Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)

Section 3.3
Derivatives of Exponential and
Logarithmic Functions
V63.0121.021, Calculus I
New York University
October 25, 2010
Announcements
Midterm is graded. Please see FAQ.
Quiz 3 next week on 2.6, 2.8, 3.1, 3.2

Announcements
Midterm is graded. Please
see FAQ.
Quiz 3 next week on 2.6,
2.8, 3.1, 3.2
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 2 / 38

Objectives
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
Recall Section 3.1–3.2
Derivative of the natural exponential function
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

Conventions on power expressions
Let a be a positive real number.
If n is a positive whole number, then an
= a · a · · · · · a
n factors
a0
= 1.
For any real number r, a−r
=
1
ar
.
For any positive whole number n, a1/n
= n
√
a.
There is only one continuous function which satisﬁes all of the above. We
call it the exponential function with base a.

Properties of exponential Functions
Theorem
If a > 0 and a = 1, then f (x) = ax
is a continuous function with domain
(−∞, ∞) and range (0, ∞). In particular, ax
> 0 for all x. For any real
numbers x and y, and positive numbers a and b we have
ax+y
= ax
ay
ax−y
=
ax
ay
(ax
)y
= axy
(ab)x
= ax
bx

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

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

Graphs of various exponential functions
x
y
y = 1x
y = 2xy = 3x
y = 10x
y = 1.5x
y = (1/2)xy = (1/3)x
y = (1/10)x
y = (2/3)x

The magic number
Deﬁnition
e = lim
n→∞
1 +
1
n
n
= lim
h→0+
(1 + h)1/h

Existence of e
See Appendix B
We can experimentally verify
that this number exists and
is
e ≈ 2.718281828459045 . . .
e is irrational
e is transcendental
n 1 +
1
n
n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106
2.71828

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
.
Facts
(i) loga(x1 · x2) = loga x1 + loga x2
(ii) loga
x1
x2
= loga x1 − loga x2
(iii) loga(xr
) = r loga x

Graphs of logarithmic functions
x
y
y = 2x
y = log2 x
(0, 1)
(1, 0)
y = 3x
y = log3 x
y = 10x
y = log10 x
y = ex
y = ln x

Change of base formula for logarithms
Fact
If a > 0 and a = 1, and the same for b, then
loga x =
logb x
logb a
Proof.
If y = loga x, then x = ay
So logb x = logb(ay
) = y logb a
Therefore
y = loga x =
logb x
logb a

Upshot of changing base
The point of the change of base formula
loga x =
logb x
logb a
=
1
logb a
· logb x = (constant) · logb x
is that all the logarithmic functions are multiples of each other. So just
pick one and call it your favorite.
Engineers like the common logarithm log = log10
Computer scientists like the binary logarithm lg = log2
Mathematicians like natural logarithm ln = loge
Naturally, we will follow the mathematicians. Just don’t pronounce it
“lawn.”

Outline
Exponential Growth
Other exponentials
Other logarithms

Derivatives of Exponential Functions
Fact
If f (x) = ax
, then f (x) = f (0)ax
.

Fact
If f (x) = ax
.
Proof.
Follow your nose:
f (x) = lim
h→0
f (x + h) − f (x)
h
= lim
h→0
ax+h − ax
h
= lim
h→0
ax ah − ax
h
= ax
· lim
h→0
ah − 1
h
= ax
· f (0).

Fact
If f (x) = ax
.
Proof.
Follow your nose:
f (x) = lim
h→0
f (x + h) − f (x)
h
= lim
h→0
ax+h − ax
h
= lim
h→0
ax ah − ax
h
= ax
· lim
h→0
ah − 1
h
= ax
· f (0).
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:
e = lim
n→∞
1 +
1
n
n
= lim
h→0
(1 + h)1/h
Question
What is lim
h→0
eh − 1
h
?

e = lim
n→∞
1 +
1
n
n
= lim
h→0
(1 + h)1/h
Question
What is lim
h→0
eh − 1
h
?
Answer
If h is small enough, e ≈ (1 + h)1/h
. So
eh − 1
h
≈
(1 + h)1/h h
− 1
h
=
(1 + h) − 1
h
=
h
h
= 1

e = lim
n→∞
1 +
1
n
n
= lim
h→0
(1 + h)1/h
Question
What is lim
h→0
eh − 1
h
?
Answer
If h is small enough, e ≈ (1 + h)1/h
. So
eh − 1
h
≈
(1 + h)1/h h
− 1
h
=
(1 + h) − 1
h
=
h
h
= 1
So in the limit we get equality: lim
h→0
eh − 1
h
= 1

From
d
dx
ax
= lim
h→0
ah − 1
h
ax
and lim
h→0
eh − 1
h
= 1
we get:
Theorem
d
dx
ex
= ex

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 derivatives of these functions:
e3x
ex2
x2
ex

Examples
Examples
e3x
ex2
x2
ex
Solution
d
dx
e3x
= 3e3x

Examples
Examples
e3x
ex2
x2
ex
Solution
d
dx
e3x
= 3e3x
d
dx
ex2
= ex2 d
dx
(x2
) = 2xex2

Examples
Examples
e3x
ex2
x2
ex
Solution
d
dx
e3x
= 3e3x
d
dx
ex2
= ex2 d
dx
(x2
) = 2xex2
d
dx
x2
ex
= 2xex
+ x2
ex

Outline
Exponential Growth
Other exponentials
Other logarithms

Let y = ln x. Then
x = ey
so

Let y = ln x. Then
x = ey
so
ey dy
dx
= 1

Let y = ln x. Then
x = ey
so
ey dy
dx
= 1
=⇒
dy
dx
=
1
ey
=
1
x

Let y = ln x. Then
x = ey
so
ey dy
dx
= 1
=⇒
dy
dx
=
1
ey
=
1
x
We have discovered:
Fact
d
dx
ln x =
1
x

Let y = ln x. Then
x = ey
so
ey dy
dx
= 1
=⇒
dy
dx
=
1
ey
=
1
x
We have discovered:
Fact
d
dx
ln x =
1
x
x
y
ln x

Let y = ln x. Then
x = ey
so
ey dy
dx
= 1
=⇒
dy
dx
=
1
ey
=
1
x
We have discovered:
Fact
d
dx
ln x =
1
x
x
y
ln x
1
x

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

The Tower of Powers
y y
x3
3x2
x2
2x1
x1
1x0
x0
0
? x−1
x−1
−1x−2
x−2
−2x−3
of one lower power
Each power function is the
derivative of another power
function, except x−1

The Tower of Powers
y y
x3
3x2
x2
2x1
x1
1x0
x0
0
ln x x−1
x−1
−1x−2
x−2
−2x−3
of one lower power
Each power function is the
derivative of another power
function, except x−1
ln x ﬁlls in this gap precisely.

Examples
Examples
ln(3x)
x ln x
ln
√
x

Examples
Example
Find
d
dx
ln(3x).

Examples
Example
Find
d
dx
ln(3x).
Solution (chain rule way)
d
dx
ln(3x) =
1
3x
· 3 =
1
x

Examples
Example
Find
d
dx
ln(3x).
d
dx
ln(3x) =
1
3x
· 3 =
1
x
Solution (properties of logarithms way)
d
dx
ln(3x) =
d
dx
(ln(3) + ln(x)) = 0 +
1
x
=
1
x

Examples
Example
Find
d
dx
ln(3x).
d
dx
ln(3x) =
1
3x
· 3 =
1
x
d
dx
ln(3x) =
d
dx
(ln(3) + ln(x)) = 0 +
1
x
=
1
x
The ﬁrst answer might be surprising until you see the second solution.

Examples
Example
Find
d
dx
x ln x

Examples
Example
Find
d
dx
x ln x
Solution
The product rule is in play here:
d
dx
x ln x =
d
dx
x ln x + x
d
dx
ln x = 1 · ln x + x ·
1
x
= ln x + 1

Examples
Example
Find
d
dx
ln
√
x.

Examples
Example
Find
d
dx
ln
√
x.
d
dx
ln
√
x =
1
√
x
d
dx
√
x =
1
√
x
1
2
√
x
=
1
2x

Examples
Example
Find
d
dx
ln
√
x.
d
dx
ln
√
x =
1
√
x
d
dx
√
x =
1
√
x
1
2
√
x
=
1
2x
d
dx
ln
√
x =
d
dx
1
2
ln x =
1
2
d
dx
ln x =
1
2
·
1
x

Examples
Example
Find
d
dx
ln
√
x.
d
dx
ln
√
x =
1
√
x
d
dx
√
x =
1
√
x
1
2
√
x
=
1
2x
d
dx
ln
√
x =
d
dx
1
2
ln x =
1
2
d
dx
ln x =
1
2
·
1
x
The ﬁrst answer might be surprising until you see the second solution.

Outline
Exponential Growth
Other exponentials
Other logarithms

Other logarithms
Example
Use implicit diﬀerentiation to ﬁnd
d
dx
ax
.

Other logarithms
Example
d
dx
ax
.
Solution
Let y = ax
, so
ln y = ln ax
= x ln a

Other logarithms
Example
d
dx
ax
.
Solution
Let y = ax
, so
ln y = ln ax
= x ln a
Diﬀerentiate implicitly:
1
y
dy
dx
= ln a =⇒
dy
dx
= (ln a)y = (ln a)ax

Other logarithms
Example
d
dx
ax
.
Solution
Let y = ax
, so
ln y = ln ax
= x ln a
Diﬀerentiate implicitly:
1
y
dy
dx
= ln a =⇒
dy
dx
= (ln a)y = (ln a)ax
Before we showed y = y (0)y, so now we know that
ln a = lim
h→0
ah − 1
h

Other logarithms
Example
Find
d
dx
loga x.

Other logarithms
Example
Find
d
dx
loga x.
Solution
Let y = loga x, so ay
= x.

Other logarithms
Example
Find
d
dx
loga x.
Solution
= x. Now diﬀerentiate implicitly:
(ln a)ay dy
dx
= 1 =⇒
dy
dx
=
1
ay ln a
=
1
x ln a

Other logarithms
Example
Find
d
dx
loga x.
Solution
= x. Now diﬀerentiate implicitly:
(ln a)ay dy
dx
= 1 =⇒
dy
dx
=
1
ay ln a
=
1
x ln a
Another way to see this is to take the natural logarithm:
ay
= x =⇒ y ln a = ln x =⇒ y =
ln x
ln a
So
dy
dx
=
1
ln a
1
x
.

More examples
Example
Find
d
dx
log2(x2
+ 1)

More examples
Example
Find
d
dx
log2(x2
+ 1)
Answer
dy
dx
=
1
ln 2
1
x2 + 1
(2x) =
2x
(ln 2)(x2 + 1)

Outline
Exponential Growth
Other exponentials
Other logarithms

A nasty derivative
Example
Let y =
(x2 + 1)
√
x + 3
x − 1
. Find y .

A nasty derivative
Example
Let y =
(x2 + 1)
√
x + 3
x − 1
. Find y .
Solution
We use the quotient rule, and the product rule in the numerator:
y =
(x − 1) 2x
√
x + 3 + (x2 + 1)1
2(x + 3)−1/2 − (x2 + 1)
√
x + 3(1)
(x − 1)2
=
2x
√
x + 3
(x − 1)
+
(x2 + 1)
2
√
x + 3(x − 1)
−
(x2 + 1)
√
x + 3
(x − 1)2

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

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

y =
2x
√
x + 3
(x − 1)
+
(x2 + 1)
2
√
x + 3(x − 1)
−
(x2 + 1)
√
x + 3
(x − 1)2
y =
2x
x2 + 1
+
1
2(x + 3)
−
1
x − 1
(x2 + 1)
√
x + 3
(x − 1)
Are these the same?

y =
2x
√
x + 3
(x − 1)
+
(x2 + 1)
2
√
x + 3(x − 1)
−
(x2 + 1)
√
x + 3
(x − 1)2
y =
2x
x2 + 1
+
1
2(x + 3)
−
1
x − 1
(x2 + 1)
√
x + 3
(x − 1)
Are these the same? Yes.

y =
2x
√
x + 3
(x − 1)
+
(x2 + 1)
2
√
x + 3(x − 1)
−
(x2 + 1)
√
x + 3
(x − 1)2
y =
2x
x2 + 1
+
1
2(x + 3)
−
1
x − 1
(x2 + 1)
√
x + 3
(x − 1)
Which do you like better?

y =
2x
√
x + 3
(x − 1)
+
(x2 + 1)
2
√
x + 3(x − 1)
−
(x2 + 1)
√
x + 3
(x − 1)2
y =
2x
x2 + 1
+
1
2(x + 3)
−
1
x − 1
(x2 + 1)
√
x + 3
(x − 1)
What kinds of expressions are well-suited for logarithmic
diﬀerentiation?

y =
2x
√
x + 3
(x − 1)
+
(x2 + 1)
2
√
x + 3(x − 1)
−
(x2 + 1)
√
x + 3
(x − 1)2
y =
2x
x2 + 1
+
1
2(x + 3)
−
1
x − 1
(x2 + 1)
√
x + 3
(x − 1)
What kinds of expressions are well-suited for logarithmic
diﬀerentiation? Products, quotients, and powers

Derivatives of powers
Question
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
x
y
1
1

Derivatives of powers
Question
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
x
y
1
1
Answer
(A) This can’t be y because xx
> 0 for all x > 0, and this function decreases at
some places
(B) This can’t be y because (ln x)xx
= 0 when x = 1, and this function does not
have a horizontal tangent at x = 1.

It’s neither! Or both?
Solution
If y = xx
, then
ln y = x ln x
1
y
dy
dx
= x ·
1
x
+ ln x = 1 + ln x
dy
dx
= (1 + ln x)xx
= xx
+ (ln x)xx

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

Solution
If y = xx
, then
ln y = x ln x
1
y
dy
dx
= x ·
1
x
+ ln x = 1 + ln x
dy
dx
= (1 + ln x)xx
= xx
+ (ln x)xx
Remarks
Each of these terms is one of the wrong answers!
y < 0 on the interval (0, e−1
)
y = 0 when x = e−1

Derivatives of power functions with any exponent
Fact (The power rule)
Let y = xr
. Then y = rxr−1
.

Derivatives of power functions with any exponent
Fact (The power rule)
Let y = xr
. Then y = rxr−1
.
Proof.
y = xr
=⇒ ln y = r ln x
Now diﬀerentiate:
1
y
dy
dx
=
r
x
=⇒
dy
dx
= r
y
x
= rxr−1

Summary
Derivatives of logarithmic and exponential functions:
y y
ex
ex
ax
(ln a) · ax
ln x
1
x
loga x
1
ln a
·
1
x
Logarithmic Diﬀerentiation can allow us to avoid the product and
quotient rules.
We are ﬁnally done with the Power Rule!

Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)

Recommended

Recommended

More Related Content

What's hot

What's hot (18)

Similar to Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)

Similar to Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides) (20)

More from Matthew Leingang

More from Matthew Leingang (20)

Recently uploaded

Recently uploaded (20)

Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)