Introduction to ArtificiaI Intelligence in Higher Education
Lesson 9: The Product and Quotient Rule
1. Section 2.4
The Product and Quotient Rules
V63.0121.034, Calculus I
September 30, 2009
Announcements
Quiz 2 is next week, covering §§1.4–2.1
Midterm I is October 14, covering §§1.1–2.4 (today)
Office Hours today 2:30–3:30, check website for current
. . . . . .
2. Outline
The Product Rule
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .
13. 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?
. .
. . . . . .
14. 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?
∆
. . . . . .
15. 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?
∆
. . . . . .
16. 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 time increase of ∆h and a wage increase of ∆w. Income is
wages times hours, so
∆I = (w + ∆w)(h + ∆h) − wh
FOIL
= w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh
= w · ∆h + ∆ w · h + ∆ w · ∆h
. . . . . .
17. A geometric argument
Draw a box:
. h
∆ w
. ∆h . w ∆h
∆
h
. w
. h . wh
∆
.
w
. . w
∆
. . . . . .
18. A geometric argument
Draw a box:
. h
∆ w
. ∆h . w ∆h
∆
h
. w
. h . wh
∆
.
w
. . w
∆
∆I = w ∆h + h ∆w + ∆ w ∆h
. . . . . .
21. Eurekamen!
We have discovered
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′ (x) = u(x)v′ (x) + u′ (x)v(x)
in Leibniz notation
d du dv
(uv) = ·v+u
dx dx dx
. . . . . .
23. 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.
. . . . . .
35. One more
Example
d
Find x sin x.
dx
Solution
( ) ( )
d d d
x sin x = x sin x + x sin x
dx dx dx
. . . . . .
36. One more
Example
d
Find x sin x.
dx
Solution
( ) ( )
d d d
x sin x = x sin x + x sin x
dx dx dx
= 1 · sin x + x · cos x
. . . . . .
37. One more
Example
d
Find x sin x.
dx
Solution
( ) ( )
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
. . . . . .
38. Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
. . . . . .
39. Musical interlude
jazz bandleader and
singer
hit song “Minnie the
Moocher” featuring “hi
de ho” chorus
played Curtis in The
Blues Brothers
Cab Calloway
1907–1994
. . . . . .
40. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
. . . . . .
41. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
Solution
(uvw)′ .
. . . . . .
42. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
Solution
(uvw)′ = ((uv)w)′ .
. . . . . .
43. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw. .
Apply the product rule
to uv and w
Solution
(uvw)′ = ((uv)w)′ .
. . . . . .
44. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw. .
Apply the product rule
to uv and w
Solution
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
. . . . . .
45. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw. .
Apply the product rule
to u and v
Solution
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
. . . . . .
46. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw. .
Apply the product rule
to u and v
Solution
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
. . . . . .
47. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
Solution
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
= u′ vw + uv′ w + uvw′
. . . . . .
48. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
Solution
(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 times, taking the derivative
of each factor once.
. . . . . .
49. Outline
The Product Rule
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .
53. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, we have
u′ = (Qv)′ = Q′ v + Qv′
u′ − Qv′ u′ u v′
=⇒ Q′ = = − ·
v v v v
. . . . . .
54. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, we have
u′ = (Qv)′ = Q′ v + Qv′
u′ − Qv′ u′ u v′
=⇒ Q′ = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2
. . . . . .
55. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, 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 Quotient Rule.
. . . . . .
56. Verifying Example
Example ( )
d x2
Verify the quotient rule by computing and comparing it
dx x
d
to (x).
dx
. . . . . .
57. Verifying Example
Example ( )
d x2
Verify the quotient rule by computing and comparing it
dx x
d
to (x).
dx
Solution
( ) d
( ) d
d x2 x dx x2 − x2 dx (x)
=
dx x x2
x · 2x − x2 · 1
=
x2
x 2 d
= 2 =1= (x)
x dx
. . . . . .
58. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x2 − 1
d t−1
3.
dt t2 + t + 2
. . . . . .
79. Solution to third example
d t−1 (t2 + t + 2)(1) − (t − 1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
. . . . . .
80. Solution to third example
d t−1 (t2 + t + 2)(1) − (t − 1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
(t2 + t + 2) − (2t2 − t − 1)
=
(t2 + t + 2)2
−t2 + 2t + 3
= 2
(t + t + 2)2
. . . . . .
81. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x2 − 1
d t−1
3.
dt t2 + t + 2
Answers
19
1. −
(3x − 2)2
( )
2 x2 + x + 1
2. −
(x2 − 1)2
−t2 + 2t + 3
3. 2
(t2 + t + 2)
. . . . . .
82. Mnemonic
Let u = “hi” and v = “lo”. Then
( u )′ vu′ − uv′
= = “lo dee hi minus hi dee lo over lo lo”
v v2
. . . . . .
83. Outline
The Product Rule
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .
85. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x
tan x =
dx dx cos x
. . . . . .
86. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
. . . . . .
87. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x
=
cos2 x
. . . . . .
88. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x 1
= 2x
=
cos cos2 x
. . . . . .
89. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x 1
= 2x
= = sec2 x
cos cos2 x
. . . . . .
93. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1
sec x =
dx dx cos x
. . . . . .
94. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
. . . . . .
95. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x
=
cos2 x
. . . . . .
96. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x 1 sin x
= 2x
= ·
cos cos x cos x
. . . . . .
97. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x 1 sin x
= 2x
= · = sec x tan x
cos cos x cos x
. . . . . .
99. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
. . . . . .
100. Recap: Derivatives of trigonometric functions
y y′
Functions come in pairs
sin x cos x
(sin/cos, tan/cot, sec/csc)
cos x − sin x Derivatives of pairs
tan x sec x 2 follow similar patterns,
with functions and
cot x − csc2 x co-functions switched
sec x sec x tan x and an extra sign.
csc x − csc x cot x
. . . . . .
101. Outline
The Product Rule
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .
104. Principle of Mathematical Induction
.
Suppose S(1) is
true and S(n + 1)
is true whenever
.
S(n) is true. Then
S(n) is true for all
n.
.
.
Image credit: Kool Skatkat
. . . . . .
105. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
d n
x = nxn−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
. . . . . .
106. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
d n
x = nxn−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
d n
Suppose for some n that x = nxn−1 . Then
dx
d n +1 d
x = (x · xn )
dx dx
. . . . . .
107. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
d n
x = nxn−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
d n
Suppose for some n that x = nxn−1 . Then
dx
d n +1 d
x = (x · xn )
dx dx )
( ( )
d n d n
= x x +x x
dx dx
. . . . . .
108. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
d n
x = nxn−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
d n
Suppose for some n that x = nxn−1 . Then
dx
d n +1 d
x = (x · xn )
dx dx )
( ( )
d n d n
= x x +x x
dx dx
= 1 · xn + x · nxn−1 = (n + 1)xn
. . . . . .
109. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
. . . . . .
110. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d −n d 1
x =
dx dx xn
. . . . . .
111. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d −n d 1
x =
dx dx xn
d d n
xn · dx 1 − 1 · dx x
=
x2n
. . . . . .
112. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d −n d 1
x =
dx dx xn
d d n
xn · dx 1 − 1 · dx x
=
x2n
0 − nx n −1
=
x2n
. . . . . .
113. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d −n d 1
x =
dx dx xn
d d
xn · dx 1 − 1 · dx xn
=
x2n
0 − nx n −1
= = −nx−n−1
x2n
. . . . . .
114. What have we learned today?
The Product Rule: (uv)′ = u′ v + uv′
( u )′ vu′ − uv′
The Quotient Rule: =
v v2
Derivatives 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 negative powers:
d n
x = nxn−1
dx
. . . . . .