Lesson 9: The Product and Quotient Rules (Section 41 slides)

Section 2.4
The Product and Quotient Rules

V63.0121.041, 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

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 4, 2010 2 / 41

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. . . . . .


Objectives

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.

. . . . . .


Outline

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

. . . . . .


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 the derivative of a product the product of the
derivatives?

. uv)′ = u′ v′ ?
( .

. . . . . .


derivatives?

. uv)′ = u′ v′ !
( .

Try this with u = x and v = x2 .

. . . . . .


derivatives?

. uv)′ = u′ v′ !
( .

Then uv = x3 =⇒ (uv)′ = 3x2 .

. . . . . .


derivatives?

. uv)′ = u′ v′ !
( .

Then uv = x3 =⇒ (uv)′ = 3x2 .
But u′ v′ = 1 · 2x = 2x.

. . . . . .


derivatives?

. uv)′ = u′ v′ !
( .

Then uv = x3 =⇒ (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?

. .

. . . . . .


Mmm...burgers


Work longer hours.

. .

. . . . . .


Mmm...burgers


Work longer hours.
Get a raise.

. .

. . . . . .


Mmm...burgers


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?

. .

. . . . . .


Mmm...burgers


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?
∆

. . . . . .


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

. . . . . .


A geometric argument

Draw a box:

. h
∆ w
. ∆h . w ∆h
∆

h
. w
. h . wh
∆

.
w
. . w
∆

. . . . . .


A geometric argument

Draw a box:

. h
∆ w
. ∆h . w ∆h
∆

h
. w
. h . wh
∆

.
w
. . w
∆

∆I = w ∆h + h ∆w + ∆w ∆h

. . . . . .


Cash flow

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

. . . . . .


Cash flow

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

. . . . . .


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

. . . . . .


Sanity Check

Example
Apply the product rule to u = x and v = x2 .

. . . . . .


Sanity Check

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.

. . . . . .


Which is better?

Example
Find this derivative two ways: first by direct multiplication and then by
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx

. . . . . .


Which is better?

Example
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx

Solution
by direct multiplication:

d [ ]
FOIL d
[ ]
(3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3
dx dx

. . . . . .


Which is better?

Example
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx

Solution
by direct multiplication:

d [ ]
FOIL d
[ ]
(3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3
dx dx
= −5x4 + 12x2 − 2x − 3

. . . . . .


Which is better?

Example
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx

Solution
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx

. . . . . .


Which is better?

Example
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx

Solution
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)

. . . . . .


Which is better?

Example
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx

Solution
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
= −5x4 + 12x2 − 2x − 3

. . . . . .


One more

Example
d
Find x sin x.
dx

. . . . . .


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

. . . . . .


One more

Example
d
Find x sin x.
dx

Solution

( ) ( )
d d d
dx dx dx
= 1 · sin x + x · cos x

. . . . . .


One more

Example
d
Find x sin x.
dx

Solution

( ) ( )
d d d
dx dx dx
= 1 · sin x + x · cos x
= sin x + x cos x

. . . . . .


Mnemonic

Let u = “hi” and v = “ho”. Then

(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”

. . . . . .


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

. . . . . .


Iterating the Product Rule

Example
Use the product rule to find the derivative of a three-fold product uvw.

. . . . . .



Example

Solution

(uvw)′ .

. . . . . .



Example

Solution

(uvw)′ = ((uv)w)′ .

. . . . . .



Example
Use the product rule to. find the derivative of a three-fold product uvw.
Apply the product rule
Solution to uv and w

(uvw)′ = ((uv)w)′ .

. . . . . .



Example
Use the product rule to. find the derivative of a three-fold product uvw.
Solution to uv and w

(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .

. . . . . .



Example
.
Solution to u and v

(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .

. . . . . .



Example
.
Solution to u and v

(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′

. . . . . .



Example

Solution

(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
= u′ vw + uv′ w + uvw′

. . . . . .



Example

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.

. . . . . .


Outline

Derivation
Examples

The Quotient Rule
Derivation
Examples



. . . . . .


The Quotient Rule

What about the derivative of a quotient?

. . . . . .


The Quotient Rule

u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv

. . . . . .


The Quotient Rule

u
v
u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q′ v + Qv′

. . . . . .


The Quotient Rule

u
v
u = Qv


u′ = (Qv)′ = Q′ v + Qv′
u′ − Qv′ u′ u v′
=⇒ Q′ = = − ·
v v v v

. . . . . .


The Quotient Rule

u
v
u = Qv


u′ = (Qv)′ = Q′ v + Qv′
u′ − Qv′ u′ u v′
=⇒ Q′ = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2

. . . . . .


The Quotient Rule

u
v
u = Qv


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.

. . . . . .


The Quotient Rule

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

. . . . . .


Verifying Example
Example
( )
d x2
Verify the quotient rule by computing and comparing it to
dx x
d
(x).
dx

. . . . . .


Verifying Example
Example
( )
d x2
Verify the quotient rule by computing and comparing it to
dx x
d
(x).
dx

Solution

( ) ( )
d x2 x dx x2 − x2 dx (x)
d d
=
dx x x2
x · 2x − x2 · 1
=
x2
x2 d
= 2 =1= (x)
x dx
. . . . . .


Mnemonic

Let u = “hi” and v = “lo”. Then
( u )′ vu′ − uv′
= = “lo dee hi minus hi dee lo over lo lo”
v v2

. . . . . .


Examples

Example
d 2x + 5
1.
dx 3x − 2
d sin x
2.
dx x2
d 1
3. 2
dt t + t + 2

. . . . . .


Solution to first example

Solution

d 2x + 5
dx 3x − 2

. . . . . .



Solution

d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2

. . . . . .



Solution

d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2

. . . . . .



Solution

d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
(6x − 4) − (6x + 15)
=
(3x − 2)2

. . . . . .



Solution

d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
(6x − 4) − (6x + 15) 19
= =−
(3x − 2) 2 (3x − 2)2

. . . . . .


Examples

Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x
2.
dx x2
d 1
3. 2
dt t + t + 2

. . . . . .


Solution to second example

Solution

d sin x
=
dx x2

. . . . . .



Solution

d sin x x2
=
dx x2

. . . . . .



Solution

d sin x x2 d sin x
= dx
dx x2

. . . . . .



Solution

d sin x x2 d sin x − sin x
= dx
dx x2

. . . . . .



Solution

d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2

. . . . . .



Solution

d
= dx
dx x2 (x2 )2

. . . . . .



Solution

d
= dx
dx x2 (x2 )2

=

. . . . . .



Solution

d
= dx
dx x2 (x2 )2
x2
=

. . . . . .



Solution

d
= dx
dx x2 (x2 )2
x2 cos x
=

. . . . . .



Solution

d
= dx
dx x2 (x2 )2
x2 cos x − 2x
=

. . . . . .



Solution

d
= dx
dx x2 (x2 )2
x2 cos x − 2x sin x
=

. . . . . .



Solution

d
= dx
dx x2 (x2 )2
=
x4

. . . . . .



Solution

d
= dx
dx x2 (x2 )2
=
x4
x cos x − 2 sin x
=
x3

. . . . . .


Another way to do it

Solution
Using the product rule this time:

d sin x d ( )
= sin x · x−2
dx x2 dx
( ) ( )
d −2 d −2
= sin x · x + sin x · x
dx dx
= cos x · x−2 + sin x · (−2x−3 )
= x−3 (x cos x − 2 sin x)

. . . . . .


Examples

Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x x cos x − 2 sin x
2. 2.
dx x2 x3
d 1
3. 2
dt t + t + 2

. . . . . .


Solution to third example

Solution

d 1
dt t2 + t + 2

. . . . . .



Solution

d 1 (t2 + t + 2)(0) − (1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2

. . . . . .



Solution

d 1 (t2 + t + 2)(0) − (1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
2t + 1
=− 2
(t + t + 2)2

. . . . . .


A nice little takeaway

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 1 v· d
dx (1) −1· d
dx v v · 0 − 1 · v′ v′
= = =− 2
dx v v2 v2 v

. . . . . .


Examples

Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x x cos x − 2 sin x
2. 2.
dx x2 x3
d 1 2t + 1
3. 2
3. − 2
dt t + t + 2 (t + t + 2)2

. . . . . .


Outline

Derivation
Examples

The Quotient Rule
Derivation
Examples



. . . . . .


Derivative of Tangent

Example
d
Find tan x
dx

. . . . . .



Example
d
Find tan x
dx

Solution

( )
d d sin x
tan x =
dx dx cos x

. . . . . .



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

. . . . . .



Example
d
Find tan x
dx

Solution

( )
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x
=
cos2 x

. . . . . .



Example
d
Find tan x
dx

Solution

( )
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x 1
= 2x
=
cos cos2 x

. . . . . .



Example
d
Find tan x
dx

Solution

( )
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x 1
= 2x
= = sec2 x
cos cos2 x

. . . . . .


Derivative of Cotangent
Example
d
Find cot x
dx

. . . . . .


Example
d
Find cot x
dx

Answer

d 1
cot x = − 2 = − csc2 x
dx sin x

. . . . . .


Example
d
Find cot x
dx

Answer

d 1
cot x = − 2 = − csc2 x
dx sin x

Solution

d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x

. . . . . .


Example
d
Find cot x
dx

Answer

d 1
cot x = − 2 = − csc2 x
dx sin x

Solution

cot x = =
dx dx sin x sin2 x
2
− sin x − cos2 x
=
sin2 x . . . . . .


Example
d
Find cot x
dx

Answer

d 1
cot x = − 2 = − csc2 x
dx sin x

Solution

cot x = =
dx dx sin x sin2 x
2
− sin x − cos2 x 1
= 2
=− 2
sin x sin x . . . . . .


Example
d
Find cot x
dx

Answer

d 1
cot x = − 2 = − csc2 x
dx sin x

Solution

cot x = =
dx dx sin x sin2 x
2
− sin x − cos2 x 1
= 2
= − 2 = − csc2 x
sin x sin x . . . . . .


Derivative of Secant

Example
d
Find sec x
dx

. . . . . .



Example
d
Find sec x
dx

Solution

( )
d d 1
sec x =
dx dx cos x

. . . . . .



Example
d
Find sec x
dx

Solution

( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x

. . . . . .



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

. . . . . .



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
= = ·
cos2 x cos x cos x

. . . . . .



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

. . . . . .


Derivative of Cosecant
Example
d
Find csc x
dx

. . . . . .


Example
d
Find csc x
dx

Answer

d
csc x = − csc x cot x
dx

. . . . . .


Example
d
Find csc x
dx

Answer

d
dx

Solution

( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x

. . . . . .


Example
d
Find csc x
dx

Answer

d
dx

Solution

( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x
=− 2
sin x . . . . . .


Example
d
Find csc x
dx

Answer

d
dx

Solution

( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x 1 cos x
=− 2 =− ·
sin x sin x sin x
. . . . . .


Example
d
Find csc x
dx

Answer

d
dx

Solution

( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x 1 cos x
=− 2 =− · = − csc x cot x
sin x sin x sin x
. . . . . .


Recap: Derivatives of trigonometric functions

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

. . . . . .


Outline

Derivation
Examples

The Quotient Rule
Derivation
Examples



. . . . . .


We will use the quotient rule to prove
Theorem

d −n
x = (−n)x−n−1
dx
for positive integers n.

. . . . . .


Theorem

d −n
x = (−n)x−n−1
dx

Proof.

. . . . . .


Theorem

d −n
x = (−n)x−n−1
dx

Proof.

d −n d 1
x =
dx dx xn

. . . . . .


Theorem

d −n
x = (−n)x−n−1
dx

Proof.

d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )

. . . . . .


Theorem

d −n
x = (−n)x−n−1
dx

Proof.

d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
nxn−1
=−
x2n

. . . . . .


Theorem

d −n
x = (−n)x−n−1
dx

Proof.

d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
nxn−1
=− = −nxn−1−2n
x2n

. . . . . .


Theorem

d −n
x = (−n)x−n−1
dx

Proof.

d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
nxn−1
=− = −nxn−1−2n = −nx−n−1
x2n

. . . . . .


Summary

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
. . . . . .


Lesson 9: The Product and Quotient Rules (Section 41 slides)

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