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# Lesson 9: The Product and Quotient Rules (handout)

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Derivatives of products and quotients

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### Lesson 9: The Product and Quotient Rules (handout)

1. 1. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Sec on 2.4 The Product and Quo ent Rules V63.0121.001: Calculus I Professor Ma hew Leingang New York University February 23, 2011 . . . Notes Announcements Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2 Midterm March 7 on all sec ons in class (covers all sec ons up to 2.5) . . Notes Help! Free resources: Math Tutoring Center (CIWW 524) College Learning Center (schedule on Blackboard) TAs’ oﬃce hours my oﬃce hours each other! . . . 1.
2. 2. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Objectives Understand and be able to use the Product Rule for the deriva ve of the product of two func ons. Understand and be able to use the Quo ent Rule for the deriva ve of the quo ent of two func ons. . . Notes Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers . . Notes Recollection and extension We have shown that if u and v are func ons, that (u + v)′ = u′ + v′ (u − v)′ = u′ − v′ What about uv? . . . 2.
3. 3. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Is the derivative of a product the Notes 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. . Notes 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? . . Notes 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 me increase of ∆h and a wage increase of ∆w. Income is wages mes hours, so ∆I = (w + ∆w)(h + ∆h) − wh FOIL = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh = w · ∆h + ∆w · h + ∆w · ∆h . . . 3.
4. 4. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h . w ∆w ∆I = w ∆h + h ∆w + ∆w ∆h . . Notes Cash ﬂow Supose wages and hours are changing con nuously over me. Over a me 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 . . Notes Eurekamen! We have discovered Theorem (The Product Rule) Let u and v be diﬀeren able at x. Then (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) in Leibniz nota on d du dv (uv) = ·v+u dx dx dx . . . 4.
5. 5. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Sanity Check Example Apply the product rule to u = x and v = x2 . Solu on (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2 This is what we get the “normal” way. . . Notes Which is better? Example Find this deriva ve two ways: ﬁrst by direct mul plica on and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx . . Notes Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by direct mul plica on: d [ ] FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx = −5x4 + 12x2 − 2x − 3 . . . 5.
6. 6. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x2 ) (x3 − x + 1) + (3 − x2 ) (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) = −5x4 + 12x2 − 2x − 3 . . Notes One more Example d Find x sin x. dx Solu on ( ) ( ) 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 . . Notes Mnemonic Let u = “hi” and v = “ho”. Then (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho” . . . 6.
7. 7. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Iterating the Product Rule Example Use the product rule to ﬁnd the deriva ve of a three-fold product uvw.product Apply the product Apply the rule to uv and w rule to u and v Solu on (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 mes, taking the deriva ve of each factor once. . . Notes Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers . . Notes The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be diﬀeren able func ons and let Q = . Then v u = Qv If Q is diﬀeren able, 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 Quo ent Rule. . . 7.
8. 8. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes The Quotient Rule We have discovered Theorem (The Quo ent Rule) u Let u and v be diﬀeren able at x, and v(x) ̸= 0. Then is v diﬀeren able at x, and ( u )′ u′ (x)v(x) − u(x)v′ (x) (x) = v v(x)2 . . Notes Verifying Example Example ( ) d x2 d Verify the quo ent rule by compu ng and comparing it to (x). dx x dx Solu on ( 2) ( ) d x x dx x2 − x2 dx (x) x · 2x − x2 · 1 d d = = dx x x2 x2 2 x d = 2 =1= (x) x dx . . Notes Mnemonic Let u = “hi” and v = “lo”. Then ( u )′ vu′ − uv′ = = “lo dee hi minus hi dee lo over lo lo” v v2 . . . 8.
9. 9. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Examples Example Answers d 2x + 5 1. dx 3x − 2 d sin x 2. dx x2 d 1 3. dt t2 + t + 2 . . Notes Solution to ﬁrst example Solu on . . Notes Solution to second example Solu on . . . 9.
10. 10. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Another way to do it Find the deriva ve with the product rule instead. Solu on No ce the technique of factoring out the largest nega ve power, leaving posi ve powers. . . Notes Solution to third example Solu on . . Notes A nice little takeaway Fact 1 Let v be diﬀeren able at x, and v(x) ̸= 0. Then is diﬀeren able at v 0, and ( )′ ′ 1 v =− 2 v v Proof. ( ) d 1 v· d dx (1)−1· d dx v v · 0 − 1 · v′ v′ = = =− 2 dx v v2 v2 v . . . 10.
11. 11. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers . . Notes Derivative of Tangent Example d Find tan x dx Solu on . . Notes Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x . . . 11.
12. 12. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Derivative of Cotangent Example d Find cot x dx Solu on . . Notes Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · = sec x tan x cos2 x cos x cos x . . Notes Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx . . . 12.
13. 13. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Derivative of Cosecant Example d Find csc x dx Solu on . . Recap: Derivatives of Notes trigonometric functions y y′ sin x cos x Func ons come in pairs (sin/cos, tan/cot, sec/csc) cos x − sin x Deriva ves of pairs follow tan x sec2 x similar pa erns, with cot x − csc2 x func ons and co-func ons switched sec x sec x tan x and an extra sign. csc x − csc x cot x . . Notes Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers . . . 13.
14. 14. . V63.0121.001: Calculus I . Sec on 2.4: Product/Quo ent Rule . February 23, 2011 Notes Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 x nxn−1 x = = − dxn 2 = − 2n = −nxn−1−2n = −nx−n−1 dx dx xn (x ) x . . Notes Summary The Product Rule: (uv)′ = u′ v + uv′ ( u )′ vu′ − uv′ The Quo ent Rule: = v v2 Deriva ves 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 nega ve powers: d n x = nxn−1 dx . . Notes . . . 14.