This document contains lecture notes from a Calculus I class covering the product and quotient rules for finding derivatives. It announces that there will be a quiz on previous sections and an upcoming midterm exam. It provides objectives for the section, which are to understand and apply the product rule for derivatives of products and the quotient rule for derivatives of quotients. The notes then cover derivation and examples of applying these rules.
Lesson 9: The Product and Quotient Rules (Section 21 slides)Matthew Leingang
The derivative of a sum of functions is the sum of the derivatives of those functions, but the derivative of a product or a quotient of functions is not so simple. We'll derive and use the product and quotient rule for these purposes. It will allow us to find the derivatives of other trigonometric functions, and derivatives of power functions with negative whole number exponents.
Lesson 9: The Product and Quotient Rules (Section 21 slides)Mel Anthony Pepito
The derivative of a sum of functions is the sum of the derivatives of those functions, but the derivative of a product or a quotient of functions is not so simple. We'll derive and use the product and quotient rule for these purposes. It will allow us to find the derivatives of other trigonometric functions, and derivatives of power functions with negative whole number exponents.
Lesson 9: The Product and Quotient Rules (Section 21 slides)Matthew Leingang
The derivative of a sum of functions is the sum of the derivatives of those functions, but the derivative of a product or a quotient of functions is not so simple. We'll derive and use the product and quotient rule for these purposes. It will allow us to find the derivatives of other trigonometric functions, and derivatives of power functions with negative whole number exponents.
Lesson 9: The Product and Quotient Rules (Section 21 slides)Mel Anthony Pepito
The derivative of a sum of functions is the sum of the derivatives of those functions, but the derivative of a product or a quotient of functions is not so simple. We'll derive and use the product and quotient rule for these purposes. It will allow us to find the derivatives of other trigonometric functions, and derivatives of power functions with negative whole number exponents.
Lesson 22: Optimization II (Section 041 handout)Matthew Leingang
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 22: Optimization II (Section 041 handout)Matthew Leingang
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Similar to Lesson 9: The Product and Quotient Rules (Section 21 handout (15)
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 9: The Product and Quotient Rules (Section 21 handout
1. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
Notes
Section 2.4
The Product and Quotient Rules
V63.0121.021, Calculus I
New York University
October 4, 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)
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 2 / 41
Help!
Notes
Free resources:
Math Tutoring Center
(CIWW 524)
College Learning Center
(schedule on Blackboard)
TAs’ office hours
my office hours
each other!
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 3 / 41
1
2. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 4 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 5 / 41
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 ?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 6 / 41
2
3. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 7 / 41
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?
∆I = 5 × $0.25 = $1.25?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 8 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 9 / 41
3
4. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 10 / 41
Cash flow
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 11 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 12 / 41
4
5. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 13 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 15 / 41
5
6. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
Mnemonic
Notes
Let u = “hi” and v = “ho”. Then
(uv ) = vu + uv = “ho dee hi plus hi dee ho”
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 16 / 41
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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 18 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 19 / 41
6
7. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 21 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 22 / 41
7
8. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 23 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 24 / 41
Solution to first example
Notes
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 25 / 41
8
9. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
Solution to second example
Notes
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
Another way to do it
Notes
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 28 / 41
Solution to third example
Notes
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 30 / 41
9
10. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
A nice little takeaway
Notes
Fact
1
Let v be differentiable at x, and v (x) = 0. Then is differentiable at 0,
v
and
1 v
=− 2
v v
Proof.
d d
d 1 v· dx (1)−1· dx v v ·0−1·v v
= = =− 2
dx v v2 v2 v
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 31 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 33 / 41
Derivative of Tangent
Notes
Example
d
Find tan x
dx
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
10
11. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
Derivative of Cotangent
Notes
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
Derivative of Cosecant
Notes
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
Solution
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
11
12. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
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
2
similar patterns, with
cot x − csc x functions and co-functions
sec x sec x tan x switched and an extra sign.
csc x − csc x cot x
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 38 / 41
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 39 / 41
Power Rule for Negative Integers
Notes
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x −n−1
dx
for positive integers n.
Proof.
d n
d −n d 1 x
x = = − dxn 2
dx dx x n (x )
nx n−1
= − 2n = −nx n−1−2n = −nx −n−1
x
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
12
13. V63.0121.021, Calculus I Section 2.4 : The Product and Quotient Rules October 4, 2010
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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 41 / 41
Notes
Notes
13