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..
Sec on 2.4The Product and Quo ent Rules
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
February 23, 2011
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Announcements
I Quiz 2 next week on§§1.5, 1.6, 2.1, 2.2
I Midterm March 7 on allsec ons in class (coversall sec ons up to 2.5)
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Help!Free resources:
I Math Tutoring Center(CIWW 524)
I College Learning Center(schedule on Blackboard)
I TAs’ office hoursI my office hoursI each other!
![Page 4: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/4.jpg)
Objectives
I Understand and be ableto use the Product Rulefor the deriva ve of theproduct of two func ons.
I Understand and be ableto use the Quo ent Rulefor the deriva ve of thequo ent of twofunc ons.
![Page 5: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/5.jpg)
OutlineDeriva ve of a Product
Deriva onExamples
The Quo ent RuleDeriva onExamples
More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant
More on the Power RulePower Rule for Nega ve Integers
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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?
![Page 7: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/7.jpg)
Is the derivative of a product theproduct of the derivatives?
..(uv)′ = u′v′?
.(uv)′ = u′v′!
Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
So we have to be more careful.
![Page 8: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/8.jpg)
Is the derivative of a product theproduct of the derivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Try this with u = x and v = x2.
I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
So we have to be more careful.
![Page 9: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/9.jpg)
Is the derivative of a product theproduct of the derivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.
I But u′v′ = 1 · 2x = 2x.So we have to be more careful.
![Page 10: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/10.jpg)
Is the derivative of a product theproduct of the derivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
So we have to be more careful.
![Page 11: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/11.jpg)
Is the derivative of a product theproduct of the derivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
So we have to be more careful.
![Page 12: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/12.jpg)
Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.I Get a raise.
Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 13: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/13.jpg)
Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.
I Get a raise.Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 14: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/14.jpg)
Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.I Get a raise.
Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 15: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/15.jpg)
Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.I Get a raise.
Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 16: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/16.jpg)
Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.I Get a raise.
Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?
..
.∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?
![Page 17: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/17.jpg)
Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.I Get a raise.
Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?
..
.∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 18: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/18.jpg)
Money money money moneyThe answer depends on how much you work already and yourcurrent wage. Suppose you work h hours and are paid w. You get ame increase of∆h and a wage increase of∆w. Income is wagesmes hours, so
∆I = (w+∆w)(h+∆h)− whFOIL= w · h+ w ·∆h+∆w · h+∆w ·∆h− wh= w ·∆h+∆w · h+∆w ·∆h
![Page 19: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/19.jpg)
A geometric argumentDraw a box:
..w
.∆w
.
h
.
∆h
.
wh
.
w∆h
.
∆wh
.
∆w∆h
∆I = w∆h+ h∆w+∆w∆h
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A geometric argumentDraw a box:
..w
.∆w
.
h
.
∆h
.
wh
.
w∆h
.
∆wh
.
∆w∆h
∆I = w∆h+ h∆w+∆w∆h
![Page 21: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/21.jpg)
Cash flowSupose wages and hours are changing con nuously over me. Overa me interval∆t, what is the average rate of change of income?
∆I∆t
=w∆h+ h∆w+∆w∆h
∆t
= w∆h∆t
+ h∆w∆t
+∆w∆h∆t
What is the instantaneous rate of change of income?
dIdt
= lim∆t→0
∆I∆t
= wdhdt
+ hdwdt
+ 0
![Page 22: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/22.jpg)
Cash flowSupose wages and hours are changing con nuously over me. Overa me interval∆t, what is the average rate of change of income?
∆I∆t
=w∆h+ h∆w+∆w∆h
∆t
= w∆h∆t
+ h∆w∆t
+∆w∆h∆t
What is the instantaneous rate of change of income?
dIdt
= lim∆t→0
∆I∆t
= wdhdt
+ hdwdt
+ 0
![Page 23: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/23.jpg)
Eurekamen!We have discoveredTheorem (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
ddx
(uv) =dudx
· v+ udvdx
![Page 24: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/24.jpg)
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.
![Page 25: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/25.jpg)
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.
![Page 26: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/26.jpg)
Which is better?
Example
Find this deriva ve two ways: first by direct mul plica on and thenby the product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
![Page 27: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/27.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]
Solu onby direct mul plica on:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]
= −5x4 + 12x2 − 2x− 3
![Page 28: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/28.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby direct mul plica on:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]
= −5x4 + 12x2 − 2x− 3
![Page 29: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/29.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby direct mul plica on:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]= −5x4 + 12x2 − 2x− 3
![Page 30: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/30.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)= −5x4 + 12x2 − 2x− 3
![Page 31: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/31.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 32: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/32.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 33: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/33.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 34: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/34.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 35: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/35.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 36: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/36.jpg)
Which is better?Example
ddx
[(3− x2)(x3 − x+ 1)
]Solu onby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)= −5x4 + 12x2 − 2x− 3
![Page 37: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/37.jpg)
One moreExample
Findddx
x sin x.
Solu on
ddx
x sin x
=
(ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x= sin x+ x cos x
![Page 38: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/38.jpg)
One moreExample
Findddx
x sin x.
Solu on
ddx
x sin x =(
ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x= sin x+ x cos x
![Page 39: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/39.jpg)
One moreExample
Findddx
x sin x.
Solu on
ddx
x sin x =(
ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x
= sin x+ x cos x
![Page 40: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/40.jpg)
One moreExample
Findddx
x sin x.
Solu on
ddx
x sin x =(
ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x= sin x+ x cos x
![Page 41: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/41.jpg)
MnemonicLet u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
![Page 42: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/42.jpg)
Musical interlude
I jazz bandleader andsinger
I hit song “Minnie theMoocher” featuring “hide ho” chorus
I played Cur s in The BluesBrothers
Cab Calloway1907–1994
![Page 43: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/43.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′
= ((uv)w)′
..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 44: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/44.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′
= ((uv)w)′
..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 45: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/45.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 46: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/46.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′...
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 47: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/47.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′...
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 48: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/48.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′...
Apply the productrule to u and v
= (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 factoronce.
![Page 49: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/49.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′...
Apply the productrule to u and v
= (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 factoronce.
![Page 50: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/50.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 51: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/51.jpg)
Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′..
.
Apply the productrule to uv and w
= (uv)′w+ (uv)w′..
.
Apply the productrule to u and v
= (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 factoronce.
![Page 52: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/52.jpg)
OutlineDeriva ve of a Product
Deriva onExamples
The Quo ent RuleDeriva onExamples
More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant
More on the Power RulePower Rule for Nega ve Integers
![Page 53: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/53.jpg)
The Quotient RuleWhat about the deriva ve of a quo ent?
Let u and v be differen able func ons and let Q =uv. Then
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′v+ Qv′
=⇒ Q′ =u′ − Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
This is called the Quo ent Rule.
![Page 54: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/54.jpg)
The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =
uv. Then
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′v+ Qv′
=⇒ Q′ =u′ − Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
This is called the Quo ent Rule.
![Page 55: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/55.jpg)
The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =
uv. Then
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′v+ Qv′
=⇒ Q′ =u′ − Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
This is called the Quo ent Rule.
![Page 56: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/56.jpg)
The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =
uv. Then
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′v+ Qv′
=⇒ Q′ =u′ − Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
This is called the Quo ent Rule.
![Page 57: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/57.jpg)
The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =
uv. Then
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′v+ Qv′
=⇒ Q′ =u′ − Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
This is called the Quo ent Rule.
![Page 58: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/58.jpg)
The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =
uv. Then
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′v+ Qv′
=⇒ Q′ =u′ − Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
This is called the Quo ent Rule.
![Page 59: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/59.jpg)
The Quotient RuleWe have discoveredTheorem (The Quo ent Rule)
Let u and v be differen able at x, and v(x) ̸= 0. Thenuvis
differen able at x, and(uv
)′(x) =
u′(x)v(x)− u(x)v′(x)v(x)2
![Page 60: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/60.jpg)
Verifying ExampleExample
Verify the quo ent rule by compu ngddx
(x2
x
)and comparing it to
ddx
(x).
Solu on
ddx
(x2
x
)=
x ddx
(x2)− x2 d
dx (x)x2
=x · 2x− x2 · 1
x2
=x2
x2= 1 =
ddx
(x)
![Page 61: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/61.jpg)
Verifying ExampleExample
Verify the quo ent rule by compu ngddx
(x2
x
)and comparing it to
ddx
(x).
Solu on
ddx
(x2
x
)=
x ddx
(x2)− x2 d
dx (x)x2
=x · 2x− x2 · 1
x2
=x2
x2= 1 =
ddx
(x)
![Page 62: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/62.jpg)
MnemonicLet u = “hi” and v = “lo”. Then(u
v
)′=
vu′ − uv′
v2= “lo dee hi minus hi dee lo over lo lo”
![Page 63: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/63.jpg)
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
sin xx2
3.ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2.x cos x− 2 sin x
x3
3. − 2t+ 1(t2 + t+ 2)2
![Page 64: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/64.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 65: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/65.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 66: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/66.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 67: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/67.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 68: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/68.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 69: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/69.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 70: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/70.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 71: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/71.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 72: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/72.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 73: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/73.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 74: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/74.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 75: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/75.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 76: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/76.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2
= − 19(3x− 2)2
![Page 77: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/77.jpg)
Solution to first exampleSolu on
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 78: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/78.jpg)
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
sin xx2
3.ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2.x cos x− 2 sin x
x3
3. − 2t+ 1(t2 + t+ 2)2
![Page 79: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/79.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=
x2 ddx sin x− sin x d
dxx2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 80: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/80.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2
ddx sin x− sin x d
dxx2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 81: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/81.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x
− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 82: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/82.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x
ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 83: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/83.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 84: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/84.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 85: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/85.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 86: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/86.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2
cos x− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 87: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/87.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x
− 2x sin xx4
=x cos x− 2 sin x
x3
![Page 88: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/88.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x− 2x
sin xx4
=x cos x− 2 sin x
x3
![Page 89: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/89.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x− 2x sin x
x4
=x cos x− 2 sin x
x3
![Page 90: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/90.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x− 2x sin x
x4
=x cos x− 2 sin x
x3
![Page 91: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/91.jpg)
Solution to second exampleSolu on
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x− 2x sin x
x4
=x cos x− 2 sin x
x3
![Page 92: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/92.jpg)
Another way to do itFind the deriva ve with the product rule instead.
Solu onddx
sin xx2
=ddx
(sin x · x−2)
=
(ddx
sin x)· x−2 + sin x ·
(ddx
x−2)
= cos x · x−2 + sin x · (−2x−3)
= x−3 (x cos x− 2 sin x)
No ce the technique of factoring out the largest nega ve power,leaving posi ve powers.
![Page 93: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/93.jpg)
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
sin xx2
3.ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2.x cos x− 2 sin x
x3
3. − 2t+ 1(t2 + t+ 2)2
![Page 94: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/94.jpg)
Solution to third exampleSolu on
ddt
1t2 + t+ 2
=(t2 + t+ 2)(0)− (1)(2t+ 1)
(t2 + t+ 2)2
= − 2t+ 1(t2 + t+ 2)2
![Page 95: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/95.jpg)
Solution to third exampleSolu on
ddt
1t2 + t+ 2
=(t2 + t+ 2)(0)− (1)(2t+ 1)
(t2 + t+ 2)2
= − 2t+ 1(t2 + t+ 2)2
![Page 96: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/96.jpg)
Solution to third exampleSolu on
ddt
1t2 + t+ 2
=(t2 + t+ 2)(0)− (1)(2t+ 1)
(t2 + t+ 2)2
= − 2t+ 1(t2 + t+ 2)2
![Page 97: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/97.jpg)
A nice little takeawayFact
Let v be differen able at x, and v(x) ̸= 0. Then1vis differen able at
0, and (1v
)′= − v′
v2
Proof.
ddx
(1v
)=
v · ddx(1)− 1 · d
dxvv2
=v · 0− 1 · v′
v2= − v′
v2
![Page 98: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/98.jpg)
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
sin xx2
3.ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2.x cos x− 2 sin x
x3
3. − 2t+ 1(t2 + t+ 2)2
![Page 99: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/99.jpg)
OutlineDeriva ve of a Product
Deriva onExamples
The Quo ent RuleDeriva onExamples
More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant
More on the Power RulePower Rule for Nega ve Integers
![Page 100: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/100.jpg)
Derivative of TangentExample
Findddx
tan x
Solu on
ddx
tan x =ddx
(sin xcos x
)
=cos x · cos x− sin x · (− sin x)
cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 101: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/101.jpg)
Derivative of TangentExample
Findddx
tan x
Solu on
ddx
tan x =ddx
(sin xcos x
)
=cos x · cos x− sin x · (− sin x)
cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 102: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/102.jpg)
Derivative of TangentExample
Findddx
tan x
Solu on
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 103: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/103.jpg)
Derivative of TangentExample
Findddx
tan x
Solu on
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x
=1
cos2 x= sec2 x
![Page 104: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/104.jpg)
Derivative of TangentExample
Findddx
tan x
Solu on
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 105: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/105.jpg)
Derivative of TangentExample
Findddx
tan x
Solu on
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 106: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/106.jpg)
Derivative of CotangentExample
Findddx
cot x
![Page 107: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/107.jpg)
Derivative of CotangentExample
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
![Page 108: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/108.jpg)
Derivative of CotangentExample
Findddx
cot x
Solu on
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x= − csc2 x
![Page 109: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/109.jpg)
Derivative of CotangentExample
Findddx
cot x
Solu on
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x
= − 1sin2 x
= − csc2 x
![Page 110: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/110.jpg)
Derivative of CotangentExample
Findddx
cot x
Solu on
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x
= − csc2 x
![Page 111: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/111.jpg)
Derivative of CotangentExample
Findddx
cot x
Solu on
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x= − csc2 x
![Page 112: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/112.jpg)
Derivative of SecantExample
Findddx
sec x
Solu on
ddx
sec x =ddx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 113: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/113.jpg)
Derivative of SecantExample
Findddx
sec x
Solu on
ddx
sec x =ddx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 114: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/114.jpg)
Derivative of SecantExample
Findddx
sec x
Solu on
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 115: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/115.jpg)
Derivative of SecantExample
Findddx
sec x
Solu on
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 116: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/116.jpg)
Derivative of SecantExample
Findddx
sec x
Solu on
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 117: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/117.jpg)
Derivative of SecantExample
Findddx
sec x
Solu on
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 118: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/118.jpg)
Derivative of CosecantExample
Findddx
csc x
![Page 119: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/119.jpg)
Derivative of CosecantExample
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
![Page 120: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/120.jpg)
Derivative of CosecantExample
Findddx
csc x
Solu on
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 121: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/121.jpg)
Derivative of CosecantExample
Findddx
csc x
Solu on
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 122: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/122.jpg)
Derivative of CosecantExample
Findddx
csc x
Solu on
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 123: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/123.jpg)
Derivative of CosecantExample
Findddx
csc x
Solu on
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 124: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/124.jpg)
Recap: Derivatives oftrigonometric functions
y y′
sin x cos x
cos x − sin x
tan x sec2 x
cot x − csc2 x
sec x sec x tan x
csc x − csc x cot x
I Func ons come in pairs(sin/cos, tan/cot, sec/csc)
I Deriva ves of pairs followsimilar pa erns, withfunc ons andco-func ons switchedand an extra sign.
![Page 125: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/125.jpg)
OutlineDeriva ve of a Product
Deriva onExamples
The Quo ent RuleDeriva onExamples
More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant
More on the Power RulePower Rule for Nega ve Integers
![Page 126: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/126.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
![Page 127: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/127.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
![Page 128: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/128.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
![Page 129: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/129.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
![Page 130: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/130.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2= −nxn−1
x2n
= −nxn−1−2n = −nx−n−1
![Page 131: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/131.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2= −nxn−1
x2n= −nxn−1−2n
= −nx−n−1
![Page 132: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/132.jpg)
Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem
ddx
x−n = (−n)x−n−1
for posi ve integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
![Page 133: Lesson 9: The Product and Quotient Rules (slides)](https://reader034.vdocuments.mx/reader034/viewer/2022042613/548ccf74b47959f44e8b45ae/html5/thumbnails/133.jpg)
SummaryI The Product Rule: (uv)′ = u′v+ uv′
I The Quo ent Rule:(uv
)′=
vu′ − uv′
v2I Deriva ves of tangent/cotangent, secant/cosecant
ddx
tan x = sec2 xddx
sec x = sec x tan x
ddx
cot x = − csc2 xddx
csc x = − csc x cot x
I The Power Rule is true for all whole number powers, includingnega ve powers:
ddx
xn = nxn−1