Transcript Differentiation - Arizona State University

Differentiation
Math 174, Spring
2004
Problems

In one-variable calculus, there are 2
basic problems
The “slope” of a curve
 Area under a curve

The 1st uses a mathematical process
called differentiation
 The 2nd uses integration
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College Algebra
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Rate of change of a linear function is
called “slope”
Denoted as m in y=mx+b
 How is it defined?

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Let’s try to get some insight into
rate of change of a line
Slope of a Line
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Let’s suppose I change x by 1 unit -- what
happens to y?
Our original equation is y0=mx+b
Now, we have
yn=m(x+1)+b
yn=mx+m+b
yn=mx+b+m
yn= y0+m
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This says that m is the amount y changes
when I change x by 1 unit
Slope of a Line
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If I change x by an arbitrary amount,  x, what
happens to y?
yn=m(x+x)+b
yn=mx+mx+b
yn=mx+b+mx
yn= y0+mx
yn- y0 = mx
(yn- y0 )/ x = m
y/ x = m
Rate of change of y with respect
to x
Now, rate of change of a line is constant -- what
happens when we generalize this idea to other
curves which are not linear?
Margins

In business, the term margin is used
to mean difference
The margin of page is the white space
beyond the printed part of the page
 A safety margin is extra space beyond
the boundaries that normally mark the
beginning of trouble or danger.
 In finance, the difference between the
selling price of an item and its cost

Example from Text
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We have a cost function that
describes the cost of preparing q
steak dinners
C(q)  9000 177q

0.633
We can use this function to calculate
the cost if we make more dinners
Marginal Cost

Let’s denote the marginal cost function,
MC(q)
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We will now look at it more closely
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This function was defined earlier as the price
for production at a given number of goods
This function measures the rate at which our
cost function is changing at q
One way to look at this is to see what
happens when we increase our quantity
by 1 -- what happens to cost?
First Plan
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Let’s take the very next dinner after the
qth dinner or (q+1)st
If q dinners are being prepared, the
marginal cost is the difference between
C(q+1) and C(q)
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This is just the cost of preparing the (q+1)st
dinner
For example, MC(1000) = $8.88 -- this
means that the 1001st dinner costs $8.88
to prepare
Graph of MC
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Now, MC(2000) = $5.93
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The marginal cost depends on q
MC(q) decreases with q
We can plot MC(q) = (C(q+1)-C(q))/1
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This is just the change in cost divided by the
change in the number of dinners
Interpretation?
Objections?
We defined marginal cost to be the
measure of the rate of change in the
cost function at a particular q
 We just arbitrarily looked at the very
next dinner -- what about the
change from (q-1) to q?
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MC(q) = C(q)-C(q-1)
Average Rate of Change
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It is not unreasonable for us to
average the two of these together
MC(1000) =

C (1001)  C (1000) C (1000)  C (999)

1
1
2
Simplifying the above expression, we
get just the change in cost over the
change in quantity
Smaller intervals
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We can only sell an integer number of
dinners -- however, we are modeling this
situation with a continuous function and
we want the rate of change at exactly
1000.
We are trying to find a model for the rate
of change at a particular quantity
Let’s let h be our change where h does
not have to be an integer
Difference Quotient
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C ( q  h)  C ( q  h)
2h
Then, we have
This is the average change over an
interval of length 2h
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Slope of a line – what line?
Usually called the difference quotient
As h gets smaller and smaller, we get
better approximation for MC(q)
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Slope of tangent line – what is a tangent line?
Instantaneous Rate of
Change
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MC(q) = instantaneous rate of change
In other words, we want h to be be
smaller and smaller (closer and closer to
0) and our quotient to stabilize
Mathematically, this is denoted as
MC(q)  lim
h 0

C ( q  h )  C ( q  h)
2h
When the quotient stabilizes, we get a
number which is called the derivative
Derivative
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Let f be a function which is defined on an
open interval containing a real number x
Suppose that f ( x  h)  f ( x  h) approaches a
2h
number m as h is taken to be smaller and
smaller
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For our situation, take h = 0.0001
f ( x  h)  f ( x  h)
2h
Thus, m = h0
m is called the derivative of f at x
lim

Denoted by f’(x) or df/dx
Tangent Line

Slope of tangent line is equal to the
derivative at every point x
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m = f’(x), where m is the slope of the tangent
line
Since we know the slope and a point on
the line, we can find the equation of the
tangent line
If the derivative at the point exists, then
the tangent line to the graph of f at the
point (a, f(a)) has the equation
y=f’(a)(x-a) +f(a)
Example

Find the slope of the line tangent to
the graph of f ( x)  x2  4x  4
at the point (3, f(3)).
 Find an equation for the tangent line
at that point.
Example #3 from text
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Let f(x) = x3 + 6
Graph both f(x) and its derivative on the
same axes on the interval (-3,3)
Calculate some values of the derivative
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First calculate a function that represents the
derivative
Luckily, you don’t have to do this by hand
every time
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Differentiating.xls
Interpretation
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The derivative is interpreted as the
instantaneous rate of change in f(x)
with respect to x
For the business community, marginal
demand is the derivative of the demand
function
 Similar for the cost, revenue, and profit
functions
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Marginal Profit
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Since we are concerned with maximum
profit, let’s look at MP(q)
P(q)=R(q)-C(q)
Profit increases with more dinners if the
increase in revenue per dinner is greater
than the increase in cost per dinner
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In other words, if the revenue per dinner is
growing faster than the cost per dinner, then
the profit increases
If MR(q)>MC(q)
Maximum Profit
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Profit decreases if increase in
revenue is less than increase in cost
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i.e., if MR(q) < MC(q )
Profit stops increasing and starts
decreasing at maximum value
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This implies that maximum profit must
occur when MR(q) = MC(q)
Example
D(q) = -0.1(q) + 150
 C(q) = 10,000 + 1100(x)1/2
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Let’s use Differentiating.xls to view
the graphs of R(q), MR(q), C(q),
MC(q), P(q) and MP(q)
Another view of Max Profit
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When P(q) is increasing, MP(q) is positive
When P(q) is decreasing, MP(q) is
negative
P(q) changes from increasing to
decreasing at maximum profit or where
MP(q) changes from positive to negative
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Thus, maximum profit must occur when
marginal profit is zero
i.e., max profit happens when MP(q) = 0
D(q) and MD(q)
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Demand function -- always
decreasing
If we increase q, then D(q+1) < D(q)
and change in q is positive
 Then, the derivative is negative
 Look at slope of tangent lines
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First general rule: If f is decreasing,
f’ is negative
Revenue
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Revenue graph is increasing to a
maximum and then decreasing
2nd general rule: When graph is
increasing, MR(q) is positive
 3rd general rule: When graph is at
maximum, MR(q) is zero
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So, what does the graph of MR(q)
look like?
Algebraic Rules
1.
2.
If f(x) = k (constant), then f’(x) = 0
If f(x) is linear, f(x) = mx + b, then
f’(x) = m

3.
4.
Why?
If f(x) = a*g(x), then f’(x) =a* g’(x)
If f(x) = g(x)  h(x), then
f’(x) = g’(x)  h’(x)
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Specifically, since P(q) = R(q) - C(q), then P’(q) =
R’(q) – C’(q) AND P’(q) = 0 when R’(q) = C’(q)
Potential Exam Questions
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Here are 2 curves, which is f(x) and which is
f’(x)?
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Calls on your knowledge of three rules that we
discussed
• Where the function is increasing, the derivative is positive,
etc.
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Here is a function, compute its derviative
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Use the limit defintion to compute
Here is a graph of f(x), sketch its derivative
graph
Here is the formal algebraic defintion of the
deriviative – explain the geometric definition
Find the equation of the tangent line
Focus on the Project
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We must consider the effect that
differentiation has on the units that we
are using
C(q) gives the cost, in millions of dollars,
for producing q thousand drives
When we calculate C’(q), this gives the
marginal cost, in millions of dollars per
thousand drives, for producing q
thousand drives
Focus on the Project
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We already have marginal cost
information (from variable cost data) that
gives us the marginal cost for q thousand
drives in dollars per drive
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However, MC(q) is giving us cost in millions of
dollars per thousand drives for producing q
thousand drives
We need to convert our units
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How?
Marginal Cost
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C’(q) is in millions of dollars per thousand drives
So, 1,000,000*C’(q) is in dollars per thousand
drives
And, (1,000,000* C’(q))/1000 is in dollars per
drive
The statement above simplifies to 1,000* C’(q)
So, where C is differentiable, 1,000*C’(q) gives
the marginal cost, MC(q) in dollars per drive, at
a production level of q thousand drives
MR(q) and MP(q)
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We can do a similar conversion for MR(q)
to get dollars per drive
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MR(q) = 1,000* R’(q) is the marginal revenue
in dollars per drive, when q thousand drives
are being sold
Since P(q) = R(q) – C(q), P’(q) = R’(q) C’(q)
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The units for C’(q) and R’(q) are both in
dollars per drive, for q thousand drives
These are the same units for MP(q)
MP(q) = 1,000* P’(q)
What should you do?
Read through slides 64-113 on
differentiation
 You should plot MR(q), MC(q), and
MP(q) for your team’s product
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Use these graphs to answer questions
1-4 for your product
 Do not forget to use the same units as
we have used for the class project
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