x = ey .

Facts

(x)
(ii) loga ′ = loga x − loga x′
x
(iii) loga (xr ) = r loga x

. . . . . .


Logarithms convert products to sums

Suppose y = loga x and y′ = loga x′
′
Then x = ay and x′ = ay
′ ′
So xx′ = ay ay = ay+y
Therefore
loga (xx′ ) = y + y′ = loga x + loga x′

. . . . . .


Example
Write as a single logarithm: 2 ln 4 − ln 3.

. . . . . .


Example

Solution
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3

. . . . . .


Example

Solution
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3

Example
3
Write as a single logarithm: ln + 4 ln 2
4

. . . . . .


Example

Solution
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3

Example
3
Write as a single logarithm: ln + 4 ln 2
4

Answer
ln 12
. . . . . .


“ .
. lawn”

.

. . . . . .
.
Image credit: Selva

Graphs of logarithmic functions
y
.
. = 2x
y

y
. = log2 x

. . 0, 1)
(

..1, 0) .
( x
.

. . . . . .

y
.
. = 3x= 2x
y . y

y
. = log2 x

y
. = log3 x
. . 0, 1)
(

..1, 0) .
( x
.

. . . . . .

y
.
. = .10x 3x= 2x
y y= . y

y
. = log2 x

y
. = log3 x
. . 0, 1)
(
y
. = log10 x
..1, 0) .
( x
.

. . . . . .

y
.
. = .10=3x= 2x
y xy
y y. = .ex

y
. = log2 x

y
. = ln x
y
. = log3 x
. . 0, 1)
(
y
. = log10 x
..1, 0) .
( x
.

. . . . . .


Change of base formula for exponentials

Fact
If a > 0 and a ̸= 1, then
ln x
loga x =
ln a

. . . . . .


Change of base formula for exponentials

Fact
If a > 0 and a ̸= 1, then
ln x
loga x =
ln a

Proof.

If y = loga x, then x = ay
So ln x = ln(ay ) = y ln a
Therefore
ln x
y = loga x =
ln a

. . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .



Fact
If f(x) = ax , then f′ (x) = f′ (0)ax .

. . . . . .



Fact

Proof.
Follow your nose:

f(x + h) − f(x) ax+h − ax
f′ (x) = lim = lim
h→0 h h→0 h
a x ah − ax a h−1
= lim = ax · lim = ax · f′ (0).
h→0 h h→0 h

. . . . . .



Fact

Proof.
Follow your nose:

f(x + h) − f(x) ax+h − ax
f′ (x) = lim = lim
h→0 h h→0 h
a x ah − ax a h−1
= lim = ax · lim = ax · f′ (0).
h→0 h h→0 h

To reiterate: the derivative of an exponential function is a constant
times that function. Much different from polynomials!
. . . . . .


The funny limit in the case of e
Remember the definition of e:
( )
1 n
e = lim 1 + = lim (1 + h)1/h
n→∞ n h→0

Question
eh − 1
What is lim ?
h→0 h

. . . . . .


( )
1 n
e = lim 1 + = lim (1 + h)1/h
n→∞ n h→0

Question
eh − 1
What is lim ?
h→0 h

Answer
[ ]h
eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h
≈ = = =1
h h h h

. . . . . .


( )
1 n
e = lim 1 + = lim (1 + h)1/h
n→∞ n h→0

Question
eh − 1
What is lim ?
h→0 h

Answer
[ ]h
eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h
≈ = = =1
h h h h
eh − 1
So in the limit we get equality: lim =1
h→0 h
. . . . . .


Derivative of the natural exponential function

From ( )
d x ah − 1 eh − 1
a = lim ax and lim =1
dx h→0 h h→0 h
we get:
Theorem

d x
e = ex
dx

. . . . . .


Exponential Growth

Commonly misused term to say something grows exponentially
It means the rate of change (derivative) is proportional to the
current value
Examples: Natural population growth, compounded interest,
social networks

. . . . . .


Examples

Examples
Find these derivatives:
e3x
2
ex
x2 ex

. . . . . .


Examples

Examples
e3x
2
ex
x2 ex

Solution
d 3x
e = 3e3x
dx

. . . . . .


Examples

Examples
e3x
2
ex
x2 ex

Solution
d 3x
e = 3e3x
dx
d x2 2 d 2
e = ex (x2 ) = 2xex
dx dx

. . . . . .


Examples

Examples
e3x
2
ex
x2 ex

Solution
d 3x
e = 3e3x
dx
d x2 2 d 2
e = ex (x2 ) = 2xex
dx dx
d 2 x
x e = 2xex + x2 ex
dx
. . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .



Let y = ln x. Then x = ey
so

. . . . . .



so
dy
ey =1
dx

. . . . . .



so
dy
ey=1
dx
dy 1 1
=⇒ = y =
dx e x

. . . . . .



so
dy
ey=1
dx
dy 1 1
=⇒ = y =
dx e x
So:
Fact

d 1
ln x =
dx x

. . . . . .



y
.
so
dy
ey=1
dx l
.n x
dy 1 1
=⇒ = y =
dx e x
. x
.
So:
Fact

d 1
ln x =
dx x

. . . . . .



y
.
so
dy
ey=1
dx l
.n x
dy 1 1 1
=⇒ = y = .
dx e x x
. x
.
So:
Fact

d 1
ln x =
dx x

. . . . . .


The Tower of Powers

y y′
x3 3x2 The derivative of a power
2 1 function is a power function
x 2x
of one lower power
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3

. . . . . .


The Tower of Powers

y y′
x 2x
of one lower power
x1 1x0 Each power function is the
x 0
0 derivative of another power
function, except x−1
? x−1
x−1 −1x−2
x−2 −2x−3

. . . . . .


The Tower of Powers

y y′
x 2x
of one lower power
x1 1x0 Each power function is the
x 0
0 derivative of another power
function, except x−1
ln x x−1
ln x fills in this gap
x−1 −1x−2 precisely.
x−2 −2x−3

. . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .


Other logarithms
Example
d x
Use implicit differentiation to find a .
dx

. . . . . .


Other logarithms
Example
d x
dx

Solution
Let y = ax , so
ln y = ln ax = x ln a

. . . . . .


Other logarithms
Example
d x
dx

Solution
Let y = ax , so
Differentiate implicitly:

1 dy dy
= ln a =⇒ = (ln a)y = (ln a)ax
y dx dx

. . . . . .


Other logarithms
Example
d x
dx

Solution
Let y = ax , so
Differentiate implicitly:

1 dy dy
= ln a =⇒ = (ln a)y = (ln a)ax
y dx dx

Before we showed y′ = y′ (0)y, so now we know that

2h − 1 3h − 1
ln 2 = lim ≈ 0.693 ln 3 = lim ≈ 1.10
h→0 h h→0 h
. . . . . .


Other logarithms

Example
d
Find loga x.
dx

. . . . . .


Other logarithms

Example
d
Find loga x.
dx

Solution
Let y = loga x, so ay = x.

. . . . . .


Other logarithms

Example
d
Find loga x.
dx

Solution
Let y = loga x, so ay = x. Now differentiate implicitly:

dy dy 1 1
(ln a)ay = 1 =⇒ = y =
dx dx a ln a x ln a

. . . . . .


Other logarithms

Example
d
Find loga x.
dx

Solution
Let y = loga x, so ay = x. Now differentiate implicitly:

dy dy 1 1
(ln a)ay = 1 =⇒ = y =
dx dx a ln a x ln a
Another way to see this is to take the natural logarithm:

ln x
ay = x =⇒ y ln a = ln x =⇒ y =
ln a
dy 1 1
So = .
dx ln a x
. . . . . .


More examples

Example
d
Find log2 (x2 + 1)
dx

. . . . . .


More examples

Example
d
Find log2 (x2 + 1)
dx

Answer

dy 1 1 2x
= 2+1
(2x) =
dx ln 2 x (ln 2)(x2 + 1)

. . . . . .


Outline
Compound Interest
The number e
A limit
Exponential Growth
Other exponentials
Other logarithms
. . . . . .


A nasty derivative

Example
√
(x2 + 1) x + 3
Let y = . Find y′ .
x−1

. . . . . .


A nasty derivative

Example
√
(x2 + 1) x + 3
Let y = . Find y′ .
x−1

Solution
We use the quotient rule, and the product rule in the numerator:
[ √ ] √
(x − 1) 2x x + 3 + (x2 + 1) 1 (x + 3)−1/2 − (x2 + 1) x + 3(1)
2
y′ =
(x − 1)2
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
= + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2

. . . . . .


Another way

√
(x2 + 1) x + 3
y=
x−1
1
ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1)
2
1 dy 2x 1 1
= 2 + −
y dx x + 1 2(x + 3) x − 1

So
( )
dy 2x 1 1
= + − y
dx x2+1 2(x + 3) x − 1
( ) √
2x 1 1 (x2 + 1) x + 3
= + −
x2 + 1 2(x + 3) x − 1 x−1

. . . . . .


Compare and contrast

Using the product, quotient, and power rules:
√ √
′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2

Using logarithmic differentiation:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 x−1

. . . . . .



√ √
′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2

( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 x−1

Are these the same?

. . . . . .



√ √
′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2

( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 x−1

Are these the same?
Which do you like better?

. . . . . .



√ √
′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2

( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 x−1

Are these the same?
Which do you like better?
What kinds of expressions are well-suited for logarithmic
differentiation?
. . . . . .


Derivatives of powers

Let y = xx . Which of these is true?
(A) Since y is a power function, y′ = x · xx−1 = xx .
(B) Since y is an exponential function, y′ = (ln x) · xx
(C) Neither

. . . . . .


It's neither! Or both?

If y = xx , then

ln y = x ln x
1 dy 1
= x · + ln x = 1 + ln x
y dx x
dy
= xx + (ln x)xx
dx
Each of these terms is one of the wrong answers!

. . . . . .


Derivative of arbitrary powers

Fact (The power rule)
Let y = xr . Then y′ = rxr−1 .

. . . . . .


Derivative of arbitrary powers

Fact (The power rule)
Let y = xr . Then y′ = rxr−1 .

Proof.

y = xr =⇒ ln y = r ln x
Now differentiate:
1 dy r
=
y dx x
dy y
=⇒ = r = rxr−1
dx x

. . . . . .


Summary

. . . . . .


Lesson 14: Derivatives of Logarithmic and Exponential Functions

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Lesson 14: Derivatives of Logarithmic and Exponential Functions