Download 3.1 Introduction Before we define derivatives, it is appropriate to first

BQT 133-Business Mathematics
Teaching Module
CHAPTER 3 : DIFFERENTIAL CALCULUS
3.1
Introduction
Before we define derivatives, it is appropriate to first define and introduce
the concept of the tangent line to a function
Definition 3.1
Consider the function y  f ( x) , and suppose a is a point in the domain of f , so
(a, f (a)) is a point on the graph y  f ( x) . The tangent line to y  f ( x) at the
point (a, f (a)) is the straight line through (a, f (a)) whose slope is
f ( x)  f (a)
provided the limit exists. In particular, if the limit m does
m  lim
x a
xa
exist, the equation of the line is y  f (a)  m( x  a) or equivalently,
y  f ( a )  m( x  a )
The slope of the straight line PQ= m 
the changein y-values
the changein x-values
43
1


62
4
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So, the equation for straight line PQ given by
1
y  mx  c gives y = x  c
4
Since straight line passing through point P  2,3  or Q  6, 4  ,
1
1) Through point P  2,3  , y = x  c
4
1
2 5
3=  2   c implies c  3  
4
4 2
1
5
y= x 
4
2
1
2) Through point Q  6, 4  , y = x  c
4
1
6 10 5
4=  6   c implies c  4  

4
4 4 2
1
5
y= x 
4
2
Example 3.1
Find the equation of the tangent line to the curve f ( x)  x 2  4 x  1 , at the
point (1, 4) .
Solution
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Activity: Locate the point (1, 4) on the graph given and draw a tangent line
to the curve at that point.
f ( x)  f (a )
f ( x)  f (1)
 lim
x a
x1
xa
x 1
2
2
f ( x)  f (1)
( x  4 x  1)  (1  4(1)  1)
m  lim
 lim
x1
x1
x 1
x 1
2
x  4x  3
( x  1)( x  3)
 lim
 lim
 lim( x  3)  2
x1
x1
x1
x 1
x 1
The slope of the tangent line is m  lim
Since the line passes through the point (1, 4) , the equation of the line is,
y  f a  m x  a
 4   2  x  1
 2 x  2
Definition 3.2
The derivative (or differentiation) of a function f ( x) at a point x  a ,
written f '(a) , is defined as:
f ( x)  f (a)
f '(a)  lim
x a
xa
provided the limit exists.
The process of getting the tangent line is called differentiation from the first
principles or differentiation using definition.
NOTE:
dy
is called the “derived function” or the
dx
“derivative” or the “differential coefficient” of y with respect to x .
dy
Then the process of finding f '( x) or
is called “differentiation”.
dx
1. If y  f ( x) , then f '( x) or
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TWO IMPORTANT FORMULA !!
The derivative (or differentiation) of a function f ( x)
1. For specific point x=a, the derivative given by
f ( x)  f (a)
x a
xa
provided the limit exists and h is a small increment.
f '(a)  lim
2. In general, the derivative given by
f ( x  h)  f ( x )
f '( x)  lim
h0
h
provided the limit exists and and h is a small increment.
Example 3.2
Find the derivative of f ( x)   x 2  4 x at x  7 using the First Principle
Method.
Solution
f ( x)  f (a)
,
x a
xa
f ( x)  f (7)
 x 2  4 x  (72  4(7))
so f '(7)  lim
 lim
x7
x7
x7
x7
f '(a)  lim
 x 2  4 x  21
( x  7)( x  3)
 lim
 lim
x7
x7
x7
x7
 lim( x  3)  10
x7
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Example 3.3
Find the derivative
Principle/Definition.
of
the
function f ( x)  x 2 ,
using
the
First
Solution
f ( x  h)  f ( x )
( x  h) 2  x 2
f '( x)  lim
 lim
h0
h0
h
h
2
2
2
x  2 xh  h  x
 lim
h0
h
2
2 xh  h
 lim
h0
h
 lim 2 x  h
h0
 2x
Exercise 3.1
Find the derivatives of the following functions using definitions.
1
x
2
(b) f ( x)  3 x
2
(c) f ( x)  2 at x  2
3x
(a) f ( x) 
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3.2 Differentiation Formulas
1.
d
(c)  0 or f (c)  0 , where c is constant function.
dx
(Derivative of a Constant Function)
2.
d n
( x )  nx n 1 or f ( x n )  nx n1 , where n is any number either integer
dx
or rational. (The Power Rule)
3.
d
cf ( x)  c d f ( x) or cf ( x)  cf ( x) , where c is constant function.
dx
dx
(The Constant Multiple Rule)
4.
d
 f ( x)  g ( x)  d f ( x)  d g ( x)
dx
dx
dx
or  f ( x)  g ( x)  f ( x)  g ( x)
(The Sum Rule)
5
d
 f ( x)  g ( x)  d f ( x)  d g ( x)
dx
dx
dx
or  f ( x)  g ( x)  f ( x)  g ( x)
(The Difference Rule)
6
d
 f ( x)  g ( x)  f ( x) d g ( x)  g ( x) d f ( x)
dx
dx
dx
or  f ( x)  g ( x)  f ( x) g ( x)  g ( x) f ( x)
(The Product Rule)
7
d  f ( x) 

dx  g ( x) 
g ( x)
d
d
f ( x)  f ( x) g ( x)
dx
dx
2
 g ( x) 

 f ( x) 
g ( x) f ( x)  f ( x) g ( x)

or 

g ( x)2
 g ( x) 
(The Quotient Rule)
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3.2.1 Differentiation of standard functions
Example 3.4
Differentiate for this function f(x) = 180
Solution
Using property 1, we get f '( x)  0
Example 3.5
Find the derivative of f ( x)  x10 .
Solution
Using property 2, we get f '( x)  10 x9
3.2.2 The sum/difference rule
d
d
d
 f ( x)  g ( x)   f ( x)  g ( x)
dx
dx
dx
d
d
d
 f ( x)  g ( x)   f ( x)  g ( x)
dx
dx
dx
Example 3.6
1
3
Differentiate x  3x  2 x 2
8
Solution
Using property 2, 3 and 4, to solve this problem
1
d 8
d
d
x  3x  2 x  ( x )  (3x 3 )  (2 x 2 )
dx
dx
dx
8
1
3
2
using property 4
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d 8
d 13
d
( x )  3 ( x )  2 ( x 2 )
dx
dx
dx
2
1
= 8 x 7  3   ( x 3 )  2(2)( x 3 )
3
=
= 8x  x
7
2
3
using property 3
 4 x 3
Exercise 3.2
Find the derivative of the following functions.
(a) y  5 x9
(b) y  3 x
(c) f ( x)  x 3  2
4
5
(d) f ( x)  3x 
3.3
1
3 x

2
1
x2
Function of a Function
3.3.1 Products Rule
d
d
d
 f ( x)  g ( x)   f ( x) g ( x)  g ( x) f ( x)
dx
dx
dx
Example 3.7
Find f ( x) if f ( x)  (6 x3 )(7 x 4 )
Solution
d
d
(7 x 4 )  7( x 4 ) (6 x3 )
dx
dx
3
3
4
= (6 x )(28 x )  7( x )(18 x 2 )
f ( x)  (6 x3 )
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=168 x6  126 x6  294 x6
Exercise 3.3
Find the derivative of the following functions.
(a) f ( x)  (5 x3  2)( x  1)
(b) f ( x)  (3x 2  7)(6  5x)
3.3.2 Quotients Rule
d  f ( x) 

dx  g ( x) 
g ( x)
d
d
f ( x)  f ( x) g ( x)
dx
dx
2
 g ( x) 
Example 3.8
x2  x  2
Find f ( x) if f ( x) 
x3  6
Solution
d  x2  x  2 
f ( x)   3

dx  x  6 
d
d
( x3  6)  x 2  x  2   ( x 2  x  2) ( x 3  6)
dx
dx
=
2
3
 x  6
=
=
( x3  6)  2 x  1  ( x 2  x  2)(3x 2 )
x
3
 6
2
( x 4  2 x3  6 x 2  12 x  6
x
3
 6
2
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Exercise 3.4
Find the derivative of the following functions.
2 x2  3
(a) f ( x) 
2x  3
(b) f ( x) 
( x  1)
( x  1)
3.3.3 Chain Rule
Suppose that we have two functions f(x) and g(x) and they are both differentiable.
1. If we define F ( x)  ( f g )( x) then the derivative of F(x) is,
F ( x)  f ( g ( x))( x) g ( x)
2. If we have y=f(u) and u=g(x) then the derivative of y is
dy dy du


dx du dx
Example 3.9
Find f ( x) if f ( x)  x 2  1
Solution
Let x 2  1  u ( x)
dy
1 21
1
 f (u )  u 
So, y (u )  f (u )  u then we get
du
2
2 u
du
Also
 u( x)  2 x .
dx
By using Chain Rule we get,
dy dy du
1
2x



 2x 

2
dx du dx 2 u
2 x 1
x
x2  1
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Example 3.10
Find the derivative of f ( x)  sin(3x 2  x) .
Solution
Let 3x 2  x  u ( x)
dy
 f (u )  cos(u )
du
So, y(u )  f (u )  sin(u ) then we get
Also
du
 u( x)  6 x  1 .
dx
By using Chain Rule we get,
dy dy du


 cos(u )  6 x  1  (6 x  1)(cos(3x 2  x))
dx du dx
Exercise 3.4
Find
dy
of the following functions using the chain rule.
dx
(a) y  ( x  3)6
y  (4 x5  3x 2  2 x  1)3
(b)
ans  3 4 x5  3x 2  2 x  1  20 x 4  6 x  2 
2
y  2 x2  1
(c)
ans 
2x
2 x2  1
3.3.4 The Power Rule
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This rule is actually another version of chain rule which is shorter and
simpler. It is stated in the following theorem and illustrated in the next
example.
If f ( x) is differentiable at the point x and y  [ f ( x)]n for any real
number n, then
dy
 n[ f ( x)]n1 f '( x)
dx
We can also write
d
[ f ( x)]n  n[ f ( x)]n1 f '( x)
dx
Example 3.11
Using the power rule differentiate the functions y  ( x  3)6
Solution
y  ( x  3)6
dy
d
 6( x  3)5 ( x  3)
dx
dx
 6( x  3)5
Exercise 3.5
Find
dy
of the following functions using the power rule.
dx
2
(a) y 
(3x  1)
2
1
5
Extra exercise
Differentiate the following functions with respect to x.
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(a) y  (8 x  5)4 ( x3  3)12
(3 x  1)5
(b) y 
(2  x)8
3.3.5 Derivatives of Trigonometric Functions
d
(sin x)  cos x (Derivative of the sine function)
dx
d
2.
(cos x)   sin x (Derivative of the cosine function)
dx
d
3.
(tan x)  sec2 x (Derivative of the tangent function)
dx
d
4.
(sec x)  sec( x) tan( x) (Derivative of the secant function)
dx
d
5.
(Derivative of the cotangent function)
(cot x)   csc2 ( x)
dx
d
6.
(Derivative of the cosecant
(csc x)   csc( x)cot( x)
dx
function)
1.
Example 3.12
Differentiate of following functions
(a)
y( x)  5sin( x)cot( x)  4csc( x)
(b)
y ( x) 
sin( x)
3  2cos( x)
Solution
(a)
d
d
d


y( x)  5 sin( x)  cot( x)   cot( x) (sin( x))   4  csc( x) 
dx
dx
dx


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= 5sin( x)( csc( x))  cot( x)(cos( x))  4   csc( x)cot( x) 
= 5cos( x)cot( x)  5csc( x)  4  csc( x)cot( x) 
y(t ) 
(b)
3  2cos(t )
d
d
 sin(t )   sin(t )  3  2cos(t ) 
dt
dt
2
 3  2cos(t ) 
=
 3  2cos(t )  cos(t )   sin(t )  2sin(t ) 
2
 3  2cos(t ) 
=
 3cos(t )   2
2
 3  2cos(t ) 
3.3.5 Logarithmic Differentiation
The basic differentiations of the logarithmic functions are as follows:
1.
d
1
(ln x) 
dx
x
2.
d
1 du
where u  f ( x)
(ln u ) 
dx
u dx
3.
d
1
1
(log a x) 
 (log a e)
dx
x ln a x
Example 3.13
Differentiate the following functions with respect to x.
(a) y  ln 7 x 2
(b) g ( x)  2 x3 ln(3x 2  2)
(c) y  ln(3x  2)6
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Solution
(a) y  ln 7 x 2
dy
1 d
 2 (7 x 2 )
dx 7 x dx
=
14 x 2

7 x2 x
(b) g ( x)  2 x3 ln(3x  2)
g '( x)  2 x3
d
d
ln(3x  2)  ln(3x  2) (2 x3 )
dx
dx
 2 x3 (
3
)  6 x 2 ln(3x  2)
3x  2
6 x3
(
)  6 x 2 ln(3 x  2)
3x  2
(using the product rule)
(c) y  ln(3x  2)6
dy
1
d

(3x  2)6
6
dx (3x  2) dx

1
18(3x  2)5
6
(3x  2)

18
(3x  2)
(using the power rule)
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Exercise 3.6
Differentiate the following functions with respect to x.
(a)
f ( x)  ln (2 x  1)
6x2
ans: f  
(2 x3  1)
(b)
 x 1 
y  ln 

 x 1
ans: y 
3.4
Application of Differentiations
3
2
( x  1)( x  1)
3.4.1 Marginal Cost, Revenue and Profit
One important use of calculus in business and economics is in
marginal analysis. In economics, the word marginal refers to rate of changes.
Thus if C ( x) is the total cost of producing x items, then C ( x) is called the
marginal cost and represents the instantaneous rate of change of total cost
with respect to the number of items produced. Similarly, the marginal
revenue is the derivative of the total revenue function and the marginal
profit is the derivative of the total profit function.
Marginal Cost, Revenue and Profit
If x is the number of units of a product produced in some time interval, then
total cost  C  x 
marginal cost  C   x 
total revenue  R  x   px , p is demand function(price per unit)
marginal revenue  R  x 
total profit  P  x   R  x   C  x 
marginal profit  P  x   R  x   C   x 
Marginal cost (or revenue or profit) is the instantaneous rate of change of
cost (or revenue or profit) relative to production at a given production level.
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Example 3.14
A company manufactures fuel tanks for automobiles. The total weekly cost
(in Ringgit) of producing x tanks is given by
C ( x)  10000  90 x  0.05x 2
a) Find the marginal cost function
b) Find the marginal cost at a production level of 500 tanks per week and
interpret the results
c) Find the exact cost of producing the 501st item.
Solution
a) C( x)  90  0.1x
b) C(500)  90  0.1 500   RM40
Interpretation
This means that at the production level of 500 fuel tanks, the cost
increasing at the rate RM 40 per fuel tank.
c) The exact cost of producing the 501st item  C (501)  C (500)
C (501)  10000  90  501  0.05  501  42539.95
2
C (500)  10000  90  500   0.05  500   42500.00
C (501)  C (500)  42539.95  42500.00  RM39.95
2
Example 3.15
The price p (in Ringgit) and the quantity demanded x for a particular clock
radio are related by the equation
x  4000  40 p
a) Express the price p in terms of the demand x.
b) Find the revenue R(x) from the sale of x clock radios.
c) Find the marginal revenue at a production level of 1600 clock radios
and interpret the results.
d) Find the marginal revenue at a production level of 2500 clock radios
and interpret the results.
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Answer
a) p  100 
x
40
x
x2

b) R  x   px  100   x  100 x 
40 
40

c) R  x   100 
2 1600 
2x
, R 1600   100 
 20
40
40
Interpretation
This means that at the production level of 1600 radios, th e revenue
increasing at the rate RM 20 per radio.
c) R  x   100 
2  2500 
2x
, R  2500   100 
 25
40
40
Interpretation
This means that at the production level of 2500 radios, the revenue
decreasing at the rate RM 25 per radio.
Example 3.16
1. The weekly demand for Pulsar DVD recorders is given by the demand
equation
p  0.02 x  300  0 x15000 
where p denotes the wholesale unit price in dollars and x denotes the
quantity demanded. The weekly total cost function associated with
manufacturing these recorders is
C ( x)  0.000003x3  0.04 x 2  200 x  70000
RM.
a) Find the revenue function R and the profit function P.
b) Find the marginal cost function, C  x  , the marginal revenue, R  x  and
the marginal profit function, P  x  .
c) Compute C  3000  , R  3000  and P  3000  . Interpret your result.
Answer
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a) R  x   0.02 x 2  300 x
P  x   R  x   C  x   0.02 x 2  300 x   0.000003x3  0.04 x 2  200 x  70000
 0.000003x3  0.02 x 2  100 x  70000
b) C   x   0.00000 x 2  0.08 x  200 ,
R  x   0.04 x  300
P  x   0.000009 x 2  0.04 x  100 .
c) C  3000   0.00000  3000   0.08  3000   200  41,
This means that, an additional cost RM 41 to produce one more DVD
recorders at any production level.
2
R  3000   0.04  3000  300  180
This means that at the production level of 3000 DVD recorders, the revenue
increasing at the rate RM 180 per DVD recorder.
Finally
P  3000   0.000009  3000   0.04 3000   100  139 .
This means that at the production levels of 3000, approximately changes in
profit per unit change in production is RM 139. That is, at the 3000 output
level, profit will be increased as the production increased.
2
Exercise
The price p (in Ringgit) and the cost function of television set are related by
the equation
x  9000  30 p and C  x   150000  30 x
a) Express the price p in terms of the demand x.
b) Find the marginal cost
c) Find the revenue R(x) from the sale of x television set.
d) Find the marginal revenue function
e) Find R  3000  and R  6000  . Interpret the result.
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