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

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We show the the derivative of the exponential function is itself! And the derivative of the natural logarithm function is the reciprocal function. We also show how logarithms can make complicated ...

We show the the derivative of the exponential function is itself! And the derivative of the natural logarithm function is the reciprocal function. We also show how logarithms can make complicated differentiation problems easier.

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

  • Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121.027, Calculus I October 22, 2009 . . 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 . . . . . . . View slide
  • 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 . . . . . . View slide
  • 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 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 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 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 . . . . . .