Chain Rule: Power Form
Marginal Analysis in Business and
Economics
The student will learn about:
the chain rule, combining different rules of
derivation, and an application.
Marginal cost, revenue, and profit as well as,
applications, andmarginal average cost,
revenue and profit.
Dr .Hayk Melikyan
Departmen of Mathematics and CS
melikyan@nccu.edu
H.Melikian
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Chain Rule: Power Rule.
We have already made extensive use of the
power rule with xn,
d n
n 1
dx
x  nx
We wish to generalize this rule to cover [u (x)]n.
That is, we already know how to find the derivative
of
f (x) = x 5
We now want to find the derivative of
f (x) = (3x 2 + 2x + 1) 5
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Chain Rule: Power Rule.
General Power Rule. [Chain Rule]
Theorem 1. If u (x) is a differential function, n
is any real number, and
y = f (x) = [u (x)]n
then
f ’ (x) = n[ u (x)]n – 1 u’ (x) = n un – 1u’
or
d n
n  1 du
u  nu
dx
dx
* * * * * VERY IMPORTANT * * * * *
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Example 1
Find the derivative of y = (x3 + 2) 5.
NOTE: If we let u = x 3 + 2, then y = u 5.
Chain Rule
d n
n  1 du
u  nu
dx
dx
Let u (x) = x3 + 2, then y = u 5 and u ‘ (x) = 3x2
d
3
5
( x  2)  5 (x3 + 2)4 3x2
dx
= 15x2(x3 + 2)4
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Example 2
Find the derivative of y = x 3  3
Rewrite as y = (x 3 + 3) 1/2
Then y’ = 1/2 (x 3 + 3) – 1/2 (3x2)
= 3/2 x2 (x3 + 3) –1/2
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Combining Rules of
Differentiation
The chain rule just developed may be used in
combination with the previous rules for taking
derivatives
Some examples follow.
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Example 3
Find f ’ (x) if f (x) =
x4
( 3x  8)
2
.
We will use a combination of the quotient rule and
the chain rule.
Let the top be t (x) = x4, then t ‘ (x) = 4x3
Let the bottom be b (x) = (3x – 8)2, then using the
chain rule b ‘ (x) = 2 (3x – 8) 3 = 6 (3x – 8)
f ' ( x) 
f ' ( x) 
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( 3x  8) 2 ( 4x 3 )  x 4 6 ( 3x  8)
(( 3x  8)2 )2
( 3 x  8) 2 ( 4x 3 )  6x 4 ( 3x  8)
( 3x  8)4
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Example 4
Find f ’ (x) and find the equation of the line tangent
to the graph of f at the indicated value of x.
f (x) = x2 (1 – x)4; at x = 2.
We will use the point-slope form. The point will
come from (2, f(2)) and the slope from f ‘ (2).
Point - When x = 2, f (x) = 22 (1 – 2)4 = (4) (1) = 4
Hence the tangent goes through the point (2,4).
f ‘ (x) = x2 4 (1 – x)3 (-1) + (1 – x)4 2x
= - 4x2 (1 – x)3 + 2x(1 – x)4 and
f ‘ (2) = (- 4) (4) (-1)3 + (2) (2) (-1)4 = 16 + 4 = 20 = slope
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Example 4 continued
Find f ’ (x) and find the equation of the line tangent
to the graph of f at the indicated value of x.
f (x) = x2 (1 – x)4; at x = 2.
We will use the point-slope form. The point is (2, 4)
and the slope is 20.
y – 4 = 20 (x – 2) = 20x – 40 or
y = 20x - 36.
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Example 4 continued
Find f ’ (x) and find the equation of the line tangent
to the graph of f at the indicated value of x.
f (x) = x2 (1 – x)4; at x = 2.
By graphing calculator!
Graph the function and use “Math”, “tangent”.
-2 ≤ x ≤ 3
-1 ≤ y ≤ 20
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Application
P. 202, #78. The number x of stereo speakers people
are willing to buy per week at a price of $p is given by
x = 1,000 - 60 p  25
for 20 ≤ p ≤ 100
1. Find dx/dp.
f ‘ (p) = - (60) (1/2) (p + 25)-1/2 (1)

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 30
p  25
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Application continued
The number x of stereo speakers people are willing to
buy per week at a price of $p is given by
x = 1,000 - 60 p  25
for 20 ≤ p ≤ 100
2. Find the demand and the instantaneous rate of
change of demand with respect to price when the
price is $75.
That is, find f (75) and f ‘ (75).
f (75) = 1,000 – 60
f ‘ (75) 
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75  25 = 1000 – 600 = 400
 30
= -30/10 = - 3
75  25
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Application continued
The number x of stereo speakers people are willing to
buy per week at a price of $p is given by
x = 1,000 - 60 p  25
for 20 ≤ p ≤ 100
3. Give a verbal interpretation of these results.
With f (75) = 400 and f ‘ (75) = - 3 that means
that the demand at a price of $75 is 400 speakers and
each time the price is raised $1, three fewer speakers
are purchased.
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Summary.
If
y = f (x) = [u (x)]n
then
H.Melikian
d n
n  1 du
u  nu
dx
dx
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Marginal Cost, Revenue, and Profit
Remember that margin refers to an
instantaneous rate of change, that is, a
derivative.
Marginal Cost
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 Cost, Revenue, and Profit
Marginal Revenue
If x is the number of units of a product sold in
some time interval, then
Total revenue = R (x)
Marginal revenue = R’ (x)
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Marginal Cost, Revenue, and Profit
Marginal Profit
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.
Theorem 1.
C (x) is the total cost of producing x items and
C (x + 1) is the cost of producing x + 1 items.
Then the exact cost of producing the x + 1st
item is
C (x + 1) – C (x)
The marginal cost is an approximation of the
exact cost. Hence,
C ’ (x) ≈ C (x + 1) – C (x).
The same is true for revenue and profit.
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Example 1
P. 210, #2. The total cost of producing x
electric guitars is
C (x) = 1,000 + 100 x – 0.25 x2
1. Find the exact cost of producing the 51st
guitar.
Exact cost is C (x + 1) – C (x)
C (51) = $5449.75
C (50) = $5375.00
Exact cost = $5449.75 - $5375 = $74.75
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Example 1 continued
The total cost of producing x electric guitars
is
C (x) = 1,000 + 100 x – 0.25 x2
2. Use the marginal cost to approximate the
cost of producing the 51st guitar.
The marginal cost is C ‘ (x)
C ‘ (x) = 100 – 0.5x
C ‘ (50) = $75.00
Exact cost = $5449.75 - $5375 = $74.75
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Marginal Average Cost
If x is the number of units of a product
produced in some time interval, then
C ( x)
Average cost per unit = C ( x) 
x
d
C ( x)
Marginal average cost = C ' ( x) 
dx
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Marginal Average Revenue
If x is the number of units of a product sold in
some time interval, then
R ( x)
Average revenue per unit = R ( x) 
x
d
Marginal average revenue = R ' ( x) 
R ( x)
dx
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Marginal Average Profit.
If x is the number of units of a product
produced and sold in some time interval, then
P ( x)
Average profit per unit = P ( x) 
x
d
Marginal average profit = P ' ( x ) 
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
P. 210, # 4. 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) 

x
20000  10x
x
20,000  10,000
C (1000) 
= $30
1000
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Example 2 continued
The total cost of printing x dictionaries is
C (x) = 20,000 + 10x
2. Find the marginal average cost at a production
level of 1,000 dictionaries, and interpret the results.
d
C ( x)
Marginal average cost = C ' ( x) 
dx
d  20000  10 x   20000
C ' ( x) 


2
x
dx 
x

 20000
C ' (1000) 
  0.02
2
1000
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What does
this mean?
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Example 2 continued
The total cost of printing x dictionaries is
C (x) = 20,000 + 10x
3. Use the results from above to estimate the
average cost per dictionary if 1,001 dictionaries are
produced.
Average cost = $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
P. 211, #14. The price-demand equation and the
cost function for the production of television sets are
given, respectively by
x
p (x) = 300 30
and C (x) = 150,000 + 30x
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.
The marginal cost is C ‘ (x) so
C ‘ (x) = $30.
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What does
this mean?
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
x
p (x) = 300 30
and C (x) = 150,000 + 30x
2. Find the revenue function in terms of x.
The revenue function is R (x) = x · p (x), so
R ( x )  300 x  x
2
30
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
2
x and C (x) = 150,000 + 30x
R ( x)  300x 
30
3. Find the marginal revenue.
The marginal revenue is R ‘ (x), so
x
R' ( x )  300 
15
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
x
R' ( x)  300 
and C (x) = 150,000 + 30x
15
4. Find R’ (1,500) and interpret the results.
1500
 $200 What does
R' (1500)  300 
15
this mean?
At a production rate of 1,500 sets, revenue is
increasing at the rate of about $200 per set.
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
2
C (x) = 150,000 + 30x
x
and R ( x)  300x 
30
5. Graph the cost function
and the revenue
function on the same
coordinate. Find the
break-even point.
0 ≤ x ≤ 9,000
(600,168000)
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0 ≤ y ≤ 700,000
(7500, 375000)
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
C (x) = 150,000 + 30x
x2
and R ( x)  300x 
30
6. Find the profit function in terms of x.
The profit is revenue minus cost, so
2
x
P ( x)  300x 
 150000  30x
30
x2
P ( x)  
 270x  150000
30
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
x2
P ( x)  
 270x  150000
30
7. Find the marginal profit.
The marginal profit is P ‘ (x), so
x
P ' ( x)  270 
15
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Example 3 continued
The price-demand equation and the cost function for
the production of television sets are given,
respectively by
x
P ' ( x)  270 
15
7. Find P’ (1,500) and interpret the results.
1500
 170
P '(1500)  270 
15
What does
this mean?
At a production level of 1500 sets, profit is
increasing at a rate of about $170 per set.
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Summary.
In business the instantaneous rate of change,
the derivative, is referred to as the margin.
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