lesson 9: the product and quotient rules (section 41 slides)
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Section 2.4The Product and Quotient Rules
V63.0121.041, Calculus I
New York University
October 4, 2010
Announcements
I Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2I 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 4, 2010 2 / 41
. . . . . .
Help!
Free resources:I Math Tutoring Center
(CIWW 524)I College Learning Center
(schedule on Blackboard)I TAs’ office hoursI my office hoursI each other!
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 3 / 41
. . . . . .
Objectives
I Understand and be able touse the Product Rule forthe derivative of theproduct of two 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 4, 2010 4 / 41
. . . . . .
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 4, 2010 5 / 41
. . . . . .
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 4, 2010 6 / 41
. . . . . .
Is the derivative of a product the product of the
derivatives?
..(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 4, 2010 7 / 41
. . . . . .
Is the derivative of a product the product of the
derivatives?
.
.(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 4, 2010 7 / 41
. . . . . .
Is the derivative of a product the product of the
derivatives?
.
.(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 4, 2010 7 / 41
. . . . . .
Is the derivative of a product the product of the
derivatives?
.
.(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 4, 2010 7 / 41
. . . . . .
Is the derivative of a product the product of the
derivatives?
.
.(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 4, 2010 7 / 41
. . . . . .
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are 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. Howmuch extra money do youmake?
..
.∆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 4, 2010 8 / 41
. . . . . .
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are 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. Howmuch extra money do youmake?
..
.∆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 4, 2010 8 / 41
. . . . . .
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are 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. Howmuch extra money do youmake?
..
.∆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 4, 2010 8 / 41
. . . . . .
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are 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. Howmuch extra money do youmake?
..
.∆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 4, 2010 8 / 41
. . . . . .
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are 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. Howmuch extra money do youmake?
..
.∆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 4, 2010 8 / 41
. . . . . .
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.What are 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. Howmuch extra money do youmake?
..
.∆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 4, 2010 8 / 41
. . . . . .
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 timeincrease of ∆h and a wage increase of ∆w. Income is wages timeshours, so
∆I = (w+∆w)(h+∆h)− whFOIL= 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 4, 2010 9 / 41
. . . . . .
A geometric argument
Draw a box:
..w .∆w
.h
.∆h
.wh
.w∆h
.∆wh
.∆w∆h
∆I = w∆h+ h∆w+∆w∆h
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 10 / 41
. . . . . .
A geometric argument
Draw a box:
..w .∆w
.h
.∆h
.wh
.w∆h
.∆wh
.∆w∆h
∆I = w∆h+ h∆w+∆w∆h
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 10 / 41
. . . . . .
Cash flow
Supose wages and hours are changing continuously over time. Over atime interval ∆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?
dIdt
= lim∆t→0
∆I∆t
= wdhdt
+ hdwdt
+ 0
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 11 / 41
. . . . . .
Cash flow
Supose wages and hours are changing continuously over time. Over atime interval ∆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?
dIdt
= lim∆t→0
∆I∆t
= wdhdt
+ hdwdt
+ 0
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 11 / 41
. . . . . .
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 notationddx
(uv) =dudx
· v+ udvdx
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 12 / 41
. . . . . .
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 4, 2010 13 / 41
. . . . . .
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 4, 2010 13 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby direct multiplication:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]
= −5x4 + 12x2 − 2x− 3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby direct multiplication:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]
= −5x4 + 12x2 − 2x− 3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby direct multiplication:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]= −5x4 + 12x2 − 2x− 3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
Which is better?
Example
Find this derivative two ways: first by direct multiplication and then bythe product rule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionby the product rule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(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 4, 2010 14 / 41
. . . . . .
One more
Example
Findddx
x sin x.
Solution
ddx
x sin x
=
(ddx
x)sin x+ x
(ddx
sin 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 4, 2010 15 / 41
. . . . . .
One more
Example
Findddx
x sin x.
Solution
ddx
x sin x =
(ddx
x)sin x+ x
(ddx
sin 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 4, 2010 15 / 41
. . . . . .
One more
Example
Findddx
x sin x.
Solution
ddx
x sin x =
(ddx
x)sin x+ x
(ddx
sin 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 4, 2010 15 / 41
. . . . . .
One more
Example
Findddx
x sin x.
Solution
ddx
x sin x =
(ddx
x)sin x+ x
(ddx
sin 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 4, 2010 15 / 41
. . . . . .
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 4, 2010 16 / 41
. . . . . .
Musical interlude
I jazz bandleader and singerI hit song “Minnie the
Moocher” featuring “hi deho” chorus
I played Curtis in The BluesBrothers
Cab Calloway1907–1994
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 17 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 rule
to uv and w
= (uv)′w+ (uv)w′..
.Apply the product rule
to 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 4, 2010 18 / 41
. . . . . .
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 4, 2010 19 / 41
. . . . . .
The Quotient Rule
What about the derivative of a quotient?
Let u and v be differentiable functions and let Q =uv. 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′ =(uv
)′=
u′v− uv′
v2
This is called the Quotient Rule.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
. . . . . .
The Quotient Rule
What about the derivative of a quotient?Let u and v be differentiable functions and let Q =
uv. 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′ =(uv
)′=
u′v− uv′
v2
This is called the Quotient Rule.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
. . . . . .
The Quotient Rule
What about the derivative of a quotient?Let u and v be differentiable functions and let Q =
uv. 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′ =(uv
)′=
u′v− uv′
v2
This is called the Quotient Rule.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
. . . . . .
The Quotient Rule
What about the derivative of a quotient?Let u and v be differentiable functions and let Q =
uv. 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′ =(uv
)′=
u′v− uv′
v2
This is called the Quotient Rule.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
. . . . . .
The Quotient Rule
What about the derivative of a quotient?Let u and v be differentiable functions and let Q =
uv. 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′ =(uv
)′=
u′v− uv′
v2
This is called the Quotient Rule.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
. . . . . .
The Quotient Rule
What about the derivative of a quotient?Let u and v be differentiable functions and let Q =
uv. 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′ =(uv
)′=
u′v− uv′
v2
This is called the Quotient Rule.
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 20 / 41
. . . . . .
The Quotient Rule
We have discovered
Theorem (The Quotient Rule)
Let u and v be differentiable at x, and v(x) ̸= 0. Thenuvis differentiable
at x, and (uv
)′(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 4, 2010 21 / 41
. . . . . .
Verifying Example
Example
Verify the quotient rule by computingddx
(x2
x
)and comparing it to
ddx
(x).
Solution
ddx
(x2
x
)=
x ddx
(x2)− x2 d
dx (x)x2
=x · 2x− x2 · 1
x2
=x2
x2= 1 =
ddx
(x)
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 22 / 41
. . . . . .
Verifying Example
Example
Verify the quotient rule by computingddx
(x2
x
)and comparing it to
ddx
(x).
Solution
ddx
(x2
x
)=
x ddx
(x2)− x2 d
dx (x)x2
=x · 2x− x2 · 1
x2
=x2
x2= 1 =
ddx
(x)
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 22 / 41
. . . . . .
Mnemonic
Let u = “hi” and v = “lo”. Then(uv
)′=
vu′ − uv′
v2= “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 4, 2010 23 / 41
. . . . . .
Examples
Example
1. ddx
2x+ 53x− 2
2. ddx
sin xx2
3. ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. x cos x− 2 sin xx3
3. − 2t+ 1(t2 + t+ 2)2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 24 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Solution to first example
Solution
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(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 4, 2010 25 / 41
. . . . . .
Examples
Example
1. ddx
2x+ 53x− 2
2. ddx
sin xx2
3. ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. x cos x− 2 sin xx3
3. − 2t+ 1(t2 + t+ 2)2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 26 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=
x2 ddx sin x− sin x d
dxx2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2
ddx sin x− sin x d
dxx2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x
− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x
ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=
x2 cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2
cos x− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x
− 2x sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(x2)2
=x2 cos x− 2x
sin xx4
=x cos x− 2 sin x
x3
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(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 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(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 4, 2010 27 / 41
. . . . . .
Solution to second example
Solution
ddx
sin xx2
=x2 d
dx sin x− sin x ddxx
2
(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 4, 2010 27 / 41
. . . . . .
Another way to do it
SolutionUsing the product rule this time:
ddx
sin xx2
=ddx
(sin x · x−2
)=
(ddx
sin x)· x−2 + sin x ·
(ddx
x−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 4, 2010 28 / 41
. . . . . .
Examples
Example
1. ddx
2x+ 53x− 2
2. ddx
sin xx2
3. ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. x cos x− 2 sin xx3
3. − 2t+ 1(t2 + t+ 2)2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 29 / 41
. . . . . .
Solution to third example
Solution
ddt
1t2 + 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 4, 2010 30 / 41
. . . . . .
Solution to third example
Solution
ddt
1t2 + 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 4, 2010 30 / 41
. . . . . .
Solution to third example
Solution
ddt
1t2 + 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 4, 2010 30 / 41
. . . . . .
A nice little takeaway
Fact
Let v be differentiable at x, and v(x) ̸= 0. Then1vis differentiable at 0,
and (1v
)′= − v′
v2
Proof.
ddx
(1v
)=
v · ddx(1)− 1 · d
dxvv2
=v · 0− 1 · v′
v2= − v′
v2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 31 / 41
. . . . . .
Examples
Example
1. ddx
2x+ 53x− 2
2. ddx
sin xx2
3. ddt
1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. x cos x− 2 sin xx3
3. − 2t+ 1(t2 + t+ 2)2
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 32 / 41
. . . . . .
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 4, 2010 33 / 41
. . . . . .
Derivative of Tangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)
=cos x · cos x− sin x · (− sin x)
cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
. . . . . .
Derivative of Tangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)
=cos x · cos x− sin x · (− sin x)
cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
. . . . . .
Derivative of Tangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
. . . . . .
Derivative of Tangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos 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 4, 2010 34 / 41
. . . . . .
Derivative of Tangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
. . . . . .
Derivative of Tangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 34 / 41
. . . . . .
Derivative of Cotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin 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 4, 2010 35 / 41
. . . . . .
Derivative of Cotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin 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 4, 2010 35 / 41
. . . . . .
Derivative of Cotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 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 4, 2010 35 / 41
. . . . . .
Derivative of Cotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x
= − 1sin2 x
= − csc2 x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 35 / 41
. . . . . .
Derivative of Cotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 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 4, 2010 35 / 41
. . . . . .
Derivative of Cotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 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 4, 2010 35 / 41
. . . . . .
Derivative of Secant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
. . . . . .
Derivative of Secant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
. . . . . .
Derivative of Secant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
. . . . . .
Derivative of Secant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
. . . . . .
Derivative of Secant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
. . . . . .
Derivative of Secant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 36 / 41
. . . . . .
Derivative of Cosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)
=sin x · 0− 1 · (cos x)
sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
. . . . . .
Derivative of Cosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)
=sin x · 0− 1 · (cos x)
sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
. . . . . .
Derivative of Cosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
. . . . . .
Derivative of Cosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
. . . . . .
Derivative of Cosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
. . . . . .
Derivative of Cosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 37 / 41
. . . . . .
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 4, 2010 38 / 41
. . . . . .
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 4, 2010 39 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n
= −nxn−1−2n = −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n
= −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
ddx
x−n = (−n)x−n−1
for positive integers n.
Proof.
ddx
x−n =ddx
1xn
= −ddxx
n
(xn)2
= −nxn−1
x2n= −nxn−1−2n = −nx−n−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 40 / 41
. . . . . .
Summary
I The Product Rule: (uv)′ = u′v+ uv′
I The Quotient Rule:(uv
)′=
vu′ − uv′
v2I Derivatives of tangent/cotangent, secant/cosecant
ddx
tan x = sec2 xddx
sec x = sec x tan x
ddx
cot x = − csc2 xddx
csc x = − csc x cot x
I The Power Rule is true for all whole number powers, includingnegative powers:
ddx
xn = nxn−1
V63.0121.041, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 4, 2010 41 / 41