Lesson 9: The Product and Quotient Rules
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Lesson 9: The Product and Quotient Rules

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The product rule or Leibniz rule is the rule by which we can differentiate products of functions!

The product rule or Leibniz rule is the rule by which we can differentiate products of functions!

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Lesson 9: The Product and Quotient Rules Lesson 9: The Product and Quotient Rules Presentation Transcript

  • Section 2.4 The Product and Quotient Rules V63.0121.006/016, Calculus I February 16, 2010 Announcements Quiz 2 is February 26, covering §§1.5–2.3 Midterm I is March 4, covering §§1.1–2.5 Office Hours W 1:30–2:30, R 9–10 do get-to-know-you survey by Thursday . . . . . .
  • Outline Grader’s Corner Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Positive Integers by Induction Power Rule for Negative Integers . . . . . .
  • Problem 1.5.20 sin x Use the theorems on continuity to show h(x) = is x+1 continuous. . . . . . .
  • Problem 1.5.20 sin x Use the theorems on continuity to show h(x) = is x+1 continuous. Solution By Theorem 6, f(x) = sin x and g(x) = x + 1 are continuous because f(x) is a trigonometric function and g(x) is a polynomial. f(x) By Theorem 4, part 5, h(x) = is continuous wherever g(x) g(x) ̸= 0. . . . . . .
  • Problem 1.5.20 sin x Use the theorems on continuity to show h(x) = is x+1 continuous. Solution By Theorem 6, f(x) = sin x and g(x) = x + 1 are continuous because f(x) is a trigonometric function and g(x) is a polynomial. f(x) By Theorem 4, part 5, h(x) = is continuous wherever g(x) g(x) ̸= 0. Note The function h is not a rational function. A rational function is the quotient of two polynomials. . . . . . .
  • Problem 1.6.20 x3 − 2x + 3 x3 1 − 2/x2 + 3/x3 lim = lim 2 · x→∞ 5 − 2x2 x→∞ x 5 /x 2 − 2 1 − 2/x2 + 3/x3 = lim x · lim x→∞ x→∞ 5 /x 2 − 2 Since the first factor tends to ∞ and the second factor tends to 1 − , the product tends to −∞. 2 Notes Make sure the “lim” is there in each stage Do not do arithmetic with ∞ on paper . . . . . .
  • Explanations Explanations are getting much better. Please (continue to) format your papers presentably. . . . . . .
  • Outline Grader’s Corner Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Positive Integers by Induction Power Rule for Negative Integers . . . . . .
  • Recollection and extension We have shown that if u and v are functions, that (u + v)′ = u′ + v′ (u − v)′ = u′ − v′ What about uv? . . . . . .
  • Is the derivative of a product the product of the derivatives? . . uv)′ = u′ v′ ? ( . . . . . .
  • Is the derivative of a product the product of the derivatives? . . uv)′ = u′ v′ ! ( Try this with u = x and v = x2 . . . . . . .
  • Is the derivative of a product the product of the derivatives? . . uv)′ = u′ v′ ! ( Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . . . . . . .
  • Is the derivative of a product the product of the derivatives? . . uv)′ = u′ v′ ! ( Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. . . . . . .
  • Is the derivative of a product the product of the derivatives? . . uv)′ = u′ v′ ! ( Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. So we have to be more careful. . . . . . .
  • Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? . . . . . . . .
  • Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. . . . . . . . .
  • Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. . . . . . . . .
  • Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . . . . . . . .
  • Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . I = 5 × $0..25 = $1.25? ∆ . . . . . .
  • Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . I = 5 × $0..25 = $1.25? ∆ . . . . . .
  • Money money money money The answer depends on how much you work already and your current wage. Suppose you work h hours and are paid w. You get a time increase of ∆h and a wage increase of ∆w. Income is wages times hours, so ∆I = (w + ∆w)(h + ∆h) − wh FOIL = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh = w · ∆h + ∆w · h + ∆w · ∆h . . . . . .
  • A geometric argument Draw a box: . h ∆ w . ∆h . w ∆h ∆ h . w . h . wh ∆ . w . . w ∆ . . . . . .
  • A geometric argument Draw a box: . h ∆ w . ∆h . w ∆h ∆ h . w . h . wh ∆ . w . . w ∆ ∆I = w ∆h + h ∆w + ∆w ∆h . . . . . .
  • Supose wages and hours are changing continuously over time. Over a time interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t . . . . . .
  • Supose wages and hours are changing continuously over time. Over a time interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t What is the instantaneous rate of change of income? dI ∆I dh dw = lim =w +h +0 dt ∆t→0 ∆t dt dt . . . . . .
  • Eurekamen! We have discovered Theorem (The Product Rule) Let u and v be differentiable at x. Then (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) in Leibniz notation d du dv (uv) = ·v+u dx dx dx . . . . . .
  • Example Apply the product rule to u = x and v = x2 . . . . . . .
  • Example Apply the product rule to u = x and v = x2 . Solution (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2 This is what we get the “normal” way. . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by direct multiplication: d [ ]FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by direct multiplication: d [ ]FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx = −5x4 + 12x2 − 2x − 3 . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) . . . . . .
  • Example Find this derivative two ways: first by direct multiplication and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx Solution by the product rule: ( ) ( ) dy d 2 3 2 d 3 = (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) = −5x4 + 12x2 − 2x − 3 . . . . . .
  • One more Example d Find x sin x. dx . . . . . .
  • One more Example d Find x sin x. dx Solution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx . . . . . .
  • One more Example d Find x sin x. dx Solution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x . . . . . .
  • One more Example d Find x sin x. dx Solution ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x = sin x + x cos x . . . . . .
  • Mnemonic Let u = “hi” and v = “ho”. Then (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho” . . . . . .
  • Musical interlude jazz bandleader and singer hit song “Minnie the Moocher” featuring “hi de ho” chorus played Curtis in The Blues Brothers Cab Calloway 1907–1994 . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ . . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ = ((uv)w)′ . . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . Apply the product rule to uv and w Solution (uvw)′ = ((uv)w)′ . . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . Apply the product rule to uv and w Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . Apply the product rule to u and v Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. . Apply the product rule to u and v Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′ . . . . . .
  • Iterating the Product Rule Example Use the product rule to find the derivative of a three-fold product uvw. Solution (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′ So we write down the product three times, taking the derivative of each factor once. . . . . . .
  • Outline Grader’s Corner Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Positive Integers by Induction Power Rule for Negative Integers . . . . . .
  • The Quotient Rule What about the derivative of a quotient? . . . . . .
  • The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv . . . . . .
  • The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ . . . . . .
  • The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v . . . . . .
  • The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2 . . . . . .
  • The Quotient Rule What about the derivative of a quotient? u Let u and v be differentiable functions and let Q = . Then v u = Qv If Q is differentiable, we have u′ = (Qv)′ = Q′ v + Qv′ u′ − Qv′ u′ u v′ =⇒ Q′ = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2 This is called the Quotient Rule. . . . . . .
  • Verifying Example Example ( ) d x2 Verify the quotient rule by computing and comparing it dx x d to (x). dx . . . . . .
  • Verifying Example Example ( ) d x2 Verify the quotient rule by computing and comparing it dx x d to (x). dx Solution ( ) d ( ) d d x2 x dx x2 − x2 dx (x) = dx x x2 x · 2x − x2 · 1 = x2 x 2 d = 2 =1= (x) x dx . . . . . .
  • Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x2 − 1 d t−1 3. dt t2 + t + 2 . . . . . .
  • Solution to first example d 2x + 5 dx 3x − 2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) = (3x − 2)2 . . . . . .
  • Solution to first example d d d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) 19 = 2 =− (3x − 2) (3x − 2)2 . . . . . .
  • Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x2 − 1 d t−1 3. dt t2 + t + 2 Answers 19 1. − (3x − 2)2 . . . . . .
  • Solution to second example d 2x + 1 dx x2 − 1 . . . . . .
  • Solution to second example d 2x + 1 (x2 − 1)(2) − (2x + 1)(2x) 2−1 = dx x (x2 − 1)2 . . . . . .
  • Solution to second example d 2x + 1 (x2 − 1)(2) − (2x + 1)(2x) 2−1 = dx x (x2 − 1)2 (2x2 − 2) − (4x2 + 2x) = (x2 − 1)2 ( 2 ) 2 x +x+1 =− (x2 − 1)2 . . . . . .
  • Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x2 − 1 d t−1 3. dt t2 + t + 2 Answers 19 1. − (3x − 2)2 ( ) 2 x2 + x + 1 2. − (x2 − 1)2 . . . . . .
  • Solution to third example d t−1 dt t2 + t + 2 . . . . . .
  • Solution to third example d t−1 (t2 + t + 2)(1) − (t − 1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 . . . . . .
  • Solution to third example d t−1 (t2 + t + 2)(1) − (t − 1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 (t2 + t + 2) − (2t2 − t − 1) = (t2 + t + 2)2 −t2 + 2t + 3 = 2 (t + t + 2)2 . . . . . .
  • Examples Example d 2x + 5 1. dx 3x − 2 d 2x + 1 2. dx x2 − 1 d t−1 3. dt t2 + t + 2 Answers 19 1. − (3x − 2)2 ( ) 2 x2 + x + 1 2. − (x2 − 1)2 −t2 + 2t + 3 3. 2 (t2 + t + 2) . . . . . .
  • Mnemonic Let u = “hi” and v = “lo”. Then ( u )′ vu′ − uv′ = = “lo dee hi minus hi dee lo over lo lo” v v2 . . . . . .
  • Outline Grader’s Corner Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Positive Integers by Induction Power Rule for Negative Integers . . . . . .
  • Derivative of Tangent Example d Find tan x dx . . . . . .
  • Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x tan x = dx dx cos x . . . . . .
  • Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x . . . . . .
  • Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x = cos2 x . . . . . .
  • Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = 2x = cos cos2 x . . . . . .
  • Derivative of Tangent Example d Find tan x dx Solution ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = 2x = = sec2 x cos cos2 x . . . . . .
  • Derivative of Cotangent Example d Find cot x dx . . . . . .
  • Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x . . . . . .
  • Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x . . . . . .
  • Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x = sin2 x . . . . . .
  • Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x 1 = 2 =− 2 sin x sin x . . . . . .
  • Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x Solution d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x 1 = 2 = − 2 = − csc2 x sin x sin x . . . . . .
  • Derivative of Secant Example d Find sec x dx . . . . . .
  • Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 sec x = dx dx cos x . . . . . .
  • Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x . . . . . .
  • Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x = cos2 x . . . . . .
  • Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = 2x = · cos cos x cos x . . . . . .
  • Derivative of Secant Example d Find sec x dx Solution ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = 2x = · = sec x tan x cos cos x cos x . . . . . .
  • Derivative of Cosecant Example d Find csc x dx . . . . . .
  • Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx . . . . . .
  • Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x . . . . . .
  • Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x =− 2 sin x . . . . . .
  • Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · sin x sin x sin x . . . . . .
  • Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx Solution ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · = − csc x cot x sin x sin x sin x . . . . . .
  • Recap: Derivatives of trigonometric functions y y′ Functions come in pairs sin x cos x (sin/cos, tan/cot, sec/csc) cos x − sin x Derivatives of pairs tan x sec x 2 follow similar patterns, with functions and cot x − csc2 x co-functions switched sec x sec x tan x and an extra sign. csc x − csc x cot x . . . . . .
  • Outline Grader’s Corner Derivative of a Product Derivation Examples The Quotient Rule Derivation Examples More derivatives of trigonometric functions Derivative of Tangent and Cotangent Derivative of Secant and Cosecant More on the Power Rule Power Rule for Positive Integers by Induction Power Rule for Negative Integers . . . . . .
  • Power Rule for Positive Integers by Induction Theorem Let n be a positive integer. Then d n x = nxn−1 dx . . . . . .
  • Power Rule for Positive Integers by Induction Theorem Let n be a positive integer. Then d n x = nxn−1 dx Proof. By induction on n. . . . . . .
  • Principle of Mathematical Induction . Suppose S(1) is true and S(n + 1) is true whenever . S(n) is true. Then S(n) is true for all n. . . Image credit: Kool Skatkat . . . . . .
  • Power Rule for Positive Integers by Induction Theorem Let n be a positive integer. Then d n x = nxn−1 dx Proof. By induction on n. We can show it to be true for n = 1 directly. . . . . . .
  • Power Rule for Positive Integers by Induction Theorem Let n be a positive integer. Then d n x = nxn−1 dx Proof. By induction on n. We can show it to be true for n = 1 directly. d n Suppose for some n that x = nxn−1 . Then dx d n +1 d x = (x · xn ) dx dx . . . . . .
  • Power Rule for Positive Integers by Induction Theorem Let n be a positive integer. Then d n x = nxn−1 dx Proof. By induction on n. We can show it to be true for n = 1 directly. d n Suppose for some n that x = nxn−1 . Then dx d n +1 d x = (x · xn ) dx dx ) ( ( ) d n d n = x x +x x dx dx . . . . . .
  • Power Rule for Positive Integers by Induction Theorem Let n be a positive integer. Then d n x = nxn−1 dx Proof. By induction on n. We can show it to be true for n = 1 directly. d n Suppose for some n that x = nxn−1 . Then dx d n +1 d x = (x · xn ) dx dx ) ( ( ) d n d n = x x +x x dx dx = 1 · xn + x · nxn−1 = (n + 1)xn . . . . . .
  • Power Rule for Negative Integers Use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. . . . . . .
  • Power Rule for Negative Integers Use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d −n d 1 x = dx dx xn . . . . . .
  • Power Rule for Negative Integers Use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d −n d 1 x = dx dx xn d d n xn · dx 1 − 1 · dx x = x2n . . . . . .
  • Power Rule for Negative Integers Use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d −n d 1 x = dx dx xn d d n xn · dx 1 − 1 · dx x = x2n 0 − nx n −1 = x2n . . . . . .
  • Power Rule for Negative Integers Use the quotient rule to prove Theorem d −n x = (−n)x−n−1 dx for positive integers n. Proof. d −n d 1 x = dx dx xn d d xn · dx 1 − 1 · dx xn = x2n 0 − nx n −1 = = −nx−n−1 x2n . . . . . .
  • What have we learned today? The Product Rule: (uv)′ = u′ v + uv′ ( u )′ vu′ − uv′ The Quotient Rule: = v v2 Derivatives of tangent/cotangent, secant/cosecant d d tan x = sec2 x sec x = sec x tan x dx dx d d cot x = − csc2 x csc x = − csc x cot x dx dx The Power Rule is true for all whole number powers, including negative powers: d n x = nxn−1 dx . . . . . .