Chapter 3
Limits and the
Derivative
Section 5
Basic Differentiation
Properties
(Part 1)
Objectives for Section 3.5
Power Rule and Differentiation Properties
■ The student will be able to:
■Calculate the derivative of a
constant function.
■Apply the power rule.
■Apply the constant multiple and
sum and difference properties.
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The Derivative
 Recall from the previous lesson:
f (x  h)  f (x)
f (x)  lim
h0
h
 f(x) is a “slope machine”. 𝑓 ′ (𝑎) will tell you the slope of
the line tangent to the graph of f(x) at x=a (if it exists).
 f(x) is also the instantaneous rate of change of f(x).
 f(x) is also the instantaneous velocity of an object.
 In this lesson, you will learn an easier way to find 𝑓 ′ 𝑥 .
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Derivative Notation
In the preceding section we defined the derivative
of a function. There are several widely used symbols
to represent the derivative. Given y = f (x), the
derivative of f at x may be represented by any of the
following:
■ f (x)
■ y
■
𝑑𝑦
𝑑𝑥
Later on, you will see how
each of these symbols has its
particular advantage in certain
situations.
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Derivative Rules
 In the next several slides, you will learn rules for finding
derivatives of:
• Constant functions
• Power functions
• Functions multiplied by constants
 These rules will enable you to find 𝑓′(𝑥) easier compared
to using a limit.
 Note that you will be tested on both methods.
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Derivative of a Constant
What is the slope of a constant function?
The graph of f (x) = C is a
horizontal line with slope 0, so
we would expect f (x) = 0.
Theorem 1. Let y = f (x) = C be a constant function, then
y = f (x) = 0.
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Example 1
Derivatives of constant functions:
 If f(x) = 8
• f (x)=0
 If y = -4
• y = 0
 If y = 
𝑑𝑦
𝑑𝑥
=0
𝑑
12
𝑑𝑥
=0
•

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Derivative of a Power Function
A function of the form f (x) = xn is called a power function.
(n is a real number)
Theorem 2. (Power Rule) Let y = xn be a power function,
then
f (x) = nxn – 1.
THEOREM 2 IS VERY IMPORTANT.
IT WILL BE USED A LOT!
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Example 2
Derivatives of power functions:
then 𝑓 ′ 𝑥 = 5𝑥 5−1 = 5𝑥 4
 If 𝑓 𝑥 = 𝑥 5
then
 If y = 𝑥
 If 𝑓 𝑥 = 𝑥
 If 𝑦 =
3
𝑦=𝑥
5 4
𝑑𝑦
𝑑𝑥
= 1𝑥 1−1 = 𝑥 0 = 1
5
4
then 𝑓 ′ 𝑥 = 𝑥 1
4
𝑥
1 3
 If 𝑓 𝑥 =
𝑓(𝑥) =
then
1
𝑥2
𝑥 −2
𝑦′
=
1 −2 3
𝑥
3
then 𝑓 ′ (𝑥) = −2𝑥 −3
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Constant Multiple Property
Theorem 3. Let y = k u(x) where k is a constant.
Then
y = k  u (x)
In words: The derivative of a constant times a function is the
constant times the derivative of the function.
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Example 3
Differentiate each function:
• f (x) = 7x4
𝑓′(𝑥) = 28𝑥 3
• 𝑦 = −3𝑥 2
• 𝑓 𝑥 = 6𝑥 2
• 𝑦=
=
𝑦 ′ = −6𝑥
3
𝑓 ′ 𝑥 = 4𝑥 −1
3
2
3𝑥 6
2𝑥 −6
3
𝑑𝑦
= −4𝑥 −7
𝑑𝑥
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Sum and Difference Properties
Theorem 5.
If
y = f (x) = u(x) + v(x),
then y = f (x) = u(x) + v(x).
(this is also true for subtraction)
■ The derivative of the sum of two differentiable functions is
the sum of the derivatives.
■ The derivative of the difference of two differentiable
functions is the difference of the derivatives.
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Example 5
Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.
Answer: f (x) = 15x4 + 4x3 – 6x2 + 10x – 7
Find
𝑑𝑦
𝑑𝑥
for 𝑦 = 𝑥 − 5x
Answer:
𝑑𝑦
𝑑𝑥
=
1 −1 2
𝑥
2
−5
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Homework
#3-5A: Pg. 185
(25-41 odd, 49)
Mammoth opens
on Nov. 7th!
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Chapter 3
Limits and the
Derivative
Section 5
Basic Differentiation
Properties
(Part 2)
Objectives for Section 3.5
Power Rule and Differentiation Properties
■ The student will be able to:
■Solve applications.
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Applications
Remember that the derivative has these meanings:
■Slope of the tangent line at a point on the graph of a
function.
■Instantaneous velocity.
■Instantaneous rate of change.
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Tangent Line Example
Let f (x) = x4 – 6x2 + 10.
(a) Find f (x)
(b) Find the equation of the tangent line at x = 1
(c) Find the values of x where the tangent line is horizontal.
Solution:
(a) f (x) = 4x3 - 12x
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Example (continued)
f (x) = x4 – 6x2 + 10.
(b) Find the equation of the tangent line at x = 1
Solution:
Slope: f (1) = 4(13) – 12(1) = -8.
Point: If x = 1, then y = f (1) = 5
Point-slope form: y – y1 = m(x – x1)
y – 5 = –8(x –1)
y = –8x + 13
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Example (continued)
Let f (x) = x4 – 6x2 + 10.
c) Find the values of x where the tangent line is horizontal.
Solution:
c) Tangent line is horizontal means slope is zero.
So set derivative = 0 and solve for x.
4𝑥 3 − 12𝑥 = 0
4𝑥 𝑥 2 − 3 = 0
𝑥 = 0, ± 3
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Barnett/Ziegler/Byleen Business Calculus 12e
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Instantaneous Velocity
 An object moves along the y-axis (marked in feet) so that
its position at time x (in seconds) is:
𝑓 𝑥 = 𝑥 3 − 15𝑥 2 + 72𝑥
a) Find the instantaneous velocity function.
b) Find the velocity at 2 and 5 seconds.
c) Find the time(s) when the velocity is 0.
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Instantaneous Velocity
 An object moves along the y-axis (marked in feet) so that
its position at time x (in seconds) is:
𝑓 𝑥 = 𝑥 3 − 15𝑥 2 + 72𝑥
• Find the instantaneous velocity function.
𝑓 ′ 𝑥 = 3𝑥 2 − 30𝑥 + 72
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Instantaneous Velocity
 An object moves along the y-axis (marked in feet) so that
its position at time x (in seconds) is:
𝑓 𝑥 = 𝑥 3 − 15𝑥 2 + 72𝑥
a) Find the instantaneous velocity function.
𝑓 ′ 𝑥 = 3𝑥 2 − 30𝑥 + 72
b) Find the velocity at 2 and 5 seconds.
𝑓 ′ 2 = 24
𝑓 ′ 5 = −3
The velocity at 2 seconds is 24 ft/sec.
The velocity at 5 seconds is − 3 ft/sec.
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Instantaneous Velocity
 An object moves along the y-axis (marked in feet) so that
its position at time x (in seconds) is:
𝑓 𝑥 = 𝑥 3 − 15𝑥 2 + 72𝑥
𝑓 ′ 𝑥 = 3𝑥 2 − 30𝑥 + 72
c) Find the time(s) when the velocity is 0.
0 = 3𝑥 2 − 30𝑥 + 72
0 = 3(𝑥 2 − 10𝑥 + 24)
0 = 3(𝑥 − 6)(𝑥 − 4)
𝑥 = 4, 6
𝑇ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 0 𝑎𝑡 4 𝑎𝑛𝑑 6 𝑠𝑒𝑐𝑜𝑛𝑑𝑠.
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Instantaneous Rate of Change
• If C(x) is the total cost of producing x items, then C(x) is
the instantaneous rate of change of cost at a production
level of x items.
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Application Example
The total cost (in dollars) of producing x
portable radios per day is
C(x) = 1000 + 100x – 0.5x2
for 0 ≤ x ≤ 100.
1. Find 𝐶 ′ 𝑥
Solution:
C(x) = 100 – x.
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Example (continued)
2. Find C(80) and 𝐶′(80) and interpret these results.
Solution:
𝐶 80 = 1000 + 100(80) – 0.5(80)2 = 5800
C(80) = 100 – 80 = 20
At a production level of 80 radios, the total cost is $5800 and is
increasing at a rate of $20 per radio.
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Homework
#3-5B: Pg. 185
(32-42 even, 50,
55, 56, 81, 87, 89)
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