1.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Notes
Section 2.4
The Product and Quotient Rules
V63.0121.041, Calculus I
New York University
October 3, 2010
Announcements
Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2
Midterm in class (covers all sections up to 2.5)
Announcements
Notes
Quiz 2 next week on §§1.5,
1.6, 2.1, 2.2
Midterm in class (covers all
sections up to 2.5)
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 2 / 40
Help!
Notes
Free resources:
Math Tutoring Center
(CIWW 524)
College Learning Center
(schedule on Blackboard)
TAs’ oﬃce hours
my oﬃce hours
each other!
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 3 / 40
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2.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Objectives
Notes
Understand and be able to
use the Product Rule for the
derivative of the product of
two functions.
Understand and be able to
use the Quotient Rule for
the derivative of the
quotient of two functions.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 4 / 40
Outline
Notes
Derivative of a Product
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 Negative Integers
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 5 / 40
Recollection and extension
Notes
We have shown that if u and v are functions, that
(u + v ) = u + v
(u − v ) = u − v
What about uv ?
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 6 / 40
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3.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Is the derivative of a product the product of the
derivatives? Notes
(uv ) = u v !
Try this with u = x and v = x 2 .
Then uv = x 3 =⇒ (uv ) = 3x 2 .
But u v = 1 · 2x = 2x.
So we have to be more careful.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 7 / 40
Mmm...burgers
Notes
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?
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 8 / 40
Money money money money
Notes
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 9 / 40
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4.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
A geometric argument
Notes
Draw a box:
∆h w ∆h ∆w ∆h
h wh ∆w h
w ∆w
∆I = w ∆h + h ∆w + ∆w ∆h
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 10 / 40
Notes
Supose wages and hours are changing continuously over time. Over a time
interval ∆t, what is the average rate of change of income?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
What is the instantaneous rate of change of income?
dI ∆I dh dw
= lim =w +h +0
dt ∆t→0 ∆t dt dt
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 11 / 40
Eurekamen!
Notes
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 12 / 40
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5.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Sanity Check
Notes
Example
Apply the product rule to u = x and v = x 2 .
Solution
(uv ) (x) = u(x)v (x) + u (x)v (x) = x · (2x) + 1 · x 2 = 3x 2
This is what we get the “normal” way.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 13 / 40
Which is better?
Notes
Example
Find this derivative two ways: ﬁrst by direct multiplication and then by the
product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d 3
= (3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)
= −5x 4 + 12x 2 − 2x − 3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 14 / 40
One more
Notes
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 15 / 40
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V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Mnemonic
Notes
Let u = “hi” and v = “ho”. Then
(uv ) = vu + uv = “ho dee hi plus hi dee ho”
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 16 / 40
Iterating the Product Rule
Notes
Example
Use the product rule to ﬁnd the derivative of a three-fold product uvw .
Apply the product rule Apply the product rule
Solution to uv and w to u and v
(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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 18 / 40
Outline
Notes
Derivative of a Product
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 Negative Integers
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 19 / 40
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7.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
The Quotient Rule
Notes
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 20 / 40
The Quotient Rule
Notes
We have discovered
Theorem (The Quotient Rule)
u
Let u and v be diﬀerentiable at x, and v (x) = 0. Then is diﬀerentiable
v
at x, and
u u (x)v (x) − u(x)v (x)
(x) =
v v (x)2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 21 / 40
Verifying Example
Notes
Example
d x2
Verify the quotient rule by computing and comparing it to
dx x
d
(x).
dx
Solution
d d
d x2 x dx x 2 − x 2 dx (x)
=
dx x x2
x · 2x − x 2 · 1
=
x2
x2 d
= 2 =1= (x)
x dx
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 22 / 40
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8.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Mnemonic
Notes
Let u = “hi” and v = “lo”. Then
u vu − uv
= = “lo dee hi minus hi dee lo over lo lo”
v v2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 23 / 40
Examples
Notes
Example Answers
d 2x + 5
1.
dx 3x − 2
d sin x
2.
dx x 2
d 1
3.
dt t 2 + t + 2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 24 / 40
Solution to ﬁrst example
Notes
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 25 / 40
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9.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Solution to second example
Notes
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 27 / 40
Another way to do it
Notes
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 28 / 40
Solution to third example
Notes
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 30 / 40
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10.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Outline
Notes
Derivative of a Product
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 Negative Integers
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 32 / 40
Derivative of Tangent
Notes
Example
d
Find tan x
dx
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 33 / 40
Derivative of Cotangent
Notes
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 34 / 40
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V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Derivative of Secant
Notes
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
= = · = sec x tan x
cos2 x cos x cos x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 35 / 40
Derivative of Cosecant
Notes
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
Solution
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 36 / 40
Recap: Derivatives of trigonometric functions
Notes
y y
sin x cos x
Functions come in pairs
cos x − sin x (sin/cos, tan/cot, sec/csc)
tan x sec2 x Derivatives of pairs follow
similar patterns, with
cot x − csc2 x functions and co-functions
sec x sec x tan x switched and an extra sign.
csc x − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 37 / 40
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12.
V63.0121.041, Calculus I Section 2.4 : The Product and Quotient Rules October 3, 2010
Outline
Notes
Derivative of a Product
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 Negative Integers
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 38 / 40
Power Rule for Negative Integers
Notes
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 x n
d d
x n · dx 1 − 1 · dx x n
=
x 2n
0 − nx n−1
= = −nx −n−1
x 2n
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 39 / 40
Summary
Notes
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 = nx n−1
dx
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 40 / 40
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