Derivatives of Exponential and Log Functions

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
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 = 2x.y = 3x
.y = 10x
.y = 1.5x
.y = (1/2)x.y = (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
, then f′
(x) = f′
(0)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
axah − ax
h
= ax
· lim
h→0
ah − 1
h
= ax
· f′
(0).

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

. . . . . .
so
ey dy
dx
= 1

. . . . . .
so
ey dy
dx
= 1
=⇒
dy
dx
=
1
ey
=
1
x

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

. . . . . .
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

. . . . . .
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.

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

. . . . . .
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.

Derivatives of Exponential and Log Functions

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