lesson 9: the product and quotient rule
DESCRIPTION
These rules allow us to differentiate the product or quotient of functions whose derivatives are themselves known.TRANSCRIPT
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Section 3.2The Product and Quotient Rules
Math 1a
February 22, 2008
Announcements
I Problem Sessions Sunday, Thursday, 7pm, SC 310
I Office hours Tuesday, Wednesday 2–4pm SC 323
I Midterm I Friday 2/29 in class (up to §3.2)
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Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More on the Power RulePower Rule for nonnegative integers by inductionPower Rule for negative integers
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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 it u′v ′?
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Is the derivative of a product the product of thederivatives?
NO!
Try this with u = x and v = x2. Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .So we have to be more careful.
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Is the derivative of a product the product of thederivatives?
NO!Try this with u = x and v = x2.
Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .So we have to be more careful.
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Is the derivative of a product the product of thederivatives?
NO!Try this with u = x and v = x2. Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .
So we have to be more careful.
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Is the derivative of a product the product of thederivatives?
NO!Try this with u = x and v = x2. Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .So we have to be more careful.
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Mmm...burgers
Say 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?
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Mmm...burgers
Say 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?
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Mmm...burgers
Say 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?
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Mmm...burgers
Say 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?
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Money money money money
The answer depends on how much you work already and yourcurrent wage. Suppose you work h hours and are paid w . You geta time increase of ∆h and a wage increase of ∆w . Income iswages times hours, so
∆I = (w + ∆w)(h + ∆h)− wh
FOIL= wh + w ∆h + ∆w h + ∆w ∆h − wh
= w ∆h + ∆w h + ∆w ∆h
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A geometric argument
Draw a box:
w ∆w
h
∆h
w h
w ∆h
∆w h
∆w ∆h
∆I = w ∆h + h ∆w + ∆w ∆h
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A geometric argument
Draw a box:
w ∆w
h
∆h
w h
w ∆h
∆w h
∆w ∆h
∆I = w ∆h + h ∆w + ∆w ∆h
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Supose wages and hours are changing continuously over time. Howdoes income change?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
SodI
dt= lim
t→0
∆I
∆t= w
dh
dt+ h
dw
dt+ 0
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
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Supose wages and hours are changing continuously over time. Howdoes income change?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
SodI
dt= lim
t→0
∆I
∆t= w
dh
dt+ h
dw
dt+ 0
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
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Supose wages and hours are changing continuously over time. Howdoes income change?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
SodI
dt= lim
t→0
∆I
∆t= w
dh
dt+ h
dw
dt+ 0
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
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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.
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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.
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Example
Find this derivative two ways: first by FOIL and then by theproduct rule:
d
dx
[(3− x2)(x3 − x + 1)
]
Solution
(i) by FOIL:
d
dx
[(3− x2)(x3 − x + 1)
] FOIL=
d
dx
[−x5 + 4x3 − x2 − 3x + 3
]= −5x4 + 12x2 − 2x − 3
(ii) by the product rule:
dy
dx=
(d
dx(3− x2)
)(x3 − x + 1) + (3− x2)
(d
dx(x3 − x + 1)
)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x − 3
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Example
Find this derivative two ways: first by FOIL and then by theproduct rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solution
(i) by FOIL:
d
dx
[(3− x2)(x3 − x + 1)
] FOIL=
d
dx
[−x5 + 4x3 − x2 − 3x + 3
]= −5x4 + 12x2 − 2x − 3
(ii) by the product rule:
dy
dx=
(d
dx(3− x2)
)(x3 − x + 1) + (3− x2)
(d
dx(x3 − x + 1)
)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x − 3
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One more
Example
Findd
dxxex
Answery ′ = ex + xex
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One more
Example
Findd
dxxex
Answery ′ = ex + xex
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Mnemonic
Let u = “ho” and v = “hi”. Then
(uv)′ = uv ′ + vu′ = “ho dee hi plus hi dee ho”
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Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More on the Power RulePower Rule for nonnegative integers by inductionPower Rule for negative integers
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The Quotient Rule
What about the derivative of a quotient?
Let u and v be differentiable and let Q =u
v. Then u = Qv . If Q
is differentiable, we have
u′ = (Qv)′ = Q ′v + Qv ′
Q ′ =u′ − Qv ′
v=
u′
v− uv ′
v2
=u′v − uv ′
v2
This is called the Quotient Rule.
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The Quotient Rule
What about the derivative of a quotient?
Let u and v be differentiable and let Q =u
v. Then u = Qv . If Q
is differentiable, we have
u′ = (Qv)′ = Q ′v + Qv ′
Q ′ =u′ − Qv ′
v=
u′
v− uv ′
v2
=u′v − uv ′
v2
This is called the Quotient Rule.
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The Quotient Rule
What about the derivative of a quotient?
Let u and v be differentiable and let Q =u
v. Then u = Qv . If Q
is differentiable, we have
u′ = (Qv)′ = Q ′v + Qv ′
Q ′ =u′ − Qv ′
v=
u′
v− uv ′
v2
=u′v − uv ′
v2
This is called the Quotient Rule.
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Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2.
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
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Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2.
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
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Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2.
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
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Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2.
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
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Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2.
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
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Mnemonic
Let u = “hi” and v = “lo”. Then(u
v
)′=
vu′ − uv ′
v2= “lo dee hi minus hi dee lo over lo lo”
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Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More on the Power RulePower Rule for nonnegative integers by inductionPower Rule for negative integers
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Power Rule for nonnegative integers by induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We have shown it to be true for n = 1.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1 =
d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
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Power Rule for nonnegative integers by induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We have shown it to be true for n = 1.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1 =
d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
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Power Rule for negative integers
Use the quotient rule to prove
Theorem
d
dxx−n = (−n)x−n−1
for positive integers n.
Proof.
d
dxx−n =
d
dx
1
xn
=xn · d
dx 1− 1 · ddx xn
x2n
=0− nxn−1
x2n= −nx−n−1
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Power Rule for negative integers
Use the quotient rule to prove
Theorem
d
dxx−n = (−n)x−n−1
for positive integers n.
Proof.
d
dxx−n =
d
dx
1
xn
=xn · d
dx 1− 1 · ddx xn
x2n
=0− nxn−1
x2n= −nx−n−1