1a
§2.1 Derivatives and Rates of Change
1b
Example. Consider 𝑦 = 𝑓(𝑥) = 𝑥 2
0 … 2 … 𝑥-, 0 … 4 … 𝑦-, 𝑓(𝑥)
Tangent Lines
axes, curve C
Consider a smooth curve C.
A line tangent to C at a point P both intersects C at P
and has the same slope as C at P. add line 𝑡
The Tangent Line Problem
Given point P on curve C, how do you find the
tangent line?
What is the equation of the line tangent to the curve
at 𝑥 = 1? add 𝑃, line
Point-slope form for a straight line passing through
𝑃(1,1)
𝑦 − 1 = 𝑚(𝑥 − 1)
2a
2b
𝑓(𝑥)−𝑓(1)
What is the slope 𝑚?
𝑚𝑃𝑄 =
What is the slope of the secant passing through
𝑃(1,1) and 𝑄1 (2,4)? add 𝑄1 , 𝑃𝑄1 
this ratio is called a difference quotient
𝑚𝑃𝑄1 =
=
=
Δ𝑦
=
=
𝑥 2 −1
𝑥−1
Δ𝑥
𝑓(2)−𝑓(1)
2−1
=
4−1
2−1
=3
What is the slope of the secant passing through
𝑃(1,1) and 𝑄2 (1.5, 𝑓(1.5))? add 𝑄2 , 𝑃𝑄2 
𝑚𝑃𝑄2 =
𝑥−1
=
(𝑥−1)(𝑥+1)
𝑥−1
𝑥 + 1, 𝑥 ≠ 1
={
undefined, 𝑥 = 1
As long as 𝑥 ≠ 1, 𝑚𝑃𝑄 = 𝑥 + 1
𝑥
𝑚𝑃𝑄
𝑓(1.5)−𝑓(1)
2
3
1.5−1
1.5
2.5
1.1
2.1
1.01
2.01
2.25−1
1.5−1
1.25
0.5
= 2.5
The slope of the tangent line is the limit of the
difference quotient as 𝑥 → 1.
𝑚 = lim(𝑥 + 1) = 2
𝑥→1
What is the slope of the secant line passing through
𝑃(1,1) and 𝑄(𝑥, 𝑥 2 )?
The equation of the tangent line is
𝑦 − 1 = 2(𝑥 − 1)
■
3a
3b
Thus
Example. Find an equation of the line tangent to the
curve 𝑔(𝑥) = √𝑥 + 1 at 𝑃(3,2).
1
𝑚 = lim
𝑥→3 √𝑥+1+2
=
1
2+2
=
1
4
Equation of tangent line
Point slope form of the tangent line
1
𝑦 − 2 = (𝑥 − 3) ■
4
𝑦 − 2 = 𝑚(𝑥 − 3)
where
𝑚 = lim
The Velocity Problem
𝑔(𝑥)−𝑔(3)
𝑥−3
𝑥→3
Drive to Spokane airport (~85 miles)
simplify the difference quotient
Start at noon
𝑔(𝑥)−𝑔(3)
𝑥−3
=
√𝑥+1−2
𝑥−3
Drive slowly through Colfax
multiply by 1 to rationalize the numerator
=
√𝑥+1−2 √𝑥+1+2
𝑥−3
√𝑥+1+2
Arrive 2pm
Average velocity =
(𝑥+1)−4
= (𝑥−3)(
√𝑥+1+2)
=
√𝑥+1+2
distance traveled
time elapsed
=
85 miles
2 hours
=
42.5 mph
cancel factors of 𝑥 − 3
1
Have lunch at Harvester
true if 𝑥 ≠ 3
The speedometer 5 miles north of Colfax reads 65
mph. This is the “instantaneous velocity.”
4a
4b
Mathematical definition of instantaneous velocity?
Galileo drops a ball off the leaning Tower of Pisa
sketch ground, tower, coordinate 𝑠 with origin at
top
What is the average velocity between 𝑡 = 1 and 𝑡 =
2?
Average velocity =
=
=
𝑠(𝑡) = 5𝑡 2
𝑠 meters,
0 … 2 … 𝑡-, 0 … 5 … 20 … 𝑠-, curve
𝑡 seconds
time elapsed
𝑠(2)−𝑠(1)
2−1
5⋅22 −5⋅12
= 15
ball falls distance 𝑠 at time 𝑡 after release.
distance traveled
1
meters
second
What is the average velocity between 𝑡 = 1 and a
variable 𝑡?
Average velocity =
=
=
=
𝑠(𝑡)−𝑠(1)
𝑡−1
5⋅𝑡 2 −5
𝑡−1
5⋅(𝑡 2 −1)
𝑡−1
5(𝑡−1)(𝑡+1)
𝑡−1
5a
5b
= 5(𝑡 + 1)
as long as 𝑡 ≠ 1.
Table
𝑡
2
1.1
1.001
average velocity (meters per sec.)
15
5 ⋅ (2.1) = 10.5
5 ⋅ (2.001) = 10.005
Derivatives
Define the derivative of a function 𝑓 at a number 𝑎,
denoted 𝑓 ′ (𝑎)
𝑓 ′ (𝑎) = lim
𝑓(𝑥)−𝑓(𝑎)
[1]
𝑥−𝑎
𝑥→𝑎
Define instantaneous velocity at 𝑡 = 1 as the limit of
average velocities over shorter and shorter time
intervals around 𝑡 = 1.
From the example above 𝑣(1) = 𝑠 ′ (1) =
Denote instantaneous velocity 𝑣(𝑡).
lim
𝑣(1) = lim
𝑡→1
𝑡−1
.
𝑠(𝑡)−𝑠(1)
Alternatively, introduce
𝑡−1
= lim 5(𝑡 + 1) = 10
𝑡→1
𝑡→1
𝑠(𝑡)−𝑠(1)
meters
second
ℎ =𝑥−𝑎
𝑥 =𝑎+ℎ
and insert in equation [1] to get
𝑓 ′ (𝑎) = lim
ℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
[2]
6a
6b
𝑓 ′ (𝑥) = lim (2𝑥 + ℎ) = 2𝑥
ℎ→0
§2.2 The Derivative as a Function
Graph and compare 𝑓(𝑥) and 𝑓 ′ (𝑥)
−2 … 0 … 2 … 𝑥-, −4 … 0 … 4 … 𝑦-, 𝑓(𝑥)
Replace the symbol 𝑎 in [2] by 𝑥. Regard 𝑥 as a
variable.
𝑓 ′ (𝑥) = lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
ℎ→0
Regard 𝑓′ as a new function.
−2 … 0 … 2 … 𝑥-, −4 … 0 … 4 … 𝑦-, 𝑓′(𝑥)
Example. Let 𝑓(𝑥) = 𝑥 2 . Find 𝑓 ′ (𝑥)
Simplify the difference quotient
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
=
=
=
(𝑥+ℎ)2 −𝑥 2
ℎ
(𝑥 2 +2𝑥ℎ+ℎ2 )−𝑥 2
ℎ
2𝑥ℎ+ℎ2
ℎ
= 2𝑥 + ℎ
Then
this step assumes ℎ ≠ 0
add tangent segments at 𝑥 = −1, 0, 1 to graph of 𝑓
7a
7b
add dots at 𝑥 = −1, 0 , 1 to graph of 𝑓′ ■
Example. Let 𝑔(𝑥) = √𝑥 + 1. Find 𝑔′ (𝑥).
?? Transparency: match 𝑓 and 𝑓′
Simplify the difference quotient
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ
√𝑥+ℎ+1−√𝑥+1
ℎ
=
rationalize numerator by multiplying by 1
√𝑥+ℎ+1−√𝑥+1 √𝑥+ℎ+1+√𝑥+1
ℎ
√𝑥+ℎ+1+√𝑥+1
=
multiply terms in numerator
ℎ(√𝑥+ℎ+1+√𝑥+1)
divide through by ℎ (assumes ℎ ≠ 0)
1
√𝑥+ℎ+1+√𝑥+1
Then
𝑔′ (𝑥) = lim
1
ℎ→0 √𝑥+ℎ+1+√𝑥+1
=
original function: 𝑦 = 𝑓(𝑥)
derivative function: 𝑓 ′ (𝑥) =
𝑑𝑦
𝑑𝑥
prime notation emphasizes idea of derivative as a
new function
(𝑥+ℎ+1)−(𝑥+1)
=
=
Notations for Derivative
1
2√𝑥+1
Show transparency comparing 𝑔 and 𝑔′
𝑓 ′ (𝑥)
the prime means differentiate with respect to
function argument
evaluate at no. 𝑎
𝑓 ′ (𝑎)
8a
8b
Leibniz notation emphasizes idea of derivative as the
limit of a ratio
… 𝑥 … 𝑥 + ℎ … , … , 𝑓, ℎ = Δ𝑥, 𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
Δ𝑦
Operator notation for derivative
Sometimes we write
𝑓 ′ (𝑥) =
𝑑
𝑑𝑥
𝑓
or
𝑓 ′ (𝑥) = 𝐷𝑓
view
𝑑
𝑑𝑥
and 𝐷 as operators: machines that convert
the functions they operate on into other functions
lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ→0
ℎ
= lim
evaluate at no. 𝑎
𝑑𝑦
|
𝑑𝑥 𝑎
Δ𝑦
ℎ→0 Δ𝑥
=
𝑑𝑦
𝑑𝑥
Differentiability
𝑓 differentiable at 𝑎 means 𝑓 ′ (𝑎) exists
𝑓 differentiable on an open interval (𝑎, 𝑏) means 𝑓 is
differentiable at every point in (𝑎, 𝑏)
9a
9b
conclude 𝑓 ′ (0) does not exist
Example (a function not differentiable at a point)
𝑓(𝑥) = |𝑥|
Geometrical Idea
Is 𝑓 differentiable at 𝑥 = 0?
axes, graph of |x|, no tangent line here (at origin)
if so
𝑓 ′ (0) = lim
𝑓(0+ℎ)−𝑓(0)
𝑓
ℎ→0
= lim
|0+ℎ|−|0|
ℎ
h→0
= lim
|ℎ|
ℎ→0 ℎ
does this limit exist?
find the limit from the right
lim
|ℎ|
ℎ→0+ ℎ
ℎ
= lim = 1
ℎ→0 ℎ
find the limit from the left
lim
|ℎ|
ℎ→0− ℎ
= lim
−ℎ
ℎ→0− ℎ
= −1
the right and left hand limit do not agree.
To be differentiable at point, the graph must have a
unique tangent line at that point. ■
10a
10b
Three ways that a function can fail to be
differentiable
(a)
(c)
at a vertical tangent
… 𝑎 … , …, function w/ a vertical tangent at 𝑎
at any discontinuity
… 𝑎 … , …, function with dcty at 𝑎
lim 𝑓(𝑥) = ∞, 𝑓 ′ (𝑎) DNE
𝑥→𝑎
lim
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
ℎ→0
(b)
DNE
at any corner or kink
… 𝑎 … , …, function with a kink at 𝑎
lim
ℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
DNE
11a
11b
Relationship between differentiability and
continuity
We have shown: if 𝑓 is not continuous then 𝑓 is not
differentiable
Let 𝐴 be the statement
Higher Derivatives
Consider
𝑦 = 𝑓(𝑥)
First derivative of 𝑓
𝑑𝑦
𝑓 is continuous at a no. 𝑥
Let 𝐵 be the statement
𝑓 is differentiable at 𝑥
We have shown
If (not 𝐴) then (not 𝐵)
This is logically equivalent to
If 𝐵 then 𝐴
𝑑𝑥
= 𝑓 ′ (𝑥)
Regarded as a function, 𝑓 may itself be differentiable.
Second derivative of 𝑓
𝑑2𝑦
𝑑𝑥 2
= 𝑓 ′′ (𝑥)
If 𝑓 ′′ (𝑥) is differentiable, form the third derivative
𝑑3𝑦
𝑑𝑥 3
= 𝑓 ′′′ (𝑥)
If 𝑓 ′′′ (𝑥) is differentiable, form the fourth derivative
𝑑4𝑦
𝑑𝑥 4
= 𝑓 (4) (𝑥)
Notation for the 𝑛𝑇ℎ derivative, with 𝑛 ≥ 4:
If 𝑓 is differentiable at 𝑥 then it is continuous at 𝑥
𝑑𝑛𝑦
𝑑𝑥 𝑛
= 𝑓 (𝑛) (𝑥)
12a
12b
Application of higher derivatives
2.3 Basic Differentiation Rules
Let 𝑠(𝑡) be the position of an object at time 𝑡.
We first consider those rules that will enable us to
differentiate polynomials.
𝑠 ′ (𝑡) =
𝑑𝑠
𝑑𝑡
= 𝑣(𝑡)
is the velocity of the object
Derivative of a Constant Function
… 𝑥-,… 0 … 𝑐 … 𝑦-, line 𝑦 = 𝑐
𝑠 ′′ (𝑡) = 𝑣 ′ (𝑡) = 𝑎(𝑡)
is the acceleration of the object
slope of tangent line?
First and second derivatives are the most important
in applications
𝑑
𝑑𝑥
𝑐=0
Derivative of 𝒇(𝒙) = 𝒙
0 … 𝑥-, 0 … 𝑦-, 𝑦 = 𝑥
Example. Mechanics
momentum = mass ⋅ velocity
force = mass ⋅ acceleration ■
slope of tangent line?
𝑑
𝑑𝑥
𝑥=1
13a
13b
Derivative of 𝒈(𝒙) = 𝒙𝟐
We have seen that
The Power Rule.
𝑔′ (𝑥) = 2𝑥
Let 𝑛 be a positive integer
𝑑
𝑑𝑥
𝑥 𝑛 = 𝑛𝑥 𝑛−1
Derivative of 𝑭(𝒙) = 𝒙𝟑
𝐹
′ (𝑥)
= lim
Example.
𝐹(𝑥+ℎ)−𝐹(𝑥)
ℎ
ℎ→0
ℎ
=
=
=
ℎ
(𝑥 − 𝑎) ⋅
(𝑥 3 +3𝑥 2 ℎ+3𝑥ℎ2 +ℎ3 )−𝑥 3
(𝑥 𝑛−1 + 𝑥 𝑛−2 𝑎 + 𝑥 𝑛−3 𝑎2 + ⋯ + 𝑥𝑎𝑛−2 + 𝑎𝑛−1 )
ℎ
3𝑥 2 ℎ+3𝑥ℎ2 +ℎ3
= 𝑥 𝑛 + 𝑥 𝑛−1 𝑎 + 𝑥 𝑛−2 𝑎2 + ⋯ + 𝑥𝑎𝑛−1
ℎ
= 3𝑥 + 3𝑥ℎ + ℎ
3
−𝑥 𝑛−1 𝑎 − 𝑥 𝑛−2 𝑎2 − ⋯ − 𝑥𝑎𝑛−1 − 𝑎𝑛
assumes ℎ ≠ 0
________________________________________
thus
𝐹
𝑥 7 = 7𝑥 6 . ■
Preliminary fact:
(𝑥+ℎ)3 −𝑥 3
2
′ (𝑥)
𝑑𝑥
Proof.
simplify the difference quotient
𝐹(𝑥+ℎ)−𝐹(𝑥)
𝑑
2
= lim 3𝑥 + 3𝑥ℎ + ℎ
ℎ→0
= 3𝑥 2
2
𝑥𝑛
− 𝑎𝑛
in other words
𝑥 𝑛 −𝑎𝑛
𝑥−𝑎
= 𝑥 𝑛−1 + 𝑥 𝑛−2 𝑎 + ⋯ 𝑥𝑎𝑛−2 + 𝑎𝑛−1
14a
14b
𝑑
notice there are 𝑛 terms on the right hand side
𝑑𝑡
Let 𝑓(𝑥) = 𝑥 𝑛
𝑓
′ (𝑎)
= lim
𝑑
𝑑𝑥
𝑥 𝑛 −𝑎𝑛
𝑥→𝑎 𝑥−𝑎
= lim (𝑥
??
𝑛−1
𝑥→𝑎
+𝑥
𝑛−2
𝑎 + ⋯ 𝑥𝑎
𝑛−2
+𝑎
= 𝑎𝑛−1 + 𝑎𝑛−2 𝑎 + ⋯ 𝑎𝑎𝑛−2 + 𝑎𝑛−1
𝑛−1
)
??
𝑡 −2 = −2𝑡 −3
−1
1
1
√𝑥 = 𝑥 2
recall √𝑥 = 𝑥 2
2
𝑑 1
𝑑𝑡 √𝑡
𝑑
𝑑𝑢
=
𝑢 √2 =
■
= 𝑛𝑎𝑛−1
Regard 𝑎 as a variable. Replace 𝑎 by 𝑥.
𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1
■
The Constant Multiple Rule
Let 𝑐 be a constant and 𝑓 a differentiable function
𝑑
𝑑𝑥
𝑐𝑓(𝑥) = 𝑐
𝑑
𝑑𝑥
𝑓(𝑥)
The Power Rule (general version)
Let 𝑛 be any real number.
𝑑
𝑑𝑥
𝑥 𝑛 = 𝑛𝑥 𝑛−1
a constant passes through the limit symbol
Examples.
??
Examples.
𝑑
𝑑𝑥
𝑥 −9 = −9𝑥 −10
??
𝑑
𝑑𝑡
𝑑
𝑑𝑤
𝑑
𝑑𝑥
5𝑥 2 = 5
10
7 √𝑡 =
√5 𝑤 𝜋 =
𝑑
𝑑𝑥
𝑥 2 = 5 ⋅ 2𝑥 = 10𝑥
15a
15b
■
■
The Sum Rule
The Difference Rule
If 𝑓 and 𝑔 are both differentiable
If 𝑓 and 𝑔 are both differentiable
𝑑
𝑑𝑥
[𝑓(𝑥) + 𝑔(𝑥)] =
𝑑
𝑑𝑥
𝑓(𝑥) +
𝑑
𝑑𝑥
𝑔(𝑥)
𝑑
𝑑𝑥
[𝑓(𝑥) − 𝑔(𝑥)] =
𝑑
𝑑𝑥
𝑓(𝑥) −
𝑑
𝑑𝑥
𝑔(𝑥)
In words: “the derivative of a sum is the sum of the
derivatives”
In words: “the derivative of a difference is the
difference of the derivatives”
prime notation
prime notation
′
′
(𝑓(𝑥) + 𝑔(𝑥)) = 𝑓 ′ (𝑥) + 𝑔′ (𝑥)
(𝑓(𝑥) − 𝑔(𝑥)) = 𝑓 ′ (𝑥) − 𝑔′ (𝑥)
shorthand
shorthand
(𝑓 + 𝑔)′ = 𝑓 ′ + 𝑔′
(𝑓 − 𝑔)′ = 𝑓 ′ − 𝑔′
the sum rule applies to the sum of any number of
functions
Example.
(𝑓 + 𝑔 + ℎ)′ = 𝑓 ′ + 𝑔′ + ℎ′
??
Example
𝑑
𝑑𝑥
[𝑥 3 + 5𝑥 2 + 𝜋] = 3𝑥 2 + 10𝑥 + 0
■
𝑑
𝑑𝑥
(𝑥 3 − 5𝑥 2 ) =
16a
16b
𝑑𝑢
𝑑𝑡
We can now differentiate any polynomial
2 −1
3 1
2 −1
3
2
3
1
= 𝑡 3 + 2 ( 𝑡 2 ) = 𝑡 3 + 3𝑡 2 ■
Example. Let
Example. Find an equation of the line tangent to the
curve
𝑔(𝑥) = 5𝑥 8 − 2𝑥 5 + 6
𝑦 = 3𝑥 2 +
then
at the point (1,7)
𝑔′ (𝑥) =
=5
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(5𝑥 8 ) −
𝑥8 − 2
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(2𝑥 5 ) +
𝑑
𝑑𝑥
6
𝑥5 + 0
4
𝑥
point slope form for straight line
𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 )
= 5(8𝑥 7 ) − 2(5𝑥 4 ) + 0
where (𝑥0 , 𝑦0 ) = (1,7)
= 40𝑥 7 − 10𝑥 4 ■
what is 𝑚?
𝑑𝑦
We can differentiate other functions too.
𝑑𝑥
= 3(2𝑥) + 4(−𝑥 −2 )
= 6𝑥 − 4𝑥 −2
Example. Let
3
𝑢 = √𝑡 2 + 2√𝑡 3
Find
𝑑𝑡
𝑢 = 𝑡 + 2𝑡
𝑑𝑦
|
𝑑𝑥 𝑥=1
=6−4=2
answer
𝑑𝑢
2
3
𝑚=
𝑦 − 7 = 2(𝑥 − 1)
3
2
■
17a
17b
marginal cost = lim
Δ𝐶
Δ𝑥→0 Δ𝑥
?? Example. A ball is thrown straight up from the
ground at 20 meters/second. Its height is given by
= 𝐶 ′ (𝑥)
Example. Jeans manufacture.
Let 𝐶(𝑥) = cost of producing 𝑥 pairs of jeans.
𝑦 = 20 𝑡 − 5 𝑡 2
= 2000 + 3𝑥 + 0.01𝑥 2
(a) Find the velocity at time 𝑡.
where
$2000
(b) Find the velocity at 𝑡 = 1sec.
capital costs (sewing machines)
3𝑥 + 0.01𝑥 2 cost of labor, materials, rent
Cost of producing 100 pairs of jeans
(c) When is the ball at rest?
𝐶(100) = 2000 + 300 + 0.01(100)2
(d) What is the average velocity between 𝑡 = 1 and
𝑡 = 2?
■
What is the cost of producing one additional pair of
jeans?
Economics – Marginal cost
𝐶(𝑥 + 1) = 2000 + 3(𝑥 + 1) + 0.01(𝑥 + 1)2
𝐶(𝑥) = cost to produce 𝑥 widgets
average rate of change of cost =
2400
𝐶(𝑥2 )−𝐶(𝑥1 )
𝑥2 −𝑥1
=
Δ𝐶
Δ𝑥
𝐶(𝑥) = 2000 + 3𝑥 + 0.01𝑥 2
___________________________________________
𝐶(𝑥 + 1) − 𝐶(𝑥) = 3 + 0.01((𝑥 + 1)2 − 𝑥 2 )
18a
18b
= 3 + 0.01(2𝑥 + 1)
lim
𝑥→0
Cost of producing the 101ST pair
cos(𝑥)−1
𝑥
=0
Recall the addition formula for cosine
𝐶(101) − 𝐶(100) = 3 + 0.01(201)
cos(𝑎 + 𝑏) = cos(𝑎) cos(𝑏) − sin(𝑎) sin(𝑏)
= $5.01
Now use the limit definition of derivative
th
Compare with the marginal cost at 100 pair
𝐶 ′ (𝑥) = 3 + 0.02𝑥
𝑑
𝑑𝑥
cos(𝑥) = lim
cos(𝑥+ℎ)−cos(𝑥)
ℎ
ℎ→0
addition formula for cosine
𝐶 ′ (100) = $5
= lim
cos(𝑥) cos(ℎ)−sin(𝑥) sin(ℎ)−cos(𝑥)
ℎ
ℎ→0
𝐶 ′ (𝑥) is often a very good approximation to the cost
of producing one additional widget. ■
difference law for limits
= lim cos(𝑥)
ℎ→0
cos(ℎ)−1
ℎ
− lim sin(𝑥)
sin(ℎ)
ℎ→0
ℎ
constant multiple law of limits
Derivatives of Sine and Cosine
Recall the limits
lim
𝑥→0
sin(𝑥)
𝑥
=1
= cos(𝑥) ⋅ lim
ℎ→0
cos(ℎ)−1
ℎ
− sin(𝑥) ⋅ lim
ℎ→0
sin(ℎ)
ℎ
recalling the limits above
= cos(𝑥) ⋅ 0 − sin(𝑥) ⋅ 1
19a
19b
= −sin(𝑥) ■
The derivative of sin(𝑥) may be found using a similar
argument (see our text).
§2.4 The Product and Quotient Rules
In summary:
Product Rule
𝑑
sin(𝑥) = cos(𝑥)
𝑑𝑥
𝑑
cos(𝑥) = −sin(𝑥)
𝑑𝑥
If 𝑓 and 𝑔 are both differentiable
𝑑
𝑑𝑥
𝑑𝑥
1.
𝑓(𝑡) = sin(𝑡) + 𝜋 cos(𝑡)
𝑑
𝑑𝑥
𝑔(𝑥) + 𝑔(𝑥)
𝑑
𝑑𝑥
𝑓(𝑥)
or alternately (as I personally prefer)
𝑑
?? Differentiate the following
[𝑓(𝑥)𝑔(𝑥)] = 𝑓(𝑥)
[𝑓(𝑥)𝑔(𝑥)] = [
𝑑
𝑑𝑥
𝑓(𝑥)] 𝑔(𝑥) + 𝑓(𝑥) [
𝑑
𝑑𝑥
𝑔(𝑥)]
prime notation
′
(𝑓(𝑥)𝑔(𝑥)) = 𝑓 ′ (𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′ (𝑥)
shorthand
(𝑓𝑔)′ = 𝑓 ′ 𝑔 + 𝑓𝑔′
2.
𝑔(𝑦) =
𝐴
𝑦 10
+ 𝐵 cos(𝑦)
WARNING: (𝑓𝑔)′ ≠ 𝑓′𝑔′ The derivative of a
product is not the product of derivatives
20a
20b
This is a common mistake!
Let 𝑓(𝑥)𝑔(𝑥) = 𝐹(𝑥)
then
Example. By the power rule
𝑑
𝑑𝑥
𝑑
𝑑𝑥
[𝑓(𝑥)𝑔(𝑥)] =
𝑥 3 = 3𝑥 2
𝐹(𝑥)
= lim
2
and 𝑔 = 𝑥
= lim
𝐹(𝑥+ℎ)−𝐹(𝑥)
ℎ
𝑓(𝑥+ℎ)𝑔(𝑥+ℎ)−𝑓(𝑥)𝑔(𝑥)
ℎ
ℎ→0
then 𝑓 ′ = 2𝑥 and 𝑔′ = 1
𝑑𝑥
𝑑𝑥
ℎ→0
Now let 𝑓 = 𝑥
𝑑
𝑑
subtract and add the same term in the numerator
3
𝑥 = (𝑓𝑔)′
= 𝑓 ′ 𝑔 + 𝑓𝑔′
= lim
𝑓(𝑥+ℎ)𝑔(𝑥+ℎ)−𝑓(𝑥+ℎ)𝑔(𝑥)+𝑓(𝑥+ℎ)𝑔(𝑥)−𝑓(𝑥)𝑔(𝑥)
ℎ
ℎ→0
algebra
= 2𝑥 ⋅ 𝑥 + 𝑥 2 ⋅ 1
= 3𝑥 2
■
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
+ 𝑔(𝑥)
]
ℎ
= lim [ 𝑓(𝑥 + ℎ) ⋅
ℎ→0
sum and product laws for limits
Proof of the Product Rule
Suppose 𝑓 and 𝑔 are both differentiable functions.
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ→0
ℎ
= lim 𝑓(𝑥 + ℎ) ⋅ lim
ℎ→0
+ lim 𝑔(𝑥) ⋅ lim
ℎ→0
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
21a
21b
continuity of 𝑓 and definition of derivative
= 𝑓(𝑥) ⋅ 𝑔′ (𝑥) + 𝑔(𝑥) ⋅ 𝑓 ′ (𝑥)
■
Then
𝐹(𝑥) = 𝑥 −1 sin(𝑥)
Product rule
Extension to a Product of Three Functions
𝐹 ′ (𝑥) = (𝑥 −1 )′ sin(𝑥) + 𝑥 −1 (sin(𝑥))′
If 𝑓, 𝑔 and ℎ are all differentiable
= (−𝑥 −2 ) sin(𝑥) + 𝑥 −1 cos(𝑥)
(𝑓 𝑔 ℎ)′ = 𝑓 ′ 𝑔 ℎ + 𝑓 𝑔′ ℎ + 𝑓 𝑔 ℎ′
=
Example. Let 𝑓(𝑥) = 𝑔(𝑥) = ℎ(𝑥) = 𝑥
then 𝑓 ′
𝑑
𝑑𝑥
= 𝑔′ =
ℎ′ = 1
𝑥 3 = 𝑓 ′ 𝑔 ℎ + 𝑓 𝑔′ ℎ + 𝑓 𝑔 ℎ′
=1⋅𝑥⋅𝑥+𝑥⋅1⋅𝑥+𝑥⋅𝑥⋅1
−sin(𝑥)
𝑥2
■
𝑥
■
If 𝑓 and 𝑔 are differentiable at a point 𝑥 where
𝑔′ (𝑥) ≠ 0 then
𝑑
(
𝑓(𝑥)
)=
𝑑
𝑑
𝑓(𝑥))𝑔(𝑥)−𝑓(𝑥) 𝑔(𝑥)
𝑑𝑥
𝑑𝑥
𝑔(𝑥)2
(
shorthand
𝑓 ′
Example. Differentiate 𝐹(𝑥) =
Law of exponents:
cos(𝑥)
Quotient Rule
𝑑𝑥 𝑔(𝑥)
= 3𝑥 2
+
1
𝑥
= 𝑥 −1
sin(𝑥)
𝑥
.
(𝑔) =
𝑓′ 𝑔−𝑓𝑔′
𝑔2
terms in numerator in same order as my product rule
(but take difference)
22a
22b
=
(𝑥−1)−(𝑥+1)
(𝑥−1)2
−2
= (𝑥−1)2 ■
Proof of quotient rule.
𝑓
Let
𝑔
=ℎ
Example. Find the equations of the tangent lines to
the curve
Then 𝑓 = 𝑔 ℎ
(𝑥+1)
By product rule
𝐺(𝑥) =
𝑓 ′ = 𝑔′ ℎ + 𝑔 ℎ′
that are parallel to the line
Solve for ℎ′
𝑥 + 2𝑦 = 2 .
𝑔 ℎ′ = 𝑓 ′ − 𝑔′ ℎ
Solution. Parallel means same slope. Slope of line?
ℎ′ =
=
𝑓′ −𝑔′ ℎ
2𝑦 − 2 − 𝑥
𝑔
𝑓′ 𝑔−𝑔′ 𝑓
𝑔2
(𝑥−1)
1
𝑦 =1− 𝑥
■
2
slope is 𝑚 =
Example. Differentiate 𝐺(𝑥) =
𝐺
′ (𝑥)
=
𝑥+1
.
where does 𝐺 have slope
−2
𝐺 ′ = (𝑥−1)2 =
(𝑥−1)2
where (𝑥 + 1) is shorthand for
2
𝑥−1
(𝑥+1)′ (𝑥−1)−(𝑥+1)(𝑥−1)′
′
−1
𝑑
𝑑𝑥
1
(𝑥 + 1)
(𝑥−1)2
=
−1
2
1
4
−1
2
?
solve for 𝑥
23a
23b
(𝑥 − 1)2 = 4
𝑥 = −1 or 𝑥 = 3
form of equation for tangent line
𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 )
Consider 𝑥0 = −1. 𝑦0 = 𝐺(−1) =
𝑦=
−1−1
=0
−1
𝑥+1
2
Consider 𝑥0 = 3. 𝑦0 = 𝐺(3) =
𝑦−2=
■
−1+1
3+1
3−1
=2
−1
(𝑥 − 3)
2
?? Class practice product and quotient rules