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dy δy , dx δx 7.1 If
f
THE CONCEPT OF DIFFERENTIATION defined as the function of
Then the derivative of
x
and can be written as
f ( x ) denoted as f ′( x ) or
‘derivative value of function
f
at
dy dx
f (x ) .
is read as
x ’. The f ′( x ) known as the FIRST
DIFFERENTIATION/ FIRST DERIVATIVE of f at x.
7.2
DIFFERENTIATION USING FIRST PRINCIPLE
(The concept of limit) Consider a function of of a function
f
, i.e
f
and assume that 2 points were at the curve
A = ( x, f ( x )) and B = ( x + h, f ( x ) + h )
y / f ( x)
Chord (AB)
((x+h), f(x+h)) = B
f(x+h)
Q
f(x)
0
Chapter 7: Differentiation
(x , f(x)) = A
Tangent (PQ)
P
x
x+h
x
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QQM1023 Managerial Mathematics From the diagram above, •
The straight line PQ is called a tangent to function
f
at the
point x . •
The slope / gradient of tangent
PQ is an approximation of
AB at that curve, i.e: f (x + h) − f (x) Slope / Gradient of AB = ⇒ (1) h When h → 0 , equation (1) can be written as chord
f (x + h ) − f (x ) ⇒ (2 ) h→0 h Since the gradient of function f at x equals to the gradient of lim
•
a tangent
PQ (equation (2 )), then equation (2 ) named as the
first principle differentiation or derivative :
f (x + h ) − f (x ) dy ′ = f ( x ) = lim h →0 dx h
Example 1:
If
f ( x ) = 6 x + 2 , find f ′( x ) by using the first principle
differentiation method. Solution: By using:
f ′( x) = lim h→0
Chapter 7: Differentiation
f ( x + h) − f ( x ) ⇒ (a) h
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Step 1:
f ( x ) = 6 x + 2 , then f ( x + h ) = 6( x + h ) + 2
Given
Step 2: Substitute
f ( x + h ) = 6( x + h ) + 2 in (a )
6( x + h) + 2 − f ( x) h→0 h
f ′(x) = lim
6(x + h) + 2 − [6x + 2] f ′(x) = lim h→0 h
6 x + 6h + 2 − 6 x − 2 h →0 h
f ′( x ) = lim
6h = lim 6 h→0 h h→0
f ′( x ) = lim
Using the concept of limits, had
f ′( x ) = 6
h →0
6 = 6 , therefore ;
Exercise: Find the first derivative ( f ′( x ) or
dy ) of the following functions by dx
using the first principle method: a)
f ( x ) = 3x − 1
b)
y = x2 − 1
Chapter 7: Differentiation
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DIFFERENTIATION RULES
7.3.1 DIFFERENTIATION RULES #1 : CONSTANT FUNCTION Constant function denoted by:
y = f (x ) = c where
c
is a constant.
Thus the derivative of a constant function at x:
dy = f ′( x ) = 0 dx Example 2: Find f ′( x ) or a)
dy for the functions below: dx
f (x) = 2 f ′( x ) = 0
b)
y = −5 dy =0 dx
7.3.2 DIFFERENTIATION RULES #2 : POWER RULES
y = f ( x ) = ax n Let
y = f ( x ) = ax n ,
where
a
is a constant. Therefore;
dy = f ′ ( x ) = nax n −1 dx
Chapter 7: Differentiation
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Example 3: Find
f ′( x )@
dy dx
y = x2
of the following functions:
answer⇒
a)
dy = 2x2−1 dx = 2x
[
f (x) = 3x 2
answer⇒ f ′( x) = 3 2x 2−1 = 3[2x]
b)
]
= 6x
y=
1 4 x 2
answer ⇒
[
dy 1 = 4 x 4 −1 dx 2 1 = 4x3 2 = 2x3
]
[ ]
c)
y=x d)
answer⇒ y = x1 dy = 1x1−1 = 1x0 dx = 1(1) = 1
f (x) = 9x e)
Chapter 7: Differentiation
[ ] [ ]
answer⇒ f ′(x) = 91x1−1 = 91x0 = 9(1) = 9
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QQM1023 Managerial Mathematics 7.3.3 DIFFERENTIATION RULES #3 : ADDITION/SUBSRACTION OF TWO FUNCTIONS If
y = f ( x ) = P[x ] ± Q[x ], hence the derivative:
dy d d = f ′( x ) = P[x ] ± Q[x ] dx dx dx Example 4: Find
f ′( x )@
dy dx
of the following functions:
f (x ) = x 2 + 5
a)
Ans:
f ′( x ) = 2 x 2 −1 + 0 = 2x
f (x ) = 3 x 2 − 4 x
b) Ans:
c)
[
y =
Ans:
] [
f ′( x ) = 3 2 x 2 −1 − 4 1 x1 −1 = 6x − 4
]
1 5 x − 2 x 3 + x − 11 5
[
] [
]
dy 1 = 5 x 5 −1 − 2 3 x 3 −1 + 1 x1−1 − 0 dx 5 = x4 − 6x2 + 1
Chapter 7: Differentiation
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7.3.4 DIFFERENTIATION RULES #4 : CHAIN RULES
(
y = f ( x ) = ax n ± b
If
)k
where
a, b, n and k
are constants,
then its derivative: (CHAIN RULE)
⇒
)
(
(
k −1 d dy = f ′( x ) = k ax n ± b × ax n ± b dx dx
Example 5: Find
f ′( x )@
dy dx
for the following functions:
f ( x ) = ( x + 1)3
a)
Ans:
f ′( x ) = 3( x + 1)3−1 ×
d ( x + 1) dx
f ′( x ) = 3( x + 1)2 × (1 + 0 ) = 3( x + 1)2
Chapter 7: Differentiation
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)
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b)
)
(
f (x) = x2 + 7 4 Ans:
(
f ′( x ) = 4 x + 7 2
(
)
4 −1
)3
×
(
d 2 x +7 dx
)
f ′( x ) = 4 x 2 + 7 × (2 x + 0 )
(
= 4(2 x ) x + 7
(
2
)
)
3
3 = 8x x2 + 7
c)
(
y = x2 + x
Ans:
)3
(
)
(
3−1 d 2 dy = 3 x2 + x × x +x dx dx
)
) 2 2 = 3(2 x + 1)(x + x ) 2 = (6 x + 3)(x 2 + x ) (
2 dy = 3 x 2 + x × (2 x + 1) dx
Chapter 7: Differentiation
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7.3.5 DIFFERENTIATION RULES #5 : PRODUCT OF TWO FUNCTIONS
y = f ( x ) = P[x ] × Q[x ], the derivative:
If
Let
u = P[x ] and v = Q[x ]
⇒
dy dv du = f ′( x ) = u + v dx dx dx
Example 6: Find a)
dy for the following functions: dx f ( x ) = ( x + 1)(3 x − 1)
f ′( x )@
Solution: let
u = ( x + 1)
, v = (3 x − 1)
du dv =1+ 0 =1 , = 3 − 0 = 3 dx dx du dv Substitute u , v, , in the formula; dx dx
Chapter 7: Differentiation
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dv du +v dx dx = ( x + 1)(3) + (3 x − 1)(1)
f ′( x ) = u
= 3x + 3 + 3x − 1 = 6x + 2 b)
)
(
y = x3 − 1 ( x + 5) Solution:
)
(
, v = ( x + 5)
u = x3 − 1
du dv = 3x 2 − 0 = 3x 2 , = 1 + 0 = 1 dx dx du dv Substitute u , v, , in the formula; dx dx dy dv du =u +v dx dx dx
(
)
( )
= x3 − 1 (1) + ( x + 5) 3 x 2 = x3 − 1 + 3 x3 + 15 x 2 = 4 x3 + 15 x 2 − 1 c)
y = ( x 2 + 3)( x − 2) 4
Chapter 7: Differentiation
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7.3.6 DIFFERENTIATION RULES #6 : QUOTIENT OF TWO FUNCTIONS (RATIONAL FUNCTION)
y = f (x ) =
If
Let
P[ x ] , then the derivative: Q[ x ]
u = P[x ] and v = Q[x ]
⇒
dy = f ′( x ) = dx
v
du dv − u dx dx v2
Example 7: Find
a)
dy for the following functions: dx (2 x − 1) y= ( x + 1)
f ′( x )@
Solution:
u = 2x − 1
,v = x +1
du dv = 2−0= 2 , =1+ 0 =1 dx dx du dv , in the formula; Substitute u , v, dx dx
Chapter 7: Differentiation
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dy = dx = =
b)
v
du dv − u dx dx
v2 ( x + 1)(2) − [(2 x − 1)](1)
( x + 1)2
=
2x + 2 − 2x + 1
( x + 1)2
3
( x + 1)2
f (x) =
x x2 + 7
Solution: Let,
u=x
, v = x2 + 7
du =1 dx
,
dv = 2x + 0 = 2x dx du dv , in the formula; Substitute u , v, dx dx f ′( x ) =
v
du dv − u dx dx v2
( x 2 + 7 )(1 ) − [x (2 x )] x 2 + 7 − 2 x 2 = = 2 2 (x + 7 ) (x 2 + 7 )2 =
Chapter 7: Differentiation
7 − x2
(x 2 + 7 )2 181
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x3
c)
y=
d)
f (x) =
( x + 1) 2 x+2 x
7.3.7 DIFFRERENTIATION RULES #7 : EXPONENTIAL FUNCTION Exponential function can be divided into two: i.
If
y = a f ( x ) where a ≠ 0 . Then the derivative: ⇒ Let u = f ( x ) then y = a u ; therefore
dy du = f ′( x ) = .au ln a dx dx Example 8: Find
a)
b)
f ′( x )@
dy dx
for the following functions:
y = 25 x
f (x) = 3
Chapter 7: Differentiation
x2
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ii.
y = e f ( x ) where e is the base to the natural log, then the
If
derivative:
⇒ Let u = f ( x ) then y = eu ; therefore
dy dx
=
f ′ (x
)=
du dx
.e u
Example 9: Find
f ′( x )@
dy dx
for the following functions:
a)
f ( x ) = e2 x
b)
y = 7e − 4 x
7.3.8 DIFFERENTIATION RULES #8 : LOGARITHMIC FUNCTION The logarithmic function can be divided into two: i.
If
y = log a f ( x ) , then the derivative:
⇒ Let u = f ( x ) then y = log a u
1 du dy = f ′( x ) = . . log a e dx u dx
Chapter 7: Differentiation
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Example 10:
a)
dy for the following functions: dx f ( x ) = log 2 3 x
b)
y = log5 x
Find
ii.
If
f ′( x )@
y = ln f ( x ) , then the derivative: ⇒ Let u = f ( x ) then y = ln u
dy 1 du ′ = f ( x ) = . dx u dx Example 11:
a)
dy dx y = ln 5 x
b)
y = ln(9 x + 4 )
c)
f ( x ) = ln x 2
Find
f ′( x )@
Chapter 7: Differentiation
for the following functions:
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7.4
HIGHER ORDER OF DIFFERENTIATION The derivative of a function
d2y dx
dy . dx
f ′( x ) is derived one more time and written as
Furthermore, if
f ′′( x )@
f ( x ) written as f ′( x )@
2
, called the second derivative of
f (x ) .
therefore:
⇒ ⇒
d2y dx
2
d3y dx
3
= f ′′( x ) =
d [ f ′( x )] dx
(Second derivative)
= f ′′′( x ) =
d [ f ′′( x )] dx
(Third Derivative)
. . .
⇒
[
dny
]
d n −1 (x) = f n (x ) = f n dx dx
(nth derivative)
Example 12: a)
Find
f ′′( x ) if f ( x ) = x5 :
Solution:
f ′( x ) = 5 x5−1 = 5 x 4 f ′′ ( x
[
d 5 x dx = 5 4 x 4 −1
)=
(
= 20 x Chapter 7: Differentiation
4
]
)
3
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b)
Find
f ′′′( x ) if f ( x ) = 4 x3 − 10 x 2
Solution:
f ′( x ) = 12 x 2 − 20 x
[
d 12 x 2 − 20 x dx = 12(2 x ) − 20(1) = 24 x − 20
f ′′( x ) =
]
d [24 x − 20 ] dx = 24 (1) − 0 = 24 ( Answer )
f ′′′( x ) =
Chapter 7: Differentiation
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7.5
CRITICAL POINT Given
y = f ( x ) = ax n + bx n −1 + ... + c , and the critical
point for this function can be obtained by:
Step 1 Find
f ′( x )@
dy dx
from the equation given then let:
f ′( x ) = 0
Step 2:
f ′( x ) = 0, find the value of x . Let x = k , by substituting the value of x in the equation of When
y = f ( x ) = ax n + bx n −1 + ... + c , we will get the of y . Let say, y = m ; Therefore, the critical point is
(x, y) = (k, m) .
Example 13: Find the critical point of the following functions: a)
f (x) = x2 + 2 x + 1
Solution:
Step 1: Find Then
f ′( x ) = 2 x + 2 and let f ′ ( x ) = 0
f ′( x ) = 0 = 2 x + 2
Chapter 7: Differentiation
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Step 2: When
f ′(x) =0, find the value of x
2x + 2 = 0 2 x = −2 ∴ x = −1
When x = −1, f (x) = y = x2 + 2x +1 = (−1) + 2(−1) +1 =0
Then, the critical point
b)
2
(x, y ) = (− 1,0)
f ( x ) = 32 x − x 4
Solution:
Step 1:
f ′( x ) = 32 − 4 x3 then let f ′(x) = 0 f ′(x ) = 32 − 4 x 3 = 0 Step 2: When
f ′( x ) = 0 , find x .
32 − 4 x3 = 0 32 = 4 x
3
8 = x3
When x = 2, f (x ) = y = 32(2) − (2)
81 / 3 = x
4
= 64 − 16 = 48
∴x = 2 Therefore, the critical point Chapter 7: Differentiation
(x, y ) = (2,48) 188
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7.5.1 TEST/PROCEDURE TO DETERMINE THE CRITICAL POINT: MINIMUM/ MAXIMUM/ INFLECTION
Step 1: Let the critical point
(x , y ) = (k , m )
Step 2: Find
f ′′( x ) . If :
f ′′( x ) > 0( positive)
i)
at
x = k , then the critical point
is ⇒ (MINIMUM )
ii)
f ′′(x ) < 0(negative)
at
x = k , then the critical point
is ⇒ (MAXIMUM)
Example 14: a)
From example 13(a); determine weather the critical point is minimum or maximum.
Solution:
Step 1: Find the critical point, ie: Critical point
(x, y ) = (− 1,0 )
Step 2: Find
f ′′( x )
f ′′ ( x ) =
d [2 x + 2 ] = 2 dx
Chapter 7: Differentiation
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Since
f ′′(x) = 2 > 0( positive)
at
x = −1, then the critical point
(x , y ) = (− 1, 0 ) is minimum.
b)
From example 13(b); determine weather the critical point is minimum or maximum.
Solution:
Step 1: Find the critical point, ie : Critical point
(x , y ) = (2 , 48 )
Step 2: Find
f ′′( x )
f ′′( x ) =
At
x = 2,
[
]
d 32 − 4 x 3 = − 12 x 2 dx f ′′( x ) = −12(2) = −48 < 0(negative) then the critical
point ( x , y ) = (2 , 48
Chapter 7: Differentiation
2
) is maximum.
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Notes: If at
x = k , we have f ′′( x ) = 0 or f ′′( x ) = ∞
(undefined), then probably the inflection point occur at that critical point where a)
for
f ′′( x ) must vary in terms of sign:
x < k , f ′′( x ) < 0 ( negative ) ⇒ (− ) to
for
x > k , f ′′( x ) > 0 ( positive ) ⇒ (+ )
f ′′( x ) vary from positive (+ ) to negative (− ) for the neighboring points
Example 15: Determine weather the curve
f ( x ) = ( x − 1)3 has
MINIMUM/MAXIMUM/inflection point. Solution:
Step 1: Find the critical point, ie :
f ′( x ) = 3( x − 1)2 when f ′( x ) = 0 , then;
3( x − 1)2 = 0
( x − 1)2 = 0 = 0 3 ( x − 1) = ± 0 = 0
If x = 1, f ( x ) = ( x − 1)
3
f (1) = (1 − 1) =0
3
∴x =1 Critical point
(x , y ) = (1, 0 )
Chapter 7: Differentiation
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Step 2: Find
f ′′( x ) .
f ′′ ( x ) = at
x = 1,
[
]
d 2 3 (x − 1 ) = 6 (x − 1 ) dx f ′′ ( x ) = 6 (1 − 1 ) = 0
probably inflection point occur at critical point
(1,0)
Test: i)
as
x < 1, let x = 0 in f ′′( x )
f ′′(0) = 6(0 − 1) = −6 < 0(negative) ⇒ (− ) ii)
as
x > 1 , let x = 2 in f ′′( x )
f ′′(2 ) = 6 (2 − 1) = 6 > 0 ( positive ) ⇒ (+ ) Since
f ′′( x ) vary from negative (− ) to positive (+ ) for the
neighboring points, then the critical point
(1,0) is said to be an
inflection point.
Chapter 7: Differentiation
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7.6
APPLICATION OF DIFFERENTIATION IN ECONOMY AND BUSINESS Remember these function??:
TC
•
Total cost function,
•
Total revenue function,
•
Profit function,
TR
and
∏
The functions above will be used again in this chapter. Besides that, several new functions will be introduced such as:
a)
If
TC
is a total cost function, then
TC = VCx + FC •
Total cost function,
•
Average cost function, C =
•
Marginal average cost function,
•
Marginal cost function,
VCq + FC
@
TC TC @ x q
C′ =
dC dC @ dx dq
(TC )′ = d (TC ) @ d (TC ) dx
dq
Example 16: Given total cost function,
x
TC = 0.2 x 2 − 60 x + 20000 where
is quantity unit been produced. Find,
C
i)
Average cost function,
ii)
Marginal average cost function,
Chapter 7: Differentiation
C′
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(TC )′
iii)
Marginal cost function,
iv)
Quantity that will minimized the total cost function.
v)
What is the minimum value for this total cost function?
vi)
Verify the answer in (d) and (e) by using the vertex formula.
b)
If
TR •
is the total revenue function, then Total revenue function, where
TR = px @ pq
p =demand function (price) R=
TR TR @ x q
•
Average revenue function,
•
Marginal average revenue function,
•
Marginal revenue function,
Chapter 7: Differentiation
R′ =
dR dR @ dx dq
(TR )′ = d (TR ) @ d (TR ) dx
dq
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Example 17: Assuming the demand function of a product given by,
p = 200 − 2q where
q
is the number of unit of the product. Find,
TR
i)
Total revenue function,
iii)
Quantity that will maximized the total revenue function.
iv)
What is the maximum value for this total revenue function?
iv)
Price that will maximized the total revenue.
c)
If
∏ is the profit function, then ∏ = TR − TC
•
Profit function,
•
Average profit function,
•
Marginal average profit function,
∏′ = •
Chapter 7: Differentiation
∏=
∏ ∏ @ x q
d∏ d∏ @ dx dq
Marginal profit function,
∏′ =
d∏ d∏ @ dx dq
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Example 18: The demand equation for the company of AXL is and the average cost function given by
p = 42 − 4q
C =2+
80 . Find, q
TR
i)
Total revenue function,
ii)
Total cost function,
iii)
Total profit function,
iv)
Quantity that will maximized the profit.
v)
What is the maximum value for this total profit function?
vi)
Price that will maximized the profit?
Chapter 7: Differentiation
TC ∏
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