lesson 9: the product and quotient rules (section 21 slides)
DESCRIPTION
The derivative of a sum of functions is the sum of the derivatives of those functions, but the derivative of a product or a quotient of functions is not so simple. We'll derive and use the product and quotient rule for these purposes. It will allow us to find the derivatives of other trigonometric functions, and derivatives of power functions with negative whole number exponents.TRANSCRIPT
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Section 2.4The Product and Quotient Rules
V63.0121.021, Calculus I
New York University
October 5, 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)
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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!
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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?
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41
. . . . . .
Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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)
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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)
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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”
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41
. . . . . .
Another way to do it
Find the derivative with the product rule instead.
Solution
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)
Notice the technique of factoring out the largest negative power,leaving positive powers.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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.
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 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
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 41 / 41