Lesson 16: Derivatives of Exponential and Logarithmic Functions
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Lesson 16: Derivatives of Exponential and Logarithmic Functions

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The derivative of an exponential is another exponential, but the derivative of a logarithm is a special "missing" power function.

The derivative of an exponential is another exponential, but the derivative of a logarithm is a special "missing" power function.

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    Lesson 16: Derivatives of Exponential and Logarithmic Functions Lesson 16: Derivatives of Exponential and Logarithmic Functions Presentation Transcript

    • Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121, Calculus I March 10/11, 2009 Announcements Quiz 3 this week: Covers Sections 2.1–2.4 Get half of all unearned ALEKS points by March 22 . . Image credit: heipei . . . . . .
    • Outline 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) ax+h − ax f′ (x) = lim = lim h h h→0 h→0 ax ah − ax ah − 1 = ax · f′ (0). = ax · lim = lim h h h→0 h→0 . . . . . .
    • Derivatives of Exponential Functions 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 h h→0 h→0 ax ah − ax ah − 1 = ax · f′ (0). = ax · lim = lim h h h→0 h→0 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: ( ) 1n = lim (1 + h)1/h e = lim 1 + n n→∞ h→0 Question eh − 1 What is lim ? h h→0 . . . . . .
    • The funny limit in the case of e Remember the definition of e: ( ) 1n = lim (1 + h)1/h e = lim 1 + n n→∞ h→0 Question eh − 1 What is lim ? h h→0 Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h (1 + h)1/h − 1 eh − 1 (1 + h) − 1 h ≈ = = =1 h h h h . . . . . .
    • The funny limit in the case of e Remember the definition of e: ( ) 1n = lim (1 + h)1/h e = lim 1 + n n→∞ h→0 Question eh − 1 What is lim ? h h→0 Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h (1 + h)1/h − 1 eh − 1 (1 + h) − 1 h ≈ = = =1 h h h h eh − 1 =1 So in the limit we get equality: lim h h→0 . . . . . .
    • Derivative of the natural exponential function From ( ) ah − 1 eh − 1 dx ax a= =1 lim and lim dx h h h→0 h→0 we get: Theorem dx 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 x2 ex . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x2 ex Solution d 3x e = 3ex dx . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x2 ex Solution d 3x e = 3ex dx d x2 2d 2 e = ex (x2 ) = 2xex dx dx . . . . . .
    • Examples Examples Find these derivatives: e3x 2 ex x2 ex Solution d 3x e = 3ex dx d x2 2d 2 e = ex (x2 ) = 2xex dx dx d 2x x e = 2xex + x2 ex dx . . . . . .
    • Outline 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 y . Let y = ln x. Then x = ey so dy ey=1 dx l .n x dy 1 1 =⇒ = y= dx e x . x . So: Fact d 1 ln x = dx x . . . . . .
    • Derivative of the natural logarithm function y . Let y = ln x. Then x = ey so dy ey=1 dx l .n x 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 power x3 3x2 function is a power function of one lower x2 2x1 power x1 1x0 x0 0 ? ? x−1 −1x−2 x−2 −2x−3 . . . . . .
    • The Tower of Powers y′ y The derivative of a power x3 3x2 function is a power function of one lower x2 2x1 power x1 1x0 Each power function is the derivative of another 0 x 0 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 power x3 3x2 function is a power function of one lower x2 2x1 power x1 1x0 Each power function is the derivative of another 0 x 0 power function, except x−1 x−1 ln x x−1 −1x−2 ln x fills in this gap precisely. x−2 −2x−3 . . . . . .
    • Outline 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 dx Use implicit differentiation to find a. dx . . . . . .
    • Other logarithms Example dx Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a . . . . . .
    • Other logarithms Example dx 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)y = (ln a)ax = ln a =⇒ y dx dx . . . . . .
    • Other logarithms Example dx 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)y = (ln a)ax = ln a =⇒ y dx dx Before we showed y′ = y′ (0)y, so now we know that 2h − 1 3h − 1 ≈ 0.693 ≈ 1.10 ln 2 = lim ln 3 = lim h h h→0 h→0 . . . . . .
    • Other logarithms Example d Find log x. dx a . . . . . .
    • Other logarithms Example d Find log x. dx a Solution Let y = loga x, so ay = x. . . . . . .
    • Other logarithms Example d Find log x. dx a 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 log x. dx a 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 11 = So . dx ln a x . . . . . .
    • More examples Example d log (x2 + 1) Find dx 2 . . . . . .
    • More examples Example d log (x2 + 1) Find dx 2 Answer dy 1 1 2x = (2x) = 2+1 (ln 2)(x2 + 1) dx ln 2 x . . . . . .
    • Outline 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 . Find y′ . Let y = x−1 . . . . . .
    • A nasty derivative Example √ (x2 + 1) x + 3 . Find y′ . Let 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 √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 +√ − = (x − 1) (x − 1)2 2 x + 3(x − 1) . . . . . .
    • 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 − =2 + x + 1 2(x + 3) x − 1 y dx So ( ) dy 2x 1 1 − = + y x2 + 1 2(x + 3) x − 1 dx √ ( ) (x2 + 1) x + 3 2x 1 1 − = + x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
    • Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − y= + x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
    • Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − y= + x2 + 1 2(x + 3) x − 1 x−1 Are these the same? . . . . . .
    • Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − 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: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − 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 . . . . . .