lesson 9: the product and quotient rules (section 41 handout)
TRANSCRIPT
Section 2.4The Product and Quotient Rules
V63.0121.041, Calculus I
New York University
October 3, 2010
Announcements
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)
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 2 / 40
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V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 3 / 40
Notes
Notes
Notes
1
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 4 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 5 / 40
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?
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 6 / 40
Notes
Notes
Notes
2
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 7 / 40
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?
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 8 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 9 / 40
Notes
Notes
Notes
3
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 10 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 11 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 12 / 40
Notes
Notes
Notes
4
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 13 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 14 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 15 / 40
Notes
Notes
Notes
5
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
Mnemonic
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
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 18 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 19 / 40
Notes
Notes
Notes
6
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 20 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 21 / 40
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)
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 22 / 40
Notes
Notes
Notes
7
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
Mnemonic
Let u = “hi” and v = “lo”. Then(u
v
)′=
vu′ − uv ′
v 2= “lo dee hi minus hi dee lo over lo lo”
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 23 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 24 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 25 / 40
Notes
Notes
Notes
8
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 27 / 40
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)
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 28 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 30 / 40
Notes
Notes
Notes
9
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 32 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 33 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 34 / 40
Notes
Notes
Notes
10
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 35 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 36 / 40
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.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 37 / 40
Notes
Notes
Notes
11
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 38 / 40
Power Rule for Negative Integers
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
=xn · d
dx 1− 1 · ddx xn
x2n
=0− nxn−1
x2n= −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 39 / 40
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
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 3, 2010 40 / 40
Notes
Notes
Notes
12
Section 2.4 : The Product and Quotient RulesV63.0121.041, Calculus I October 3, 2010