Lesson 14: Derivatives of Logarithmic and Exponential Functions (handout)
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The document contains lecture notes from a Calculus I course focused on derivatives of exponential and logarithmic functions. It covers the definitions, properties, and rules for calculating derivatives of these functions, including examples and the application of logarithmic differentiation. Key concepts include the differentiation of natural exponential functions, exponential growth, and the change of base formula for logarithms.
![. V63.0121.001: Calculus I
. Sec on 3.3: Deriva ve of Exp and Log
.
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
.
.
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
.
.
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
.
.
. 6
.](https://image.slidesharecdn.com/lesson14-derivativesoflogarithmicandexponentialfunctions001handout-110318134233-phpapp01/85/Lesson-14-Derivatives-of-Logarithmic-and-Exponential-Functions-handout-6-320.jpg)
![. 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
.
.
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
.
.
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?
.
.
. 14
.](https://image.slidesharecdn.com/lesson14-derivativesoflogarithmicandexponentialfunctions001handout-110318134233-phpapp01/85/Lesson-14-Derivatives-of-Logarithmic-and-Exponential-Functions-handout-14-320.jpg)
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 . 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 . . Notes Announcements Quiz 3 next week on 2.6, 2.8, 3.1, 3.2 . . 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. . . . 1 .
- 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 . . 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. . . 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 . . . 2 .
- 3. . V63.0121.001: Calculus I . Sec on 3.3: Deriva ve of Exp and Log . 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 . . Notes The magic number Defini on ( )n 1 e = lim 1+ = lim+ (1 + h)1/h n→∞ n h→0 . . 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 . . . 3 .
- 4. . V63.0121.001: Calculus I . Sec on 3.3: Deriva ve of Exp and Log . 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 . . . 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 . . 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 . . . 4 .
- 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 . . 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.” . . 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 . . . 5 .
- 6. . V63.0121.001: Calculus I . Sec on 3.3: Deriva ve of Exp and Log . 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 . . 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 . . 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 . . . 6 .
- 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 . . 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 . . Notes Examples Example d Find e3x . dx Solu on d 3x d e = e3x (3x) = 3e3x dx dx . . . 7 .
- 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 . . Notes Examples Example d Find x2 ex . dx Solu on . . 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 . . . 8 .
- 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 . . 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 . . Notes Examples Examples Find deriva ves of these func ons: ln(3x) x ln x √ ln x . . . 9 .
- 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 . . 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. . . 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 . . . 10 .
- 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 . . 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. . . 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 . . . 11 .
- 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 . . 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 . . 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 . . . 12 .
- 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 . . Notes More examples Example d Find log (x2 + 1) dx 2 Answer . . 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 . . . 13 .
- 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 . . 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 . . 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? . . . 14 .
- 15. . V63.0121.001: Calculus I . Sec on 3.3: Deriva ve of Exp and Log . 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 . . 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 . . Notes Solu on . . . 15 .
- 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 . . 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 . . Notes . . . 16 .