Lesson 13: Derivatives of Logarithmic and Exponential Functions
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Lesson 13: Derivatives of Logarithmic and Exponential Functions

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    Lesson 13: Derivatives of Logarithmic and Exponential Functions Lesson 13: Derivatives of Logarithmic and Exponential Functions Presentation Transcript

    • Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121.006/016, Calculus I March 2, 2010 Announcements Review sessions: tonight, 7:30 in CIWW 202 and 517; tomorrow, 7:00 in CIWW 109 Midterm is March 4, covering §§1.1–2.5 Recitation this week will cover §§3.1–3.2 . . . . . .
    • Announcements Review sessions: tonight, 7:30 in CIWW 202 and 517; tomorrow, 7:00 in CIWW 109 Midterm is March 4, covering §§1.1–2.5 Recitation this week will cover §§3.1–3.2 . . . . . .
    • 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 exponential functions Let a be a positive real number. If n is a positive whole number, then an = a · a · · · · · a n factors a0 = 1. 1 For any real number r, a−r = . 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 R and range (0, ∞). In particular, ax > 0 for all x. If a, b > 0 and x, y ∈ R, then ax+y = ax ay ax ax−y = y a (ax )y = axy (ab)x = ax bx . . . . . .
    • Graphs of various exponential functions y . . x . . . . . . .
    • Graphs of various exponential functions y . . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . . = 2x y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . . = 3x. = 2x y y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . . = 10x= 3x. = 2x y y . y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . . = 10x= 3x. = 2x y y . y . = 1.5x y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . . = (1/2)x y . = 10x= 3x. = 2x y y . y . = 1.5x y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . y y . = x . = (1/2)x (1/3) . = 10x= 3x. = 2x y y . y . = 1.5x y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . y y . = x . = (1/2)x (1/3) . = (1/10)x. = 10x= 3x. = 2x y y y . y . = 1.5x y . = 1x y . x . . . . . . .
    • Graphs of various exponential functions y . yy = 213)x . . = ((//2)x (1/3)x y . = . = (1/10)x. = 10x= 3x. = 2x y y y . y . = 1.5x y . = 1x y . x . . . . . . .
    • The magic number Definition ( ) 1 n e = lim 1 + = lim+ (1 + h)1/h n→∞ n h →0 . . . . . .
    • Existence of e See Appendix B ( )n 1 n 1+ n 1 2 2 2.25 . . . . . .
    • Existence of e See Appendix B ( )n 1 n 1+ n 1 2 2 2.25 3 2.37037 . . . . . .
    • Existence of e See Appendix B ( )n 1 n 1+ n 1 2 2 2.25 3 2.37037 10 2.59374 . . . . . .
    • Existence of e See Appendix B ( )n 1 n 1+ n 1 2 2 2.25 3 2.37037 10 2.59374 100 2.70481 . . . . . .
    • Existence of e See Appendix B ( )n 1 n 1+ n 1 2 2 2.25 3 2.37037 10 2.59374 100 2.70481 1000 2.71692 . . . . . .
    • Existence of e See Appendix B ( )n 1 n 1+ n 1 2 2 2.25 3 2.37037 10 2.59374 100 2.70481 1000 2.71692 106 2.71828 . . . . . .
    • Existence of e See Appendix B ( )n 1 We can experimentally n 1+ n verify that this number 1 2 exists and is 2 2.25 e ≈ 2.718281828459045 . . . 3 2.37037 10 2.59374 100 2.70481 1000 2.71692 106 2.71828 . . . . . .
    • Existence of e See Appendix B ( )n 1 We can experimentally n 1+ n verify that this number 1 2 exists and is 2 2.25 e ≈ 2.718281828459045 . . . 3 2.37037 10 2.59374 e is irrational 100 2.70481 1000 2.71692 106 2.71828 . . . . . .
    • Existence of e See Appendix B ( )n 1 We can experimentally n 1+ n verify that this number 1 2 exists and is 2 2.25 e ≈ 2.718281828459045 . . . 3 2.37037 10 2.59374 e is irrational 100 2.70481 e is transcendental 1000 2.71692 106 2.71828 . . . . . .
    • 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 . . . . . . .
    • 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 (x · x′ ) = loga x + loga x′ . . . . . .
    • 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 (x · x′ ) = loga x + loga x′ (x) (ii) loga ′ = loga x − loga x′ x . . . . . .
    • 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 (x · x′ ) = loga x + loga x′ (x) (ii) loga ′ = loga x − loga x′ x (iii) loga (xr ) = r loga x . . . . . .
    • Logarithms convert products to sums Suppose y = loga x and y′ = loga x′ ′ Then x = ay and x′ = ay ′ ′ So xx′ = ay ay = ay+y Therefore loga (xx′ ) = y + y′ = loga x + loga x′ . . . . . .
    • Graphs of logarithmic functions y . . = 2x y y . = log2 x . . 0 , 1) ( ..1, 0) . ( x . . . . . . .
    • Graphs of logarithmic functions y . . = 3x= 2x y . y y . = log2 x y . = log3 x . . 0 , 1) ( ..1, 0) . ( x . . . . . . .
    • Graphs of logarithmic functions y . . = .10x 3x= 2x y y=. y y . = log2 x y . = log3 x . . 0 , 1) ( y . = log10 x ..1, 0) . ( x . . . . . . .
    • Graphs of logarithmic functions y . . = .10=3xx 2x y xy y y. = .e = y . = log2 x y . = ln x y . = log3 x . . 0 , 1) ( y . = log10 x ..1, 0) . ( x . . . . . . .
    • Change of base formula for exponentials Fact If a > 0 and a ̸= 1, then ln x loga x = ln a . . . . . .
    • Change of base formula for exponentials Fact If a > 0 and a ̸= 1, then ln x loga x = ln a Proof. If y = loga x, then x = ay So ln x = ln(ay ) = y ln a Therefore ln x y = loga x = ln a . . . . . .
    • 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 . . . . . .
    • Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . . . . . . .
    • Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . Proof. Follow your nose: f(x + h) − f(x) a x+h − a x f′ (x) = lim = lim h →0 h h →0 h a x a h − ax a h−1 = lim = ax · lim = ax · f′ (0). h →0 h h →0 h . . . . . .
    • Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . Proof. Follow your nose: f(x + h) − f(x) a x+h − a x f′ (x) = lim = lim h →0 h h →0 h a x a h − ax a h−1 = lim = ax · lim = ax · f′ (0). h →0 h h →0 h To reiterate: the derivative of an exponential function is a constant times that function. Much different from polynomials! . . . . . .
    • The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h →0 Question eh − 1 What is lim ? h →0 h . . . . . .
    • The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h →0 Question eh − 1 What is lim ? h →0 h Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h . . . . . .
    • The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h →0 Question eh − 1 What is lim ? h →0 h Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h eh − 1 So in the limit we get equality: lim =1 h→0 h . . . . . .
    • Derivative of the natural 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 . . . . . .
    • 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 . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x 2 ex . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx d x2 2 d 2 e = ex (x2 ) = 2xex dx dx . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx d x2 2 d 2 e = ex (x2 ) = 2xex dx dx d 2 x x e = 2xex + x2 ex dx . . . . . .
    • 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 . . . . . .
    • Derivative of the natural logarithm function Let y = ln x. Then x = ey so . . . . . .
    • Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx . . . . . .
    • Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx dy 1 1 =⇒ = y = dx e x . . . . . .
    • Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx dy 1 1 =⇒ = y = dx e x So: Fact d 1 ln x = dx x . . . . . .
    • Derivative of the natural logarithm function Let y = ln x. Then y . x = ey so dy ey =1 dx .n x l dy 1 1 =⇒ = y = dx e x . x . So: Fact d 1 ln x = dx x . . . . . .
    • Derivative of the natural logarithm function Let y = ln x. Then y . x = ey so dy ey =1 dx .n x l dy 1 1 1 =⇒ = y = . dx e x x . x . So: Fact d 1 ln x = dx x . . . . . .
    • The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 x0 0 ? ? x−1 −1x−2 x−2 −2x−3 . . . . . .
    • The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 Each power function is x 0 0 the derivative of another power function, except ? x −1 x−1 x−1 −1x−2 x−2 −2x−3 . . . . . .
    • The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 Each power function is x 0 0 the derivative of another power function, except ln x x −1 x−1 x−1 −1x−2 ln x fills in this gap precisely. x−2 −2x−3 . . . . . .
    • 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 . . . . . .
    • Other logarithms Example d x Use implicit differentiation to find a. dx . . . . . .
    • Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a . . . . . .
    • Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = ln a =⇒ = (ln a)y = (ln a)ax y dx dx . . . . . .
    • Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = ln a =⇒ = (ln a)y = (ln a)ax y dx dx Before we showed y′ = y′ (0)y, so now we know that 2h − 1 3h − 1 ln 2 = lim ≈ 0.693 ln 3 = lim ≈ 1.10 h →0 h h →0 h . . . . . .
    • Other logarithms Example d Find loga x. dx . . . . . .
    • Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. . . . . . .
    • Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. Now differentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ = y = dx dx a ln a x ln a . . . . . .
    • Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. Now differentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ = y = dx dx a ln a x ln a Another way to see this is to take the natural logarithm: ln x ay = x =⇒ y ln a = ln x =⇒ y = ln a dy 1 1 So = . dx ln a x . . . . . .
    • More examples Example d Find log2 (x2 + 1) dx . . . . . .
    • More examples Example d Find log2 (x2 + 1) dx Answer dy 1 1 2x = 2+1 (2x) = dx ln 2 x (ln 2)(x2 + 1) . . . . . .
    • 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 . . . . . .
    • A nasty derivative Example √ (x2 + 1) x + 3 Let y = . Find y′ . x−1 . . . . . .
    • A nasty derivative Example √ (x2 + 1) x + 3 Let y = . Find y′ . x−1 Solution We use the quotient 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) 2 y = (x − 1)2 √ √ 2x x + 3 (x2 + 1) (x 2 + 1 ) x + 3 = + √ − (x − 1 ) 2 x + 3(x − 1) (x − 1)2 . . . . . .
    • Another way √ (x 2 + 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 ( ) √ 2x 1 1 (x2 + 1) x + 3 = + − x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
    • Compare and contrast Using the product, quotient, 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 differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
    • Compare and contrast Using the product, quotient, 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 differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 Are these the same? . . . . . .
    • Compare and contrast Using the product, quotient, 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 differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 Are these the same? Which do you like better? . . . . . .
    • Compare and contrast Using the product, quotient, 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 differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 Are these the same? Which do you like better? What kinds of expressions are well-suited for logarithmic differentiation? . . . . . .
    • 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 . . . . . .
    • 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 . . . . . .
    • It’s neither! Or both? If y = xx , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = xx + (ln x)xx dx Each of these terms is one of the wrong answers! . . . . . .
    • Derivative of arbitrary powers Fact (The power rule) Let y = xr . Then y′ = rxr−1 . . . . . . .
    • 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: 1 dy r = y dx x dy y =⇒ = r = rxr−1 dx x . . . . . .