Date post: | 11-May-2015 |
Category: |
Technology |
Upload: | mel-anthony-pepito |
View: | 352 times |
Download: | 2 times |
Sections 3.1–3.3Derivatives of Exponential and
Logarithmic Functions
V63.0121.002.2010Su, Calculus I
New York University
June 1, 2010
Announcements
I Today: Homework 2 dueI Tomorrow: Section 3.4, reviewI Thursday: Midterm in class
. . . . . .
. . . . . .
Announcements
I Today: Homework 2 dueI Tomorrow: Section 3.4,
reviewI 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
I Know the definition of anexponential function
I Know the properties ofexponential functions
I Understand and apply thelaws of logarithms,including the change ofbase formula.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 3 / 54
. . . . . .
Objectives for Section 3.3
I Know the derivatives of theexponential functions (withany base)
I Know the derivatives of thelogarithmic functions (withany base)
I Use the technique oflogarithmic differentiationto find derivatives offunctions involving roducts,quotients, and/orexponentials.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 4 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 5 / 54
. . . . . .
Derivation of exponential functions
DefinitionIf a is a real number and n is a positive whole number, then
an = a · a · · · · · a︸ ︷︷ ︸n factors
Examples
I 23 = 2 · 2 · 2 = 8I 34 = 3 · 3 · 3 · 3 = 81I (−1)5 = (−1)(−1)(−1)(−1)(−1) = −1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 6 / 54
. . . . . .
Derivation of exponential functions
DefinitionIf a is a real number and n is a positive whole number, then
an = a · a · · · · · a︸ ︷︷ ︸n factors
Examples
I 23 = 2 · 2 · 2 = 8I 34 = 3 · 3 · 3 · 3 = 81I (−1)5 = (−1)(−1)(−1)(−1)(−1) = −1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 6 / 54
. . . . . .
FactIf a is a real number, then
I ax+y = axay
I ax−y =ax
ayI (ax)y = axy
I (ab)x = axbx
whenever all exponents are positive whole numbers.
Proof.Check for yourself:
ax+y = a · a · · · · · a︸ ︷︷ ︸x+ y factors
= a · a · · · · · a︸ ︷︷ ︸x factors
·a · a · · · · · a︸ ︷︷ ︸y factors
= axay
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 7 / 54
. . . . . .
FactIf a is a real number, then
I ax+y = axay
I ax−y =ax
ayI (ax)y = axy
I (ab)x = axbx
whenever all exponents are positive whole numbers.
Proof.Check for yourself:
ax+y = a · a · · · · · a︸ ︷︷ ︸x+ y factors
= a · a · · · · · a︸ ︷︷ ︸x factors
·a · a · · · · · a︸ ︷︷ ︸y factors
= axay
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 7 / 54
. . . . . .
Let's be conventional
I The desire that these properties remain true gives us conventionsfor ax when x is not a positive whole number.
I For example:an = an+0 !
= ana0
DefinitionIf a ̸= 0, we define a0 = 1.
I Notice 00 remains undefined (as a limit form, it’s indeterminate).
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 8 / 54
. . . . . .
Let's be conventional
I The desire that these properties remain true gives us conventionsfor ax when x is not a positive whole number.
I For example:an = an+0 !
= ana0
DefinitionIf a ̸= 0, we define a0 = 1.
I Notice 00 remains undefined (as a limit form, it’s indeterminate).
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 8 / 54
. . . . . .
Let's be conventional
I The desire that these properties remain true gives us conventionsfor ax when x is not a positive whole number.
I For example:an = an+0 !
= ana0
DefinitionIf a ̸= 0, we define a0 = 1.
I Notice 00 remains undefined (as a limit form, it’s indeterminate).
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 8 / 54
. . . . . .
Let's be conventional
I The desire that these properties remain true gives us conventionsfor ax when x is not a positive whole number.
I For example:an = an+0 !
= ana0
DefinitionIf a ̸= 0, we define a0 = 1.
I Notice 00 remains undefined (as a limit form, it’s indeterminate).
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 8 / 54
. . . . . .
Conventions for negative exponents
If n ≥ 0, we wantan · a−n !
= an+(−n) = a0 = 1
Definition
If n is a positive integer, we define a−n =1an
.
Fact
I The convention that a−n =1an
“works” for negative n as well.
I If m and n are any integers, then am−n =am
an.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 9 / 54
. . . . . .
Conventions for negative exponents
If n ≥ 0, we wantan · a−n !
= an+(−n) = a0 = 1
Definition
If n is a positive integer, we define a−n =1an
.
Fact
I The convention that a−n =1an
“works” for negative n as well.
I If m and n are any integers, then am−n =am
an.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 9 / 54
. . . . . .
Conventions for negative exponents
If n ≥ 0, we wantan · a−n !
= an+(−n) = a0 = 1
Definition
If n is a positive integer, we define a−n =1an
.
Fact
I The convention that a−n =1an
“works” for negative n as well.
I If m and n are any integers, then am−n =am
an.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 9 / 54
. . . . . .
Conventions for fractional exponents
If q is a positive integer, we want
(a1/q)q != a1 = a
DefinitionIf q is a positive integer, we define a1/q = q
√a. We must have a ≥ 0 if q
is even.
Notice that q√ap =( q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 10 / 54
. . . . . .
Conventions for fractional exponents
If q is a positive integer, we want
(a1/q)q != a1 = a
DefinitionIf q is a positive integer, we define a1/q = q
√a. We must have a ≥ 0 if q
is even.
Notice that q√ap =( q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 10 / 54
. . . . . .
Conventions for fractional exponents
If q is a positive integer, we want
(a1/q)q != a1 = a
DefinitionIf q is a positive integer, we define a1/q = q
√a. We must have a ≥ 0 if q
is even.
Notice that q√ap =( q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 10 / 54
. . . . . .
Conventions for irrational powers
I So ax is well-defined if x is rational.I What about irrational powers?
DefinitionLet a > 0. Then
ax = limr→x
r rational
ar
In other words, to approximate ax for irrational x, take r close to x butrational and compute ar.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 11 / 54
. . . . . .
Conventions for irrational powers
I So ax is well-defined if x is rational.I What about irrational powers?
DefinitionLet a > 0. Then
ax = limr→x
r rational
ar
In other words, to approximate ax for irrational x, take r close to x butrational and compute ar.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 11 / 54
. . . . . .
Conventions for irrational powers
I So ax is well-defined if x is rational.I What about irrational powers?
DefinitionLet a > 0. Then
ax = limr→x
r rational
ar
In other words, to approximate ax for irrational x, take r close to x butrational and compute ar.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 11 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 12 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 13 / 54
. . . . . .
Properties of exponential Functions.
.
TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a continuous function with domain R andrange (0,∞). In particular, ax > 0 for all x. If a,b > 0 and x, y ∈ R, then
I ax+y = axay
I ax−y =ax
ay
negative exponents mean reciprocals.
I (ax)y = axy
fractional exponents mean roots
I (ab)x = axbx
Proof.
I This is true for positive integer exponents by natural definitionI Our conventional definitions make these true for rational exponentsI Our limit definition make these for irrational exponents, too
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 14 / 54
. . . . . .
Properties of exponential Functions.
.
TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a continuous function with domain R andrange (0,∞). In particular, ax > 0 for all x. If a,b > 0 and x, y ∈ R, then
I ax+y = axay
I ax−y =ax
aynegative exponents mean reciprocals.
I (ax)y = axy
fractional exponents mean roots
I (ab)x = axbx
Proof.
I This is true for positive integer exponents by natural definitionI Our conventional definitions make these true for rational exponentsI Our limit definition make these for irrational exponents, too
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 14 / 54
. . . . . .
Properties of exponential Functions.
.
TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a continuous function with domain R andrange (0,∞). In particular, ax > 0 for all x. If a,b > 0 and x, y ∈ R, then
I ax+y = axay
I ax−y =ax
aynegative exponents mean reciprocals.
I (ax)y = axy fractional exponents mean rootsI (ab)x = axbx
Proof.
I This is true for positive integer exponents by natural definitionI Our conventional definitions make these true for rational exponentsI Our limit definition make these for irrational exponents, too
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 14 / 54
. . . . . .
Simplifying exponential expressions
Example
Simplify: 82/3
Solution
I 82/3 =3√82 =
3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Example
Simplify:√8
21/2
Answer2
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 15 / 54
. . . . . .
Simplifying exponential expressions
Example
Simplify: 82/3
Solution
I 82/3 =3√82 =
3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Example
Simplify:√8
21/2
Answer2
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 15 / 54
. . . . . .
Simplifying exponential expressions
Example
Simplify: 82/3
Solution
I 82/3 =3√82 =
3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Example
Simplify:√8
21/2
Answer2
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 15 / 54
. . . . . .
Simplifying exponential expressions
Example
Simplify: 82/3
Solution
I 82/3 =3√82 =
3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Example
Simplify:√8
21/2
Answer2
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 15 / 54
. . . . . .
Simplifying exponential expressions
Example
Simplify: 82/3
Solution
I 82/3 =3√82 =
3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Example
Simplify:√8
21/2
Answer2
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 15 / 54
. . . . . .
Limits of exponential functions
Fact (Limits of exponentialfunctions)
I If a > 1, then limx→∞
ax = ∞and lim
x→−∞ax = 0
I If 0 < a < 1, thenlimx→∞
ax = 0 andlim
x→−∞ax = ∞ . .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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 16 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 17 / 54
. . . . . .
Compounded Interest
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 18 / 54
. . . . . .
Compounded Interest
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100+ 10% = $110
I $110+ 10% = $110+ $11 = $121I $100(1.1)t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 18 / 54
. . . . . .
Compounded Interest
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121
I $100(1.1)t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 18 / 54
. . . . . .
Compounded Interest
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 18 / 54
. . . . . .
Compounded Interest: quarterly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 19 / 54
. . . . . .
Compounded Interest: quarterly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100(1.025)4 = $110.38,
not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 19 / 54
. . . . . .
Compounded Interest: quarterly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100(1.025)4 = $110.38, not $100(1.1)4!
I $100(1.025)8 = $121.84I $100(1.025)4t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 19 / 54
. . . . . .
Compounded Interest: quarterly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84
I $100(1.025)4t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 19 / 54
. . . . . .
Compounded Interest: quarterly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have
I After one year?I After two years?I after t years?
Answer
I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 19 / 54
. . . . . .
Compounded Interest: monthly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded twelve times a year. How much do you have after tyears?
Answer$100(1+ 10%/12)12t
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 20 / 54
. . . . . .
Compounded Interest: monthly
QuestionSuppose you save $100 at 10% annual interest, with interestcompounded twelve times a year. How much do you have after tyears?
Answer$100(1+ 10%/12)12t
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 20 / 54
. . . . . .
Compounded Interest: general
QuestionSuppose you save P at interest rate r, with interest compounded ntimes a year. How much do you have after t years?
Answer
B(t) = P(1+
rn
)nt
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 21 / 54
. . . . . .
Compounded Interest: general
QuestionSuppose you save P at interest rate r, with interest compounded ntimes a year. How much do you have after t years?
Answer
B(t) = P(1+
rn
)nt
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 21 / 54
. . . . . .
Compounded Interest: continuous
QuestionSuppose you save P at interest rate r, with interest compounded everyinstant. How much do you have after t years?
Answer
B(t) = limn→∞
P(1+
rn
)nt= lim
n→∞P(1+
1n
)rnt
= P[
limn→∞
(1+
1n
)n
︸ ︷︷ ︸independent of P, r, or t
]rt
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 22 / 54
. . . . . .
Compounded Interest: continuous
QuestionSuppose you save P at interest rate r, with interest compounded everyinstant. How much do you have after t years?
Answer
B(t) = limn→∞
P(1+
rn
)nt= lim
n→∞P(1+
1n
)rnt
= P[
limn→∞
(1+
1n
)n
︸ ︷︷ ︸independent of P, r, or t
]rt
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 22 / 54
. . . . . .
The magic number
Definition
e = limn→∞
(1+
1n
)n
So now continuously-compounded interest can be expressed as
B(t) = Pert.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 23 / 54
. . . . . .
The magic number
Definition
e = limn→∞
(1+
1n
)n
So now continuously-compounded interest can be expressed as
B(t) = Pert.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 23 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.25
3 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.37037
10 2.59374100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374
100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.70481
1000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692
106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrational
I e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 24 / 54
. . . . . .
Meet the Mathematician: Leonhard Euler
I Born in Switzerland, livedin Prussia (Germany) andRussia
I Eyesight trouble all his life,blind from 1766 onward
I Hundreds of contributionsto calculus, number theory,graph theory, fluidmechanics, optics, andastronomy
Leonhard Paul EulerSwiss, 1707–1783
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 25 / 54
. . . . . .
A limit.
.
Question
What is limh→0
eh − 1h
?
Answer
I If h is small enough, e ≈ (1+ h)1/h. So
eh − 1h
≈ 1
I In fact, limh→0
eh − 1h
= 1.
I This can be used to characterize e: limh→0
2h − 1h
= 0.693 · · · < 1 and
limh→0
3h − 1h
= 1.099 · · · > 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 26 / 54
. . . . . .
A limit.
.
Question
What is limh→0
eh − 1h
?
Answer
I If h is small enough, e ≈ (1+ h)1/h. So
eh − 1h
≈ 1
I In fact, limh→0
eh − 1h
= 1.
I This can be used to characterize e: limh→0
2h − 1h
= 0.693 · · · < 1 and
limh→0
3h − 1h
= 1.099 · · · > 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 26 / 54
. . . . . .
A limit.
.
Question
What is limh→0
eh − 1h
?
Answer
I If h is small enough, e ≈ (1+ h)1/h. So
eh − 1h
≈ 1
I In fact, limh→0
eh − 1h
= 1.
I This can be used to characterize e: limh→0
2h − 1h
= 0.693 · · · < 1 and
limh→0
3h − 1h
= 1.099 · · · > 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 26 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 27 / 54
. . . . . .
Logarithms
Definition
I The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x′
(ii) loga( xx′)= loga x− loga x
′
(iii) loga(xr) = r loga x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 28 / 54
. . . . . .
Logarithms
Definition
I The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x′
(ii) loga( xx′)= loga x− loga x
′
(iii) loga(xr) = r loga x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 28 / 54
. . . . . .
Logarithms
Definition
I The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x′
(ii) loga( xx′)= loga x− loga x
′
(iii) loga(xr) = r loga x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 28 / 54
. . . . . .
Logarithms
Definition
I The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x′
(ii) loga( xx′)= loga x− loga x
′
(iii) loga(xr) = r loga x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 28 / 54
. . . . . .
Logarithms convert products to sums
I Suppose y = loga x and y′ = loga x′
I Then x = ay and x′ = ay′
I So xx′ = ayay′= ay+y′
I Thereforeloga(xx
′) = y+ y′ = loga x+ loga x′
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 29 / 54
. . . . . .
Example
Write as a single logarithm: 2 ln 4− ln 3.
Solution
I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42
3
I notln 42
ln 3!
Example
Write as a single logarithm: ln34+ 4 ln 2
Answerln 12
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 30 / 54
. . . . . .
Example
Write as a single logarithm: 2 ln 4− ln 3.
Solution
I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42
3
I notln 42
ln 3!
Example
Write as a single logarithm: ln34+ 4 ln 2
Answerln 12
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 30 / 54
. . . . . .
Example
Write as a single logarithm: 2 ln 4− ln 3.
Solution
I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42
3
I notln 42
ln 3!
Example
Write as a single logarithm: ln34+ 4 ln 2
Answerln 12
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 30 / 54
. . . . . .
Example
Write as a single logarithm: 2 ln 4− ln 3.
Solution
I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42
3
I notln 42
ln 3!
Example
Write as a single logarithm: ln34+ 4 ln 2
Answerln 12
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 30 / 54
. . . . . .
..“lawn”
.
.Image credit: SelvaV63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 31 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 32 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 32 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 32 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 32 / 54
. . . . . .
Change of base formula for exponentials
FactIf a > 0 and a ̸= 1, then
loga x =ln xln a
Proof.
I If y = loga x, then x = ay
I So ln x = ln(ay) = y ln aI Therefore
y = loga x =ln xln a
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 33 / 54
. . . . . .
Change of base formula for exponentials
FactIf a > 0 and a ̸= 1, then
loga x =ln xln a
Proof.
I If y = loga x, then x = ay
I So ln x = ln(ay) = y ln aI Therefore
y = loga x =ln xln a
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 33 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 34 / 54
. . . . . .
Derivatives of Exponential Functions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Follow your nose:
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1h
= ax · f′(0).
To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 35 / 54
. . . . . .
Derivatives of Exponential Functions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Follow your nose:
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1h
= ax · f′(0).
To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 35 / 54
. . . . . .
Derivatives of Exponential Functions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Follow your nose:
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1h
= ax · f′(0).
To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 35 / 54
. . . . . .
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1+
1n
)n= lim
h→0(1+ h)1/h
Question
What is limh→0
eh − 1h
?
AnswerIf h is small enough, e ≈ (1+ h)1/h. So
eh − 1h
≈[(1+ h)1/h
]h − 1h
=(1+ h)− 1
h=
hh= 1
So in the limit we get equality: limh→0
eh − 1h
= 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 36 / 54
. . . . . .
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1+
1n
)n= lim
h→0(1+ h)1/h
Question
What is limh→0
eh − 1h
?
AnswerIf h is small enough, e ≈ (1+ h)1/h. So
eh − 1h
≈[(1+ h)1/h
]h − 1h
=(1+ h)− 1
h=
hh= 1
So in the limit we get equality: limh→0
eh − 1h
= 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 36 / 54
. . . . . .
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1+
1n
)n= lim
h→0(1+ h)1/h
Question
What is limh→0
eh − 1h
?
AnswerIf h is small enough, e ≈ (1+ h)1/h. So
eh − 1h
≈[(1+ h)1/h
]h − 1h
=(1+ h)− 1
h=
hh= 1
So in the limit we get equality: limh→0
eh − 1h
= 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 36 / 54
. . . . . .
Derivative of the natural exponential function
Fromddx
ax =(limh→0
ah − 1h
)ax and lim
h→0
eh − 1h
= 1
we get:
Theorem
ddx
ex = ex
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 37 / 54
. . . . . .
Exponential Growth
I Commonly misused term to say something grows exponentiallyI It means the rate of change (derivative) is proportional to the
current valueI Examples: Natural population growth, compounded interest,
social networks
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 38 / 54
. . . . . .
Examples
Examples
Find these derivatives:I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 39 / 54
. . . . . .
Examples
Examples
Find these derivatives:I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 39 / 54
. . . . . .
Examples
Examples
Find these derivatives:I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 39 / 54
. . . . . .
Examples
Examples
Find these derivatives:I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 39 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 40 / 54
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Then x = ey
so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 41 / 54
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Then x = ey
so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 41 / 54
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Then x = ey
so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 41 / 54
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Then x = ey
so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 41 / 54
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Then x = ey
so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 41 / 54
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Then x = ey
so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 41 / 54
. . . . . .
The Tower of Powers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a power functionof one lower power
I Each power function is thederivative of another powerfunction, except x−1
I ln x fills in this gapprecisely.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 42 / 54
. . . . . .
The Tower of Powers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? x−1
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a power functionof one lower power
I Each power function is thederivative of another powerfunction, except x−1
I ln x fills in this gapprecisely.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 42 / 54
. . . . . .
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
I The derivative of a powerfunction is a power functionof one lower power
I Each power function is thederivative of another powerfunction, except x−1
I ln x fills in this gapprecisely.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 42 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 43 / 54
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = lnax = x ln a
Differentiate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 44 / 54
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = lnax = x ln a
Differentiate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 44 / 54
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = lnax = x ln a
Differentiate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 44 / 54
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = lnax = x ln a
Differentiate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 44 / 54
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Now differentiate implicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 45 / 54
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x.
Now differentiate implicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 45 / 54
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Now differentiate implicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 45 / 54
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Now differentiate implicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 45 / 54
. . . . . .
More examples
Example
Findddx
log2(x2 + 1)
Answer
dydx
=1ln 2
1x2 + 1
(2x) =2x
(ln 2)(x2 + 1)
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 46 / 54
. . . . . .
More examples
Example
Findddx
log2(x2 + 1)
Answer
dydx
=1ln 2
1x2 + 1
(2x) =2x
(ln 2)(x2 + 1)
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 46 / 54
. . . . . .
Outline
Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function
Compound InterestThe number eA limit
Logarithmic FunctionsDerivatives of Exponential Functions
Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms
Other exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 47 / 54
. . . . . .
A nasty derivative
Example
Let y =(x2 + 1)
√x+ 3
x− 1. Find y′.
SolutionWe use the quotient rule, and the product rule in the numerator:
y′ =(x− 1)
[2x
√x+ 3+ (x2 + 1)12(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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 48 / 54
. . . . . .
A nasty derivative
Example
Let y =(x2 + 1)
√x+ 3
x− 1. Find y′.
SolutionWe use the quotient rule, and the product rule in the numerator:
y′ =(x− 1)
[2x
√x+ 3+ (x2 + 1)12(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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 48 / 54
. . . . . .
Another way
y =(x2 + 1)
√x+ 3
x− 1
ln y = ln(x2 + 1) +12ln(x+ 3)− ln(x− 1)
1ydydx
=2x
x2 + 1+
12(x+ 3)
− 1x− 1
So
dydx
=
(2x
x2 + 1+
12(x+ 3)
− 1x− 1
)y
=
(2x
x2 + 1+
12(x+ 3)
− 1x− 1
)(x2 + 1)
√x+ 3
x− 1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 49 / 54
. . . . . .
Compare and contrast
I 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
I Using logarithmic differentiation:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
x− 1
I Are these the same?I Which do you like better?I What kinds of expressions are well-suited for logarithmic
differentiation?
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 50 / 54
. . . . . .
Compare and contrast
I 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
I Using logarithmic differentiation:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
x− 1
I Are these the same?
I Which do you like better?I What kinds of expressions are well-suited for logarithmic
differentiation?
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 50 / 54
. . . . . .
Compare and contrast
I 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
I Using logarithmic differentiation:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
x− 1
I Are these the same?I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 50 / 54
. . . . . .
Compare and contrast
I 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
I Using logarithmic differentiation:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
x− 1
I Are these the same?I Which do you like better?I What kinds of expressions are well-suited for logarithmic
differentiation?
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 50 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 51 / 54
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 51 / 54
. . . . . .
It's neither! Or both?
If y = xx, then
ln y = x ln x1ydydx
= x · 1x+ ln x = 1+ ln x
dydx
= xx + (ln x)xx
Each of these terms is one of the wrong answers!
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 52 / 54
. . . . . .
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:
1ydydx
=rx
=⇒ dydx
= ryx= rxr−1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 53 / 54
. . . . . .
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:
1ydydx
=rx
=⇒ dydx
= ryx= rxr−1
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 53 / 54
. . . . . .
Summary
V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 54 / 54