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Lesson 12 (Section 3.2)The Product and Quotient Rules
Math 1a
October 22, 2007
Announcements
I Midterm I: 10/24, 7-9pm. A-G in Hall A, H-Z in Hall C
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 ′?
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.
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.
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.
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.
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 in your hourly wages and work 5 hoursmore per week. How much extra money do you make?
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.
I Get a raise.
Say you get a 25 cent raise in your hourly wages and work 5 hoursmore per week. How much extra money do you make?
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.
I Get a raise.
Say you get a 25 cent raise in your hourly wages and work 5 hoursmore per week. How much extra money do you make?
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are your choices?
I Work longer hours.
I Get a raise.
Say you get a 25 cent raise in your hourly wages and work 5 hoursmore per week. How much extra money do you make?
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
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
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
Supose wages and hours are changing over time. How does incomechange?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
→ wdh
dt+ h
dw
dt+ 0
as t → 0.
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
Math 1a - October 22, 2007.GWBMonday, Oct 22, 2007
Page6of14
Supose wages and hours are changing over time. How does incomechange?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
→ wdh
dt+ h
dw
dt+ 0
as t → 0.
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
Math 1a - October 22, 2007.GWBMonday, Oct 22, 2007
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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).
Math 1a - October 22, 2007.GWBMonday, Oct 22, 2007
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Math 1a - October 22, 2007.GWBMonday, Oct 22, 2007
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Math 1a - October 22, 2007.GWBMonday, Oct 22, 2007
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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.
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.
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.
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
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
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
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
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
Example (Quadratic Tangent to identity function)
The curve y = ax2 + bx + c passes through the point (1, 2) and istangent to the line y = x at the point (0, 0). Find a, b, and c .
Answera = 1, b = 1, c = 0.
Example (Quadratic Tangent to identity function)
The curve y = ax2 + bx + c passes through the point (1, 2) and istangent to the line y = x at the point (0, 0). Find a, b, and c .
Answera = 1, b = 1, c = 0.
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.
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.
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.
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.
Math 1a - October 22, 2007.GWBMonday, Oct 22, 2007
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