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

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

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Notes

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1


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.


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


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 ?


2


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.


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?


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


3


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


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


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


4


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.


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


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


5


Mnemonic
Notes

Let u = “hi” and v = “ho”. Then

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


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.


Outline
Notes

Derivation
Examples

The Quotient Rule
Derivation
Examples




6


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.


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


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


7


Mnemonic
Notes
Let u = “hi” and v = “lo”. Then
u vu − uv
= = “lo dee hi minus hi dee lo over lo lo”
v v2


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


Solution to ﬁrst example
Notes

Solution


8


Solution to second example
Notes

Solution


Another way to do it
Notes

Solution


Solution to third example
Notes

Solution


9


Outline
Notes

Derivation
Examples

The Quotient Rule
Derivation
Examples




Derivative of Tangent
Notes

Example
d
Find tan x
dx

Solution


Derivative of Cotangent
Notes
Example
d
Find cot x
dx

Answer

d 1
cot x = − 2 = − csc2 x
dx sin x

Solution


10


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


Derivative of Cosecant
Notes
Example
d
Find csc x
dx

Answer

d
csc x = − csc x cot x
dx

Solution


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


11


Outline
Notes

Derivation
Examples

The Quotient Rule
Derivation
Examples




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

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


12

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

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