Section 2.4The Product and Quotient Rules
V63.0121, Calculus I
February 17–18, 2009
Announcements
I Quiz 2 is this week, covering 1.3–1.6
I Midterm I is March 4/5, covering 1.1–2.4 (today)
I ALEKS is due February 27, 11:59pm
Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant
More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers
Calculus
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?
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.
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.
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.
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.
Is the derivative of a product the product of thederivatives?
(uv)′ = u′v ′?
(uv)′ = u′v ′!
Try this with u = x and v = x2.
I Then uv = x3 =⇒ (uv)′ = 3x2.
I But u′v ′ = 1 · 2x = 2x .
So we have to be more careful.
Mmm...burgers
Say you work in a fast-food joint. You want to make more money.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?
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?
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?
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?
Money money money money
The answer depends on how much you work already and yourcurrent wage. Suppose you work h hours and are paid w . You geta time increase of ∆h and a wage increase of ∆w . Income iswages times hours, so
∆I = (w + ∆w)(h + ∆h)− wh
FOIL= wh + w ∆h + ∆w h + ∆w ∆h − wh
= w ∆h + ∆w h + ∆w ∆h
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
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
Supose wages and hours are changing continuously over time. Howdoes income change?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
SodI
dt= lim
t→0
∆I
∆t= w
dh
dt+ h
dw
dt+ 0
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
Supose wages and hours are changing continuously over time. Howdoes income change?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
SodI
dt= lim
t→0
∆I
∆t= w
dh
dt+ h
dw
dt+ 0
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
Supose wages and hours are changing continuously over time. Howdoes income change?
∆I
∆t=
w ∆h + h ∆w + ∆w ∆h
∆t
= w∆h
∆t+ h
∆w
∆t+ ∆w
∆h
∆t
SodI
dt= lim
t→0
∆I
∆t= w
dh
dt+ h
dw
dt+ 0
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′(x) = u(x)v ′(x) + u′(x)v(x)
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.
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.
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]
Solutionby direct multiplication:
d
dx
[(3− x2)(x3 − x + 1)
]FOIL=
d
dx
[−x5 + 4x3 − x2 − 3x + 3
]
= −5x4 + 12x2 − 2x − 3
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby direct multiplication:
d
dx
[(3− x2)(x3 − x + 1)
]FOIL=
d
dx
[−x5 + 4x3 − x2 − 3x + 3
]
= −5x4 + 12x2 − 2x − 3
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby direct multiplication:
d
dx
[(3− x2)(x3 − x + 1)
]FOIL=
d
dx
[−x5 + 4x3 − x2 − 3x + 3
]= −5x4 + 12x2 − 2x − 3
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby 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
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby 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
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby 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
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby 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
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby 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
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby 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
Example
Find this derivative two ways: first by direct multiplication andthen by the product rule:
d
dx
[(3− x2)(x3 − x + 1)
]Solutionby the product rule:
dy
dx=
(d
dx(3− x2)
)(x3 − x + 1) + (3− x2)
(d
dx(x3 − x + 1)
)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x − 3
One more
Example
Findd
dxx sin x .
Solution
d
dxx sin x
=
(d
dxx
)sin x + x
(d
dxsin x
)= 1 · sin x + x · cos x
= sin x + x cos x
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
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
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
Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv ′ = “ho dee hi plus hi dee ho”
Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold productuvw .
Solution
(uvw)′ = ((uv)w)′ = (uv)′w + (uv)w ′
= (u′v + uv ′)w + (uv)w ′
= u′vw + uv ′w + uvw ′
So we write down the product three times, taking the derivative ofeach factor once.
Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold productuvw .
Solution
(uvw)′ = ((uv)w)′ = (uv)′w + (uv)w ′
= (u′v + uv ′)w + (uv)w ′
= u′vw + uv ′w + uvw ′
So we write down the product three times, taking the derivative ofeach factor once.
Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold productuvw .
Solution
(uvw)′ = ((uv)w)′ = (uv)′w + (uv)w ′
= (u′v + uv ′)w + (uv)w ′
= u′vw + uv ′w + uvw ′
So we write down the product three times, taking the derivative ofeach factor once.
Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant
More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers
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 ′
v2
This is called the Quotient Rule.
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 ′
v2
This is called the Quotient Rule.
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 ′
v2
This is called the Quotient Rule.
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 ′
v2
This is called the Quotient Rule.
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 ′
v2
This is called the Quotient Rule.
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 ′
v2
This is called the Quotient Rule.
Verifying Example
Example
Verify the quotient rule by computingd
dx
(x2
x
)and comparing it
tod
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)
Verifying Example
Example
Verify the quotient rule by computingd
dx
(x2
x
)and comparing it
tod
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)
Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
Solution to first example
d
dx
2x + 5
3x − 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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Solution to first example
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
Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
Solution to second example
d
dx
2x + 1
x2 − 1
=(x2 − 1)(2)− (2x + 1)(2x)
(x2 − 1)2
=(2x2 − 2)− (4x2 + 2x)
(x2 − 1)2
= −2
(x2 + x + 1
)(x2 − 1)2
Solution to second example
d
dx
2x + 1
x2 − 1=
(x2 − 1)(2)− (2x + 1)(2x)
(x2 − 1)2
=(2x2 − 2)− (4x2 + 2x)
(x2 − 1)2
= −2
(x2 + x + 1
)(x2 − 1)2
Solution to second example
d
dx
2x + 1
x2 − 1=
(x2 − 1)(2)− (2x + 1)(2x)
(x2 − 1)2
=(2x2 − 2)− (4x2 + 2x)
(x2 − 1)2
= −2
(x2 + x + 1
)(x2 − 1)2
Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
Solution to third example
d
dt
t − 1
t2 + t + 2
=(t2 + t + 2)(1)− (t − 1)(2t + 1)
(t2 + t + 2)2
=(t2 + t + 2)− (2t2 − t − 1)
(t2 + t + 2)2
=−t2 + 2t + 3
(t2 + t + 2)2
Solution to third example
d
dt
t − 1
t2 + t + 2=
(t2 + t + 2)(1)− (t − 1)(2t + 1)
(t2 + t + 2)2
=(t2 + t + 2)− (2t2 − t − 1)
(t2 + t + 2)2
=−t2 + 2t + 3
(t2 + t + 2)2
Solution to third example
d
dt
t − 1
t2 + t + 2=
(t2 + t + 2)(1)− (t − 1)(2t + 1)
(t2 + t + 2)2
=(t2 + t + 2)− (2t2 − t − 1)
(t2 + t + 2)2
=−t2 + 2t + 3
(t2 + t + 2)2
Examples
Example
1.d
dx
2x + 5
3x − 2
2.d
dx
2x + 1
x2 − 1
3.d
dt
t − 1
t2 + t + 2
Answers
1. − 19
(3x − 2)2
2. −2
(x2 + x + 1
)(x2 − 1)2
3.−t2 + 2t + 3
(t2 + t + 2)2
Mnemonic
Let u = “hi” and v = “lo”. Then(u
v
)′=
vu′ − uv ′
v2= “lo dee hi minus hi dee lo over lo lo”
Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant
More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers
Derivative of Tangent
Example
Findd
dxtan x
Solution
d
dxtan x =
d
dx
(sin x
cos x
)
=cos x · sin x − sin x · (− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x=
1
cos2 x= sec2 x
Derivative of Tangent
Example
Findd
dxtan x
Solution
d
dxtan x =
d
dx
(sin x
cos x
)
=cos x · sin x − sin x · (− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x=
1
cos2 x= sec2 x
Derivative of Tangent
Example
Findd
dxtan x
Solution
d
dxtan x =
d
dx
(sin x
cos x
)=
cos x · sin x − sin x · (− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x=
1
cos2 x= sec2 x
Derivative of Tangent
Example
Findd
dxtan x
Solution
d
dxtan x =
d
dx
(sin x
cos x
)=
cos x · sin x − sin x · (− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x
=1
cos2 x= sec2 x
Derivative of Tangent
Example
Findd
dxtan x
Solution
d
dxtan x =
d
dx
(sin x
cos x
)=
cos x · sin x − sin x · (− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x=
1
cos2 x
= sec2 x
Derivative of Tangent
Example
Findd
dxtan x
Solution
d
dxtan x =
d
dx
(sin x
cos x
)=
cos x · sin x − sin x · (− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x=
1
cos2 x= sec2 x
Derivative of Cotangent
Example
Findd
dxcot x
Answer
d
dxcot x = − 1
sin2 x= − csc2 x
Derivative of Cotangent
Example
Findd
dxcot x
Answer
d
dxcot x = − 1
sin2 x= − csc2 x
Derivative of Secant
Example
Findd
dxsec x
Solution
d
dxsec x =
d
dx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin x
cos2 x=
1
cos x· sin x
cos x= sec x tan x
Derivative of Secant
Example
Findd
dxsec x
Solution
d
dxsec x =
d
dx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin x
cos2 x=
1
cos x· sin x
cos x= sec x tan x
Derivative of Secant
Example
Findd
dxsec x
Solution
d
dxsec x =
d
dx
(1
cos x
)=
cos x · 0− 1 · (− sin x)
cos2 x
=sin x
cos2 x=
1
cos x· sin x
cos x= sec x tan x
Derivative of Secant
Example
Findd
dxsec x
Solution
d
dxsec x =
d
dx
(1
cos x
)=
cos x · 0− 1 · (− sin x)
cos2 x
=sin x
cos2 x
=1
cos x· sin x
cos x= sec x tan x
Derivative of Secant
Example
Findd
dxsec x
Solution
d
dxsec x =
d
dx
(1
cos x
)=
cos x · 0− 1 · (− sin x)
cos2 x
=sin x
cos2 x=
1
cos x· sin x
cos x
= sec x tan x
Derivative of Secant
Example
Findd
dxsec x
Solution
d
dxsec x =
d
dx
(1
cos x
)=
cos x · 0− 1 · (− sin x)
cos2 x
=sin x
cos2 x=
1
cos x· sin x
cos x= sec x tan x
Derivative of Cosecant
Example
Findd
dxcsc x
Answer
d
dxcsc x = − csc x cot x
Derivative of Cosecant
Example
Findd
dxcsc x
Answer
d
dxcsc x = − csc x cot x
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 pairsfollow similar patterns,with functions andco-functions switchedand an extra sign.
Outline
The Product RuleDerivation of the product ruleExamples
The Quotient RuleDerivationExamples
More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant
More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers
Power Rule for Positive Integers by Induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We can show it to be true for n = 1 directly.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1
=d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
Power Rule for Positive Integers by Induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n.
We can show it to be true for n = 1 directly.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1
=d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
Principle of Mathematical Induction
Suppose S(1) istrue and S(n + 1)is true when-ever S(n) is true.Then S(n) is truefor all n.
Image credit: Kool Skatkat
Power Rule for Positive Integers by Induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We can show it to be true for n = 1 directly.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1
=d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
Power Rule for Positive Integers by Induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We can show it to be true for n = 1 directly.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1 =
d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
Power Rule for Positive Integers by Induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We can show it to be true for n = 1 directly.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1 =
d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)
= 1 · xn + x · nxn−1 = (n + 1)xn
Power Rule for Positive Integers by Induction
TheoremLet n be a positive integer. Then
d
dxxn = nxn−1
Proof.By induction on n. We can show it to be true for n = 1 directly.
Suppose for some n thatd
dxxn = nxn−1. Then
d
dxxn+1 =
d
dx(x · xn)
=
(d
dxx
)xn + x
(d
dxxn
)= 1 · xn + x · nxn−1 = (n + 1)xn
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
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
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
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
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