Lesson 9: The Product and Quotient Rules (Section 21 handout

Section 2.4The Product and Quotient Rules

V63.0121.021, Calculus I

New York University

October 4, 2010

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I Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2

I Midterm in class (covers all sections up to 2.5)

Announcements

I Quiz 2 next week on §§1.5,1.6, 2.1, 2.2

I Midterm in class (covers allsections up to 2.5)

V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 2 / 41

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Notes

Notes

Notes

1

Section 2.4 : The Product and Quotient RulesV63.0121.021, Calculus I October 4, 2010

http://www.math.nyu.edu/degree/undergrad/tutor_schedule.html

Objectives

I Understand and be able touse the Product Rule for thederivative of the product oftwo functions.

I Understand and be able touse the Quotient Rule forthe derivative of thequotient of two functions.


Outline

Derivative of a ProductDerivationExamples

The Quotient RuleDerivationExamples

More derivatives of trigonometric functionsDerivative of Tangent and CotangentDerivative of Secant and Cosecant

More on the Power RulePower 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?


Notes

Notes

Notes

2


Is the derivative of a product the product of thederivatives?

(uv)′ = u′v ′?

(uv)′ = u′v ′!

Try this with u = x and v = x2.

I Then uv = x3 =⇒ (uv)′ = 3x2.

I 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. Whatare your choices?

I Work longer hours.

I Get a raise.

Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. How muchextra money do you make?

∆I = 5× $0.25 = $1.25?

∆I = 5× $0.25 = $1.25?∆I = 5× $0.25 = $1.25?


Money money money money

The answer depends on how much you work already and your currentwage. Suppose you work h hours and are paid w . You get a time increaseof ∆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


Notes

Notes

Notes

3


A geometric argument

Draw a box:

w ∆w

h

∆h

w h

w ∆h

∆w h

∆w ∆h

∆I = w ∆h + h ∆w + ∆w ∆h


Cash flow

Supose wages and hours are changing continuously over time. Over a timeinterval ∆t, what is the average rate of change of income?

∆I

∆t=

w ∆h + h ∆w + ∆w ∆h

∆t

= w∆h

∆t+ h

∆w

∆t+ ∆w

∆h

∆t

What is the instantaneous rate of change of income?

dI

dt= lim

∆t→0

∆I

∆t= w

dh

dt+ h

dw

dt+ 0


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 notationd

dx(uv) =

du

dx· v + u

dv

dx


Notes

Notes

Notes

4


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 theproduct rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solution

by the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3


One more

Example

Findd

dxx sin x .

Solution

d

dxx sin x =

(d

dxx

)sin x + x

(d

dxsin x

)= 1 · sin x + x · cos x

= sin x + x cos x


Notes

Notes

Notes

5


Mnemonic

Let u = “hi” and v = “ho”. Then

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


Iterating the Product Rule

Example

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

Solution

(uvw)′ = ((uv)w)′

Apply the product ruleto uv and w

= (uv)′w + (uv)w ′

Apply the product ruleto u and v

= (u′v + uv ′)w + (uv)w ′

= u′vw + uv ′w + uvw ′

So we write down the product three times, taking the derivative of eachfactor once.


Outline






Notes

Notes

Notes

6


The Quotient Rule

What about the derivative of a quotient?

Let u and v be differentiable functions and let Q =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v 2

This is called the Quotient Rule.


The Quotient Rule

We have discovered

Theorem (The Quotient Rule)

Let u and v be differentiable at x, and v(x) 6= 0. Thenu

vis differentiable

at x, and (u

v

)′(x) =

u′(x)v(x)− u(x)v ′(x)

v(x)2


Verifying Example

Example

Verify the quotient rule by computingd

dx

(x2

x

)and comparing it to

d

dx(x).

Solution

d

dx

(x2

x

)=

x ddx

(x2)− x2 d

dx (x)

x2

=x · 2x − x2 · 1

x2

=x2

x2= 1 =

d

dx(x)


Notes

Notes

Notes

7


Mnemonic

Let u = “hi” and v = “lo”. Then(u

v

)′=

vu′ − uv ′

v 2= “lo dee hi minus hi dee lo over lo lo”


Examples

Example

1.d

dx

2x + 5

3x − 2

2.d

dx

sin x

x2

3.d

dt

1

t2 + t + 2

Answers

1. − 19

(3x − 2)2

2.x cos x − 2 sin x

x3

3. − 2t + 1

(t2 + t + 2)2


Solution to first example

Solution

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (3x − 2)

(3x − 2)2

=(3x − 2)(2)− (2x + 5)(3)

(3x − 2)2

=(6x − 4)− (6x + 15)

(3x − 2)2= − 19

(3x − 2)2


Notes

Notes

Notes

8


Solution to second example

Solution

d

dx

sin x

x2=

x2 ddx sin x − sin x d

dx x2

(x2)2

=x2 cos x − 2x sin x

x4

=x cos x − 2 sin x

x3


Another way to do it

Solution

Using the product rule this time:

d

dx

sin x

x2=

d

dx

(sin x · x−2

)=

(d

dxsin x

)· x−2 + sin x ·

(d

dxx−2

)= cos x · x−2 + sin x · (−2x−3)

= x−3 (x cos x − 2 sin x)


Solution to third example

Solution

d

dt

1

t2 + t + 2=

(t2 + t + 2)(0)− (1)(2t + 1)

(t2 + t + 2)2

= − 2t + 1

(t2 + t + 2)2


Notes

Notes

Notes

9


A nice little takeaway

Fact

Let v be differentiable at x, and v(x) 6= 0. Then1

vis differentiable at 0,

and (1

v

)′= − v ′

v 2

Proof.

d

dx

(1

v

)=

v · ddx (1)− 1 · d

dx v

v 2=

v · 0− 1 · v ′

v 2= − v ′

v 2


Outline






Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)=

cos x · cos x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x= sec2 x


Notes

Notes

Notes

10


Derivative of Cotangent

Example

Findd

dxcot x

Answer

d

dxcot x = − 1

sin2 x= − csc2 x

Solution

d

dxcot x =

d

dx

(cos x

sin x

)=

sin x · (− sin x)− cos x · cos x

sin2 x

=− sin2 x − cos2 x

sin2 x= − 1

sin2 x= − csc2 x


Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)=

cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x=

1

cos x· sin x

cos x= sec x tan x


Derivative of Cosecant

Example

Findd

dxcsc x

Answer

d

dxcsc x = − csc x cot x

Solution

d

dxcsc x =

d

dx

(1

sin x

)=

sin x · 0− 1 · (cos x)

sin2 x

= − cos x

sin2 x= − 1

sin x· cos x

sin x= − csc x cot x


Notes

Notes

Notes

11


Recap: Derivatives of trigonometric functions

y y ′

sin x cos x

cos x − sin x

tan x sec2 x

cot x − csc2 x

sec x sec x tan x

csc x − csc x cot x

I Functions come in pairs(sin/cos, tan/cot, sec/csc)

I Derivatives of pairs followsimilar patterns, withfunctions and co-functionsswitched and an extra sign.


Outline






Power Rule for Negative Integers

We will use the quotient rule to prove

Theorem

d

dxx−n = (−n)x−n−1

for positive integers n.

Proof.

d

dxx−n =

d

dx

1

xn= −

ddx xn

(xn)2

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1


Notes

Notes

Notes

12


Summary

I The Product Rule: (uv)′ = u′v + uv ′

I The Quotient Rule:(u

v

)′=

vu′ − uv ′

v 2

I Derivatives of tangent/cotangent, secant/cosecant

d

dxtan x = sec2 x

d

dxsec x = sec x tan x

d

dxcot x = − csc2 x

d

dxcsc x = − csc x cot x

I The Power Rule is true for all whole number powers, includingnegative powers:

d

dxxn = nxn−1


Notes

Notes

Notes

13


Lesson 9: The Product and Quotient Rules (Section 21 handout

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