Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 slides)

Section 3.3
Derivatives of Exponential and
Logarithmic Functions

V63.0121.041, 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

. . . . . .

V63.0121.041, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 2 / 34

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
0
a = 1.
1
For any real number r, a−r = .
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
ax−y = y
a
(a ) = axy
x y

(ab)x = ax bx

. . . . . .



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

(ab)x = ax bx

. . . . . .



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

(ab)x = ax bx

. . . . . .


Graphs of various exponential functions
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
.
. . . . . .


The magic number

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

. . . . . .


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
e is irrational 100 2.70481
1000 2.71692
e is transcendental
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
( )
x1
(ii) loga = loga x1 − loga x2
x2
(iii) loga (xr ) = r loga x
. . . . . .


Graphs of logarithmic functions
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 logarithms

Fact
If a > 0 and a ̸= 1, and the same for b, then

logb x
loga x =
logb a

Proof.

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

. . . . . .


Upshot of changing base

The point of the change of base formula

logb x 1
loga x = = · logb x = (constant) · logb x
logb a logb a

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

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



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 derivatives of these functions:
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


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
We have discovered:
Fact

d 1
ln x =
dx x

. . . . . .



y
.
so
dy
ey
=1
dx l
.n x
dy 1 1
=⇒ = y =
dx e x
. x
.
We have discovered:
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
.
We have discovered:
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


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
ah − 1
ln a = lim
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


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
.
y
.
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 . .
1
function, y′ = (ln x) · xx
(C) Neither . .
x
.
1
.

. . . . . . .


Derivatives of powers
.
y
.
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 . .
1
function, y′ = (ln x) · xx
(C) Neither . .
x
.
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 dy 1
= x · + ln x = 1 + ln x
y dx x
dy
= (1 + ln x)xx = xx + (ln x)xx
dx

. . . . . .


Solution
If y = xx , then

ln y = x ln x
1 dy 1
= x · + ln x = 1 + ln x
y dx x
dy
dx

Remarks

Each of these terms is one of the wrong answers!

. . . . . .


Solution
If y = xx , then

ln y = x ln x
1 dy 1
= x · + ln x = 1 + ln x
y dx x
dy
dx

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 dy r
=
y dx x
dy y
=⇒ = r = rxr−1
dx x

. . . . . .


Summary
Derivatives of logarithmic and exponential functions:

y y′

ex ex

ax (ln a) · ax

1
ln x
x
1 1
loga x ·
ln a x
Logarithmic Differentiation can allow us to avoid the product and
quotient rules. . . . . . .


Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 slides)

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