Transcript (x).

§ 10.5 Limits and the Derivative
Derivatives of Constants, Power
Forms, and Sums
The student will learn about:
the derivative of a constant function, the power rule,
a constant times f (x),
derivatives of sums and differences, and
an application.
Dr .Hayk Melikyan
Departmen of Mathematics and CS
[email protected]
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WARM UP EXERCISE
The ozone level (in parts per billion) on a summer day at R
University is given by
P(x) = 80 + 12x – x 2
Where t is hours after 9 am.
1. Use the two-step process to find P’(x).
2. Find P(3) and P’(3).
3. Write an interpretation.
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Objectives for Section 10.5
Power Rule and Differentiation Properties
■ The student will be able to calculate the derivative of a
constant function.
■ The student will be able to apply the power rule.
■ The student will be able to apply the constant multiple
and sum and difference properties.
■ The student will be able to solve applications.
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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
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Example 1
What is the slope of a constant function?
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Barnett/Ziegler/Byleen College
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Example 1
(continued)
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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Barnett/Ziegler/Byleen College
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Power Rule
A function of the form f (x) = xn is called a power function.
This includes f (x) = x (where n = 1) and radical functions
(fractional n).
Theorem 2. (Power Rule) Let y = xn be a power function,
then
y = f (x) = dy/dx = n xn – 1.
THEOREM 2 IS VERY IMPORTANT.
IT WILL BE USED A LOT!
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Example 2
Differentiate f (x) = x5.
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Example 2
Differentiate f (x) = x5.
Solution:
By the power rule, the derivative of xn is n xn–1.
In our case n = 5, so we get f (x) = 5 x4.
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Example 3
Differentiate f ( x ) 
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3
x.
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Example 3
Differentiate f ( x ) 
3
x.
Solution:
Rewrite f (x) as a power function, and apply the power rule:
f ( x)  x
f ( x ) 
1
3
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1/ 3
x
2 / 3
1

3
3 x
2
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Constant Multiple Property
Theorem 3. Let y = f (x) = k u(x) be a constant k times a
function u(x). Then
y = f (x) = 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 4
Differentiate f (x) = 7x4.
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Example 4
Differentiate f (x) = 7x4.
Solution:
Apply the constant multiple property and the power rule.
f (x) = 7(4x3) = 28 x3.
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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).
In words:
■ 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.
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Example 5
Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.
Solution:
Apply the sum and difference rules, as well as the constant
multiple property and the power rule.
f (x) = 15x4 + 4x3 – 6x2 + 10x – 7.
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Applications
Remember that the derivative gives the instantaneous rate of
change of the function with respect to x. That might be:
■ Instantaneous velocity.
■ Tangent line slope at a point on the curve of the function.
■ Marginal Cost. If C(x) is the cost function, that is, the total cost
of producing x items, then C(x) approximates the cost of
producing one more item at a production level of x items. C(x) is
called the marginal cost.
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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
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Tangent Line Example(continued)
Let f (x) = x4 – 6x2 + 10.
(a) Find f (x)
(b) Find the equation of the tangent line at x = 1
Solution:
(a) f (x) = 4x3 - 12x
(b) Slope: f (1) = 4(13) – 12(1) = -8.
Point: If x = 1, then y = f (1) = 1 – 6 + 10 = 5.
Point-slope form: y – y1 = m(x – x1)
y – 5 = –8(x –1)
y = –8x + 13
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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 the marginal cost at a production
level of x radios.
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Example
(continued)
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 the marginal cost at a production
level of x radios.
Solution: The marginal cost will be
C(x) = 100 – x.
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Example(continued)
2. Find the marginal cost at a production level of 80 radios and
interpret the result.
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Example(continued)
2. Find the marginal cost at a production level of 80 radios and
interpret the result.
Solution: C(80) = 100 – 80 = 20.
It will cost approximately $20 to produce the 81st radio.
3. Find the actual cost of producing the 81st radio and compare
this with the marginal cost.
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Example(continued)
2. Find the marginal cost at a production level of 80 radios and
interpret the result.
Solution: C(80) = 100 – 80 = 20.
It will cost approximately $20 to produce the 81st radio.
3. Find the actual cost of producing the 81st radio and compare
this with the marginal cost.
Solution: The actual cost of the 81st radio will be
C(81) – C(80) = $5819.50 – $5800 = $19.50.
This is approximately equal to the marginal cost.
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Summary

If f (x) = C, then f (x) = 0

If f (x) = xn, then f (x) = n xn-1

If f (x) = ku(x), then f (x) = ku(x)

If f (x) = u(x) ± v(x), then f (x) = u(x) ± v(x).
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Practice Problems
§3.4
1, 5, 9, 13, 17, 19, 23, 25, 29, 33, 37, 41, 45, 49,
53, 57, 61, 65, 69, 73, 77, 81, 83, 87, 89.
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Learning Objectives for Section 10.6
Differentials
The student will be able to apply the concept of
increments.
 The student will be able to compute
differentials.
 The student will be able to calculate
approximations using differentials.

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Hayk Melikyan/ MATH2000/
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Increments
In a previous section we defined the derivative of f at x as
the limit of the difference quotient:
f ' ( x )  lim
h 0
f ( x  h)  f ( x)
h
Increment notation will enable us to interpret the
numerator and the denominator of the difference quotient
separately.
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Example
Let y = f (x) = x3. If x changes from 2 to 2.1, then y will
change from y = f (2) = 8 to y = f (2.1) = 9.261.
We can write this using increment notation. The change in x is
called the increment in x and is denoted by x.  is the
Greek letter “delta”, which often stands for a difference or
change. Similarly, the change in y is called the increment in
y and is denoted by y.
In our example,
x = 2.1 – 2 = 0.1
y = f (2.1) – f (2) = 9.261 – 8 = 1.261.
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Graphical Illustration of Increments
For y = f (x)
x = x2 - x1
y = y2 - y1
x2 = x1 + x = f (x2) – f (x1) = f (x1 + x) – f (x1)
■ y represents the
change in y
corresponding to a
x change in x.
■ x can be either
(x2, f (x2))
y
(x1, f (x1))
x1
x
x2
positive or negative.
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Differentials
Assume that the limit
For small x,
f ' ( x )  lim
x  0
f '( x) 
y
x
exists.
y
x
Multiplying both sides of this equation by x gives us
y  f ’(x) x.
Here the increments x and y represent the actual changes
in x and y.
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Differentials(continued)
One of the notations for the derivative is f ' ( x ) 
dy
dx
If we pretend that dx and dy are actual quantities, we get
dy  f ' ( x ) dx
We treat this equation as a definition, and call dx and dy
differentials.
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Interpretation of Differentials
x and dx are the same, and represent the change in x.
The increment y stands for the actual change in y
resulting from the change in x.
The differential dy stands for the approximate change in y,
estimated by using derivatives.
 y  dy  f ' ( x ) dx
In applications, we use dy (which is easy to calculate) to
estimate y (which is what we want).
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Example 1
Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and
dx = 0.1.
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Example 1
Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and
dx = 0.1.
Solution:
dy = f ’(x) dx = (2x + 3) dx
When x = 2 and dx = 0.1, dy = [2(2) + 3] 0.1 = 0.7.
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Example 2 Cost-Revenue
A company manufactures and sells x transistor radios per
week. If the weekly cost and revenue equations are
C ( x )  5 , 000  2 x
R ( x )  10 x 
x
2
1, 000
0  x  8 , 000
find the approximate changes in revenue and profit if
production is increased from 2,000 to 2,010 units/week.
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Example 2 Solution
The profit is
x
P ( x)  R ( x)  C ( x)  8 x 
2
 5 , 000
1, 000
We will approximate R and P with dR and dP,
respectively, using x = 2,000 and dx = 2,010 – 2,000 = 10.
x
dR  R ' ( x ) dx  (10 
) dx
500
 (10 
2 , 000
) 10  $ 60
500
x
dP  P ' ( x ) dx  ( 8 
per week
) dx
500
 (8 
2 , 000
) 10  $ 40
per week
500
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Objectives for Section-10.7
Marginal Analysis
The student will be able to compute:
■ Marginal cost, revenue and profit
■ Marginal average cost, revenue and profit
■ The student will be able to solve applications
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Marginal Cost
Remember that marginal refers to an instantaneous rate of
change, that is, a derivative.
Definition:
If x is the number of units of a product produced in some
time interval, then
Total cost = C(x)
Marginal cost = C’(x)
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Marginal Revenue and Marginal Profit
Definition:
If x is the number of units of a product sold in some time interval,
then
Total revenue = R(x)
Marginal revenue = R’(x)
If x is the number of units of a product produced and sold in some
time interval, then
Total profit = P(x) = R(x) – C(x)
Marginal profit = P’(x) = R’(x) – C’(x)
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Marginal Cost and Exact Cost
Assume C(x) is the total cost of producing x items. Then the exact
cost of producing the (x + 1)st item is
C(x + 1) – C(x).
The marginal cost is an approximation of the exact cost.
C’(x) ≈ C(x + 1) – C(x).
Similar statements are true for revenue and profit.
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Example 1
The total cost of producing x electric guitars is
C(x) = 1,000 + 100x – 0.25x2.
1. Find the exact cost of producing the 51st guitar.
2. Use the marginal cost to approximate the cost
of producing the 51st guitar.
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Example 1(continued)
The total cost of producing x electric guitars is
C(x) = 1,000 + 100x – 0.25x2.
1. Find the exact cost of producing the 51st guitar.
The exact cost is C(x + 1) – C(x).
C(51) – C(50) = 5,449.75 – 5375 = $74.75.
2. Use the marginal cost to approximate the cost
of producing the 51st guitar.
The marginal cost is C’(x) = 100 – 0.5x
C’(50) = $75.
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Marginal Average Cost
Definition:
If x is the number of units of a product produced in some time
interval, then
Average cost per unit
C ( x) 
C ( x)
x
Marginal average cost C ' ( x ) 
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d
C ( x)
dx
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Marginal Average Revenue
Marginal Average Profit
If x is the number of units of a product sold in some time
interval, then
Average revenue per unit R ( x ) 
R ( x)
Marginal average revenue R ' ( x ) 
x
d
R ( x)
dx
If x is the number of units of a product produced and sold in
some time interval, then
Average profit per unit
Marginal average profit
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P ( x) 
P ( x)
P '( x) 
x
d
P ( x)
dx
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Warning!
To calculate the marginal averages you must
calculate the average first (divide by x), and
then the derivative. If you change this order
you will get no useful economic
interpretations.
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Example 2
The total cost of printing x dictionaries is
C(x) = 20,000 + 10x
1. Find the average cost per unit if 1,000
dictionaries are produced.
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Example 2(continued)
The total cost of printing x dictionaries is
C(x) = 20,000 + 10x
1. Find the average cost per unit if 1,000
dictionaries are produced.
C ( x) 
C ( x)

20 , 000  10 x
x
C (1, 000 ) 
x
20 , 000  10 , 000
1, 000
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= $30
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Example 2(continued)
2. Find the marginal average cost at a production level of 1,000
dictionaries, and interpret the results.
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Example 2(continued)
2. Find the marginal average cost at a production level of 1,000
dictionaries, and interpret the results.
Marginal average cost =
C '( x) 
d
C ( x)
dx
 20000
d  20000  10 x 
C '( x) 

 
2
dx 
x
x

C ' (1000 ) 
 20000
1000
2
  0 . 02
This means that if you raise production from 1,000 to 1,001
dictionaries, the price per book will fall approximately 2 cents.
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Example 2(continued)
3. Use the results from above to estimate the average cost per
dictionary if 1,001 dictionaries are produced.
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Example 2(continued)
3. Use the results from above to estimate the average cost per
dictionary if 1,001 dictionaries are produced.
Average cost for 1000 dictionaries = $30.00
Marginal average cost = - 0.02
The average cost per dictionary for 1001 dictionaries would be
the average for 1000, plus the marginal average cost, or
$30.00 + $(- 0.02) = $29.98
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Example 3
The price-demand equation and the cost function for the
production of television sets are given by
p ( x )  300 
x
and
C ( x )  150 , 000  30 x
30
where x is the number of sets that can be sold at a price of $p
per set, and C(x) is the total cost of producing x sets.
1.
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Find the marginal cost.
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Example 3(continued)
The price-demand equation and the cost function for the
production of television sets are given by
p ( x )  300 
x
and
C ( x )  150 , 000  30 x
30
where x is the number of sets that can be sold at a price of $p
per set, and C(x) is the total cost of producing x sets.
1.
Find the marginal cost.
Solution: The marginal cost is C’(x) = $30.
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Example 3(continued)
2. Find the revenue function in terms of x.
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Example 3(continued)
2. Find the revenue function in terms of x.
The revenue function is
R ( x )  x  p ( x )  300 x 
x
2
30
3. Find the marginal revenue.
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Example 3(continued)
2. Find the revenue function in terms of x.
The revenue function is
R ( x )  x  p ( x )  300 x 
x
2
30
3. Find the marginal revenue.
The marginal revenue is R ' ( x )  300 
x
15
4. Find R’(1500) and interpret the results.
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Example 3(continued)
2. Find the revenue function in terms of x.
The revenue function is
R ( x )  x  p ( x )  300 x 
x
2
30
3. Find the marginal revenue.
The marginal revenue is R ' ( x )  300 
x
15
4. Find R’(1500) and interpret the results.
R ' (1500 )  300 
1500
 $ 200
15
At a production rate of 1,500, each additional set increases
revenue by approximately $200.
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Example 3(continued)
5. Graph the cost function and the revenue function on the
same coordinate. Find the break-even point.
0 < x < 9,000
0 < y < 700,000
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Example 3(continued)
5. Graph the cost function and the revenue function on the
same coordinate. Find the break-even point.
0 < x < 9,000
R(x)
0 < y < 700,000
Solution: There are two
break-even points.
C(x)
(600,168,000)
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(7500, 375,000)
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Example 3(continued)
6. Find the profit function in terms of x.
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Example 3(continued)
6. Find the profit function in terms of x.
The profit is revenue minus cost, so
7. Find the marginal profit.
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P ( x)  
x
2
 270 x  150000
30
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Example 3(continued)
6. Find the profit function in terms of x.
The profit is revenue minus cost, so P ( x )  
x
2
 270 x  150000
30
7. Find the marginal profit.
P ' ( x )  270 
x
15
8. Find P’(1500) and interpret the results.
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Example 3(continued)
6. Find the profit function in terms of x.
The profit is revenue minus cost, so P ( x )  
x
2
 270 x  150000
30
7. Find the marginal profit.
x
P ' ( x )  270 
15
8. Find P’(1500) and interpret the results.
P ' (1500 )  270 
1500
 170
15
At a production level of 1500 sets, profit is increasing at a rate of
about $170 per set.
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QUIZZ Time
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§ 9.5 Derivates of Products and Quotients
The student will learn about:
the derivative of a product of two functions, and
the derivative of a quotient of two functions.
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Derivates of Products
The derivative of the product of two functions is the first
function times the derivative of the second function plus the
second function times the derivative of the first function.
Theorem 1 - Product Rule
If
f (x) = F (x) • S (x),
Then f’(x) = F (x)’ • S (x) + F (x) • S’ (x), OR
f’(x) = F
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dS
dx
+ S
dF
dx
68
Example
Find the derivative of y = 5x2(x3 + 2).
If
Product Rule
f (x) = F (x) • S (x),
Then f ’ (x) = F (x)’ • S (x) + F (x) • S’ (x).
Let F (x) = 5x2 then F ‘ (x) = 10x
Let S (x) = x3 + 2 then S ‘ (x) = 3x2, and
f ‘ (x) = 5x2 • 3x2 + (x3 + 2) • 10x
= 15x4 + 10x4 + 20x = 25x4 + 20x
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Derivatives of Quotients
The derivative of the quotient of two functions is the
bottom function times the derivative of the top function
minus the top function times the derivative of the bottom
function, all over the bottom function squared.
Theorem 2. Quotient Rule:
If y = f (x) = T (x) / B (x),
Then
f '(x) 
B (x)  T '(x)  T (x)  B '(x)
[ B (x) ]
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Derivatives of Quotients
f '(x) 
B (x)  T '(x)  T (x)  B '(x)
[ B (x) ]
2
May also be expressed as dy
dx
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B

dT
 T
dx
dB
dx
B
2
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Example
3x
Find the derivative of y 
f '(x) 
2x  5
.
B (x)  T '(x)  T (x)  B '(x)
[ B (x) ]
2
Let T (x) = 3x and then T ‘ (x) = 3.
Let B (x) = 2x + 5 and then B ‘ (x) = 2.
f '(x ) 
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(2x  5) 3  3x  2
(2x  5)
2

15
(2x  5)
2
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Summary.
Product Rule. If F (x) and S (x), then
d
dx
F ( x )  S ( x )   F (x) • S‘(x) + S (x) • F‘(x).
Quotient Rule. If T(x) and B(x), then
d  T (x) 

 
dx  B ( x ) 
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B
dT
T
dx
dB
dx
B
2
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Practice Problems
§9.5
1, 5, 9, 13, 17, 19, 23, 27, 31, 35, 39, 43, 47, 51,
55, 65, 69, 71.
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