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

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 / 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.
Notes
Notes
Notes
1
Section 3.3 : Derivs of Exp and LogV63.0121.021, Calculus I October 25, 2010

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
(negative exponents mean reciprocals)
(ax
)y
= axy
(fractional exponents mean roots)
(ab)x
= ax
bx
Notes
Notes
Notes
2

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
Notes
Notes
Notes
3

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
Notes
Notes
Notes
4

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
.
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!
Notes
Notes
Notes
5

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
?
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
Notes
Notes
Notes
6

Examples
Examples
Find derivatives of these functions:
e3x
ex2
x2
ex
Solution
d
dx
e3x
= 3e3x
Outline
Exponential Growth
Other exponentials
Other logarithms
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
Notes
Notes
Notes
7

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
The derivative of a power
function is a power function
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
.
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
Notes
Notes
Notes
8

Other logarithms
Example
Find
d
dx
loga x.
Solution
Let y = loga x, so ay
= 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)
Answer
Outline
Exponential Growth
Other exponentials
Other logarithms
Notes
Notes
Notes
9

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
Are these the same?
Which do you like better?
What kinds of expressions are well-suited for logarithmic
diﬀerentiation?
Notes
Notes
Notes
10

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.
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.
Notes
Notes
Notes
11

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

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