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Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)

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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.

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Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)

1. 1. Sec on 3.3 Deriva ves of Logarithmic and Exponen al Func ons V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 21, 2011.
2. 2. Announcements Quiz 3 next week on 2.6, 2.8, 3.1, 3.2
3. 3. 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.
4. 4. 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. 5. 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 0 a = 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.
6. 6. 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 x−y ax a = y a (a ) = axy x y (ab)x = ax bx
7. 7. 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 x−y ax a = y (nega ve exponents mean reciprocals) a (a ) = axy x y (ab)x = ax bx
8. 8. 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 x−y ax a = y (nega ve exponents mean reciprocals) a (a ) = axy (frac onal exponents mean roots) x y (ab)x = ax bx
9. 9. 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
10. 10. The magic number Deﬁni on ( )n 1 e = lim 1+ = lim+ (1 + h)1/h n→∞ n h→0
11. 11. Existence of eSee 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
12. 12. 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 .
13. 13. 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
14. 14. 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
15. 15. Change of base formula Fact If a > 0 and a ̸= 1, and the same for b, then logb x loga x = logb a
16. 16. 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.”
17. 17. 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
18. 18. Derivatives of Exponentials Fact If f(x) = ax , then f′ (x) = f′ (0)ax .
19. 19. 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 a a −a x h x a −1 h = lim = ax · lim = ax · f′ (0). h→0 h h→0 h
20. 20. The funny limit in the case of e Ques on eh − 1 What is lim ? h→0 h Solu on
21. 21. 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) . So 1/h [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h
22. 22. 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
23. 23. Derivative of the naturalexponential 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
24. 24. 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
25. 25. Examples Example d Find e3x . dx
26. 26. Examples Example d Find e3x . dx Solu on d 3x d e = e3x (3x) = 3e3x dx dx
27. 27. Examples Example d 2 Find ex . dx
28. 28. Examples Example d 2 Find ex . dx Solu on d x2 2 d 2 e = ex (x2 ) = 2xex dx dx
29. 29. Examples Example d Find x2 ex . dx
30. 30. Examples Example d Find x2 ex . dx Solu on d 2 x x e = 2xex + x2 ex dx
31. 31. 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
32. 32. Derivative ofy the natural logarithm Let y = ln x. Then x = e so
33. 33. Derivative ofy the natural logarithm Let y = ln x. Then x = e so dy ey =1 dx
34. 34. Derivative ofy the natural logarithm Let y = ln x. Then x = e so dy ey =1 dx dy 1 1 =⇒ = y= dx e x
35. 35. Derivative ofy the natural logarithm Let y = ln x. Then x = e so dy ey =1 dx dy 1 1 =⇒ = y= dx e x We have discovered: Fact d 1 ln x = dx x
36. 36. Derivative ofy the natural logarithm Let y = ln x. Then x = e so y dy ey =1 dx dy 1 1 =⇒ = y= ln x dx e x We have discovered: . x Fact d 1 ln x = dx x
37. 37. Derivative ofy the natural logarithm Let y = ln x. Then x = e so y dy ey =1 dx dy 1 1 =⇒ = y= ln x 1 dx e x x We have discovered: . x Fact d 1 ln x = dx x
38. 38. The Tower of Powers y y′ x3 3x2 x2 2x1 The deriva ve of a power func on is a power func on of one lower power x1 1x0 x0 0 ? ? x−1 −1x−2 x−2 −2x−3
39. 39. The Tower of Powers y y′ x3 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 x0 0 another power func on, except x−1 ? x−1 x−1 −1x−2 x−2 −2x−3
40. 40. The Tower of Powers y y′ x3 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 x0 0 another power func on, except x−1 ln x x−1 ln x ﬁlls in this gap precisely. x−1 −1x−2 x−2 −2x−3
41. 41. Examples Examples Find deriva ves of these func ons: ln(3x) x ln x √ ln x
42. 42. Examples Example d Find ln(3x). dx
43. 43. Examples Example d Find ln(3x). dx Solu on (chain rule way) d 1 1 ln(3x) = ·3= dx 3x x
44. 44. 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.
45. 45. Examples Example d Find x ln x dx
46. 46. 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
47. 47. Examples Example d √ Find ln x. dx
48. 48. 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
49. 49. 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.
50. 50. 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
51. 51. Other logarithms Example d x Use implicit diﬀeren a on to ﬁnd a. dx
52. 52. 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
53. 53. 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
54. 54. 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
55. 55. Other logarithms Example d Find log x. dx a
56. 56. Other logarithms Example d Find log x. dx a Solu on Let y = loga x, so ay = x.
57. 57. 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 =⇒ = y = dx dx a ln a x ln a
58. 58. 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
59. 59. More examples Example d Find log2 (x2 + 1) dx
60. 60. More examples Example d Find log2 (x2 + 1) dx Answer dy 1 1 2x = (2x) = dx ln 2 x2 + 1 (ln 2)(x2 + 1)
61. 61. 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
62. 62. A nasty derivative Example √ (x2 + 1) x + 3 Let y = . Find y′ . x−1
63. 63. 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 √ √ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2
64. 64. Another way √ (x2 + 1) x + 3 y= x−1
65. 65. Another way √ (x2 + 1) x + 3 y= x−1 1 ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1) 2
66. 66. Another way √ (x2 + 1) x + 3 y= x−1 1 ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1) 2 1 dy 2x 1 1 = 2 + − y dx x + 1 2(x + 3) x − 1
67. 67. Another way √ (x2 + 1) x + 3 y= x−1 1 ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1) 2 1 dy 2x 1 1 = 2 + − y dx x + 1 2(x + 3) x − 1 So ( ) dy 2x 1 1 = + − y dx x2 + 1 2(x + 3) x − 1
68. 68. Another way √ (x2 + 1) x + 3 y= x−1 1 ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1) 2 1 dy 2x 1 1 = 2 + − y dx x + 1 2(x + 3) x − 1 So ( ) √ dy 2x 1 1 (x2 + 1) x + 3 = + − dx x 2+1 2(x + 3) x − 1 x−1
69. 69. 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)
70. 70. 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?
71. 71. 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?
72. 72. 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?
73. 73. 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?
74. 74. 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?
75. 75. 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?
76. 76. 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?
77. 77. 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?
78. 78. 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?
79. 79. 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?
80. 80. 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?
81. 81. 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?
82. 82. 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? Yes.
83. 83. Derivatives of powers Ques on y Let y = xx . 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
84. 84. Derivatives of powers Ques on y Let y = xx . 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
85. 85. Why not? Answer y (A) y′ ̸= xx because xx > 0 for all x > 0, and this func on decreases at some places 1 . x 1
86. 86. Why not? Answer y (A) y′ ̸= xx because xx > 0 for all 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
87. 87. It’s neither! Solu on If y = xx , then
88. 88. It’s neither! Solu on If y = xx , then ln y = x ln x
89. 89. It’s neither! Solu on If y = xx , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x
90. 90. It’s neither! Solu on If y = xx , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = (1 + ln x)xx = xx + (ln x)xx dx
91. 91. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx 1 . x 1
92. 92. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx Remarks Each of these terms is one of the 1 wrong answers! . x 1
93. 93. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx Remarks Each of these terms is one of the 1 wrong answers! . x 1
94. 94. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx Remarks Each of these terms is one of the 1 wrong answers! . x 1
95. 95. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx Remarks Each of these terms is one of the 1 wrong answers! . x 1
96. 96. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx Remarks Each of these terms is one of the 1 wrong answers! . y′ < 0 on the interval (0, e−1 ) x 1
97. 97. Or both? Solu on d x y x = xx + (ln x)xx = (1 + ln x)xx dx Remarks Each of these terms is one of the 1 wrong answers! . y′ < 0 on the interval (0, e−1 ) x 1 y′ = 0 when x = e−1
98. 98. Derivatives of power functionswith any exponent Fact (The power rule) Let y = xr . Then y′ = rxr−1 .
99. 99. Derivatives of power functionswith 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
100. 100. 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