BT1093 Matematik Perniagaan (Business Mathematics)

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BT1093 Matematik Perniagaan(Business
Mathematics)

• DIFFERENTIATION
• Week 9, Semester 1, 2006/2007
• School of Business Economics, Universiti
  Malaysia Sabah

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Slopes tangents
y
tangents
x
200
The slope of tangents to any curve can be
calculated using differentiation.
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Rules for Differentiation
Notation The derivative of a function is denoted
by
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Rules for Differentiation
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Rules for Differentiation (cont.)
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Rules for Differentiation (cont.)
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Rules for Differentiation (cont.)
Special Cases
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The introductory example has
represents the slope of the tangent to the
curve y.
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When the tangent is horizontal (slope 0) then
The maximum occurs at x 200.
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Using differentiation to find the maximum and
minimum points is a more general method in that
it works for all curves.
Using is a shortcut method that
works for quadratic equations (parabolas) only.
The equation of the tangent is not difficult to
determine. At a point on a curve, the
slope is calculated using differentiation and the
result is used in the equation
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Example Find the equation of the tangent to the
curve when x 2.
When x 2, and the point on the curve is (2,9)
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So using
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Example Find all the points on the curve
where the slope is 6.
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Example Find the derivative of the following
function
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Example Find the derivative of the following
function
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Example Find the derivative of the following
function
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The Derivative as a Measure of Rate of Change
The derivative as a measure of the gradient of a
tangent to a curve has been discussed.
A second interpretation of this process describes
the rate of change of some variable.
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Consider the following physics example
A cars position (s) is described by the
equation where s is in metres and t in seconds
After 2 seconds, (t 2),
The car is 4 metres from some starting point.
At t 10, s 102 100 and so on.
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Looking more closely at the case when
t 10, s 100, the question is asked How
fast was the car travelling?
One answer is 100m in 10 seconds equals
100/10 10 m/s. This describes the average
velocity (velocity is a measure of the rate of
change of position(s) ). But as with any car
journey, the velocity is always changing.
Differentiation allows us to calculate the
velocity at any time instant.
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t (secs) s (m)
1 1 2
3 9 6
5 25 10
7 49 14
10 100 20
The derivative describes the rate of change of
position (instantaneous velocity) at any time
In summary
Average velocity 100/10 10 m/s
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Consider the following situation
The total cost function is given by
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The average cost for producing q items is given by
( is the usual symbol for average)
That is,
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How can the derivative be used and interpreted?
describes how quickly cost is changing
A comparison shows that cost is changing at a
faster rate when q 10 than when q 1000.
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These figures also give an approximate cost to
produce the next item.
The 11th item will cost approximately 9990
to produce.
The 1001st item will cost approximately
9000 to produce.
When C describes the total cost, then represents
the marginal cost .
is called the marginal cost function.
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Example
Marginal cost function
Marginal cost
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Example
Total cost function
Marginal cost function
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Marginal cost
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There is an equivalent interpretation for the
revenue function.
If R represents total revenue then
is called the marginal revenue.
Marginal revenue describes two things
(1). How quickly revenue is changing.
(2). Approximate revenue received by selling the
next unit.
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Example
Marginal revenue
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The following conclusions can be drawn from these
results
(1). Revenue is changing faster at q 10 than at
q 20.
(2). The 11th item will generate approximately
56 and the 21st item will earn 52
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Derivates of Products
The derivative of the product of 2 functions is
the 1st function times the derivative of the 2nd
function plus the 2nd function times the
derivative of the 1st function.
Theorem 1 - Product Rule
If f (x) F (x) S (x),
Then f (x) F (x) S (x) F (x) S (x),
OR
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Example
Find the derivative of y 5x2(x3 2).
Product Rule
If f (x) F (x) S (x),
Then f (x) F (x) S (x) F (x) S (x).
10x
Let F (x) 5x2 then F (x)
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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Example
Use the product rule to find the slope of graph
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Derivatives of Quotients
The derivative of the quotient of 2 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
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Derivatives of Quotients
May also be expressed as -
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Example
Find the derivative of .
Let T (x) 3x and then T(x) 3
Let B (x) 2x 5 and then B(x) 2
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Chain Rule
1. Chain Rule If y f (u) and u g (x),
define the composite function y f (u) f g
(x), then
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Example
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Example
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Power Rule
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
VERY IMPORTANT
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Example 1
Find the derivative of y (x3 2) 5.
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Example 2
Find the derivative of y
Rewrite as y (x 3 3) 1/2
Then y 1/2 (x 3 3) 1 1/2 (3x2)
3/2 x2 (x3 3) 1/2
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Example 3
Find f(x) if f (x)
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)
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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 (Contd.)
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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Application
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
for 20 p 100

1. Find dx/dp.

(1/2)
(1)
f (p)
- (60)
(p 25)-1/2
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Application continued
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
1000 600 400
f (75)
-30/10 - 3
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Application continued
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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Tutorial Questions
11.2 ?3,7,17,23,31,32,75,79 11.3
?9,19,21,23,27,41 11.5 ?1,19,27,31,49,52,69,71,77
11.6 ?1,5,9,19,27,39,41,57,69,70,71