(chapter 6) Differentiation

QQM1023 Managerial Mathematics

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

170

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

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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

176

)

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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

QQM1023 Managerial Mathematics

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 ∏

196