The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Lesson 14: Derivatives of Logarithmic and Exponential Functions (handout)
1. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
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Notes
Sec on 3.3
Deriva ves of Logarithmic and
Exponen al Func ons
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
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Notes
Announcements
Quiz 3 next week on 2.6,
2.8, 3.1, 3.2
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Notes
Objectives
Know the deriva ves of the
exponen al func ons (with any
base)
Know the deriva ves of the
logarithmic func ons (with any
base)
Use the technique of logarithmic
differen a on to find deriva ves
of func ons involving roducts,
quo ents, and/or exponen als.
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. 1
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2. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers
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Notes
Conventions on power expressions
Let a be a posi ve real number.
If n is a posi ve whole number, then an = a · a · · · · · a
n factors
a0 = 1.
1
For any real number r, a−r = .
ar √
For any posi ve whole number n, a1/n = n a.
There is only one con nuous func on which sa sfies all of the
above. We call it the exponen al func on with base a.
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Notes
Properties of exponentials
Theorem
If a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on with
domain (−∞, ∞) and range (0, ∞). In par cular, ax > 0 for all x.
For any real numbers x and y, and posi ve numbers a and b we have
ax+y = ax ay
ax
ax−y = y (nega ve exponents mean reciprocals)
a
(ax )y = axy (frac onal exponents mean roots)
(ab)x = ax bx
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. 2
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3. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
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Notes
Graphs of exponential functions
y
y y =y/=3(1/3)x
= (1(2/x)x
2) y = (1/10y x= 10x 3x = 2x
) y= y y = 1.5x
y = 1x
. x
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Notes
The magic number
Defini on
( )n
1
e = lim 1+ = lim+ (1 + h)1/h
n→∞ n h→0
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Existence of e Notes
See Appendix B
( )n
1
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 irra onal 100 2.70481
1000 2.71692
e is transcendental
106 2.71828
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4. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
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Notes
Logarithms
Defini on
The base a logarithm loga x is the inverse of the func on ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .
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Notes
Facts about Logarithms
Facts
(i) loga (x1 · x2 ) = loga x1 + loga x2
( )
x1
(ii) loga = loga x1 − loga x2
x2
(iii) loga (xr ) = r loga x
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Notes
Graphs of logarithmic functions
y
y =x ex
y =y10y3= 2x
= x
y = log2 x
yy= log3 x
= ln x
(0, 1)
y = log10 x
.
(1, 0) x
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. 4
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5. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Change of base formula
Fact
If a > 0 and a ̸= 1, and the same for b, then
logb x
loga x =
logb a
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Notes
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 func ons are mul ples of each other. So
just pick one and call it your favorite.
Engineers like the common logarithm log = log10
Computer scien sts like the binary logarithm lg = log2
Mathema cians like natural logarithm ln = loge
Naturally, we will follow the mathema cians. Just don’t pronounce
it “lawn.”
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Notes
Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers
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. 5
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6. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
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Notes
Derivatives of Exponentials
Fact
If f(x) = ax , then f′ (x) = f′ (0)ax .
Proof.
Follow your nose:
f(x + h) − f(x) ax+h − ax
f′ (x) = lim = lim
h→0 h h→0 h
ax ah − ax ah − 1
= lim = a · lim
x
= ax · f′ (0).
h→0 h h→0 h
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Notes
The funny limit in the case of e
Ques on
eh − 1
What is lim ?
h→0 h
Solu on
( )n
1
Recall e = lim 1+ = lim (1 + h)1/h . If h is small enough,
n→∞ n h→0
e ≈ (1 + h)1/h . So
[ ]h
eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h
≈ = = =1
h h h h
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Notes
The funny limit in the case of e
Ques on
eh − 1
What is lim ?
h→0 h
Solu on
So in the limit we get equality:
eh − 1
lim =1
h→0 h
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. 6
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7. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Derivative of the natural Notes
exponential function
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
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Notes
Exponential Growth
Commonly misused term to say something grows exponen ally
It means the rate of change (deriva ve) is propor onal to the
current value
Examples: Natural popula on growth, compounded interest,
social networks
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Notes
Examples
Example
d
Find e3x .
dx
Solu on
d 3x d
e = e3x (3x) = 3e3x
dx dx
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. 7
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8. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Examples
Example
d 2
Find ex .
dx
Solu on
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Notes
Examples
Example
d
Find x2 ex .
dx
Solu on
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Notes
Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers
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. 8
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9. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Derivative ofy the natural logarithm
Let y = ln x. Then x = e so
y
y dy
e =1
dx
dy 1 1
=⇒ = = ln x
1
dx ey x
x
We have discovered: . x
Fact
d 1
ln x =
dx x
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Notes
The Tower of Powers
y y′
3
x 3x2
x2 2x1 The deriva ve of a power func on is a
power func on of one lower power
x1 1x0 Each power func on is the deriva ve of
x 0
0 another power func on, except x−1
ln x x−1 ln x fills in this gap precisely.
−1
x −1x−2
x−2 −2x−3
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Notes
Examples
Examples
Find deriva ves of these func ons:
ln(3x)
x ln x
√
ln x
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. 9
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10. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Examples
Example
d
Find ln(3x).
dx
Solu on (chain rule way)
d 1 1
ln(3x) = ·3=
dx 3x x
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Notes
Examples
Example
d
Find ln(3x).
dx
Solu on (proper es of logarithms way)
d d 1 1
ln(3x) = (ln(3) + ln(x)) = 0 + =
dx dx x x
The first answer might be surprising un l you see the second solu on.
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Notes
Examples
Example
d
Find x ln x
dx
Solu on
The product rule is in play here:
( ) ( )
d d d 1
x ln x = x ln x + x ln x = 1 · ln x + x · = ln x + 1
dx dx dx x
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11. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Examples
Example
d √
Find ln x.
dx
Solu on (chain rule way)
d √ 1 d√ 1 1 1
ln x = √ x=√ √ =
dx x dx x 2 x 2x
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Notes
Examples
Example
d √
Find ln x.
dx
Solu on (proper es of logarithms way)
( )
d √ d 1 1d 1 1
ln x = ln x = ln x = ·
dx dx 2 2 dx 2 x
The first answer might be surprising un l you see the second solu on.
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Notes
Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers
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. 11
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12. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Other logarithms
Example
d x
Use implicit differen a on to find a.
dx
Solu on
Let y = ax , so
ln y = ln ax = x ln a
Differen ate implicitly:
1 dy dy
= ln a =⇒ = (ln a)y = (ln a)ax
y dx dx
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Notes
The funny limit in the case of a
x ′ ′
Let y = e . Before we showed y = y (0)y, and now we know
y′ = (ln a)y. So
Corollary
ah − 1
lim = ln a
h→0 h
In par cular
2h − 1 3h − 1
ln 2 = lim ≈ 0.693 ln 3 = lim ≈ 1.10
h→0 h h→0 h
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Notes
Other logarithms
Example
d
Find log x.
dx a
Solu on
Let y = loga x, so ay = x. Now differen ate implicitly:
dy dy 1 1
(ln a)ay = 1 =⇒ = =
dx dx ay ln a x ln a
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. 12
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13. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
Other logarithms
Example
d
Find log x.
dx a
Solu on
Or we can use the change of base formula:
ln x dy 1 1
y= =⇒ =
ln a dx ln a x
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Notes
More examples
Example
d
Find log (x2 + 1)
dx 2
Answer
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Notes
Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers
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. 13
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14. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Notes
A nasty derivative
Example
√
(x2 + 1) x + 3
Let y = . Find y′ .
x−1
Solu on
We use the quo ent 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)
y′ = 2
(x − 1)2
√ 2
√
2x x + 3 (x + 1) (x2 + 1) x + 3
= + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
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Notes
Another way
√
(x2 + 1) x + 3
y=
x−1
1
ln y = ln(x + 1) + ln(x + 3) − ln(x − 1)
2
2
1 dy 2x 1 1
= + −
y dx x2 + 1 2(x + 3) x − 1
So
( ) √
dy 2x 1 1 (x2 + 1) x + 3
= + −
dx x2 + 1 2(x + 3) x − 1 x−1
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Notes
Compare and contrast
Using the product, quo ent, 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 differen a on:
( ) 2 √
2x 1 1 (x + 1) x + 3
y′ = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?
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. 14
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15. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
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Notes
Derivatives of powers
Ques on y
x
Let y = x . Which of these is true?
(A) Since y is a power func on,
y′ = x · xx−1 = xx .
(B) Since y is an exponen al 1
func on, y′ = (ln x) · xx
.
(C) Neither x
1
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Notes
Why not?
Answer y
′
(A) y ̸= x because x > 0 for all
x x
x > 0, and this func on
decreases at some places
(B) y′ ̸= (ln x)xx because (ln x)xx = 0
when x = 1, and this func on 1
does not have a horizontal .
tangent at x = 1. x
1
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Notes
Solu on
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. 15
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16. . V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
Derivatives of power functions Notes
with any exponent
Fact (The power rule)
Let y = xr . Then y′ = rxr−1 .
Proof.
y = xr =⇒ ln y = r ln x
Now differen ate:
1 dy r dy y
= =⇒ = r = rxr−1
y dx x dx x
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Notes
Summary
Deriva ves of y y′
Logarithmic and
Exponen al Func ons ex ex
Logarithmic
Differen a on can allow ax (ln a) · ax
us to avoid the product 1
and quo ent rules. ln x
x
We are finally done with 1 1
loga x ·
the Power Rule! ln a x
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Notes
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. 16
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