Lesson 9: The Product and Quotient Rules (slides)
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Transcript of Lesson 9: The Product and Quotient Rules (slides)
..
Sec on 2.4The Product and Quo ent Rules
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
February 23, 2011
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)
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!
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.
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
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?
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.
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.
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.
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.
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.
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?
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?
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?
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?
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?
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?
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
A geometric argumentDraw a box:
..w
.∆w
.
h
.
∆h
.
wh
.
w∆h
.
∆wh
.
∆w∆h
∆I = w∆h+ h∆w+∆w∆h
A geometric argumentDraw a box:
..w
.∆w
.
h
.
∆h
.
wh
.
w∆h
.
∆wh
.
∆w∆h
∆I = w∆h+ h∆w+∆w∆h
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
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
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
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.
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.
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)
]
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
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
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
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
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
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
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
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
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
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
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
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
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
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
MnemonicLet u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
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
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
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.
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.
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.
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.
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.
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.
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
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)
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)
MnemonicLet u = “hi” and v = “lo”. Then(u
v
)′=
vu′ − uv′
v2= “lo dee hi minus hi dee lo over lo lo”
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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
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
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
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
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
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
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
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
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
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
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
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
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
Derivative of CotangentExample
Findddx
cot x
Derivative of CotangentExample
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
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
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
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
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
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
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
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
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
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
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
Derivative of CosecantExample
Findddx
csc x
Derivative of CosecantExample
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
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
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
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
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
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.
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
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
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
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
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
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
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
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
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