Pascal's Triangle & Binomial Theorem

Differentiation (Part – 1)

Department of GED

Daffodil International University

Function

2

If and are two variables related to one another in such a way that each value of determines exactly one value of , then we say that is a function of and it is simply written as , where is an independent variable and is a dependent variable. The value of or is called functional value.

 

is a function For

We get,

  is not a function

For

We get,

 

1 1

-1

x y

4 2

-2 9

3 -3 1 1

-1

y x

2 4 -2 3 9 -3

Concept of Derivative

Differentiation Instantaneous rate of change of a function with respect to one of its variables

to find the slope of the tangent line to the function at a point.

 

����� = �h���� �� ��������

�h���� �� ���� = Δ � Δ �

 

Δ �

Δ �

�

�

 

5 20 25

15

0 10

Position

�

�

 

5 20 25 15

0 10

Position

Geometrical Representation of Derivative

4

 

�

   

� + h �

 

�

 

( ^{�+h , � (�+h)})

( �,� (�))

� (�+h)

 

� (�)

 h

Secant Line

Geometrical Representation of Derivative

�

   

� + h �

 

�

 

( �+h , � (�+h)) ( ^{�,� (�)})

� (�+h)

 

� (�)

 h

 

Secant Line

Geometrical Representation of Derivative

6

�

   

� + h

 

�

�

 

( �+h , � (�+h)) ( �,� (�))

� (�+h)

 

� (�)

 h

 

Secant Line

Geometrical Representation of Derivative

First Principle: A function is differentiable at a point if exists (is finite).

 

∴ ��

�� = � ^′

(

_�

)

₌

lim

h →0

�

(

�+h

)

− � (�) h

�

 

 

�

 

�

 

( ^{�,� (�)})

� (�)

h →  0

  Secant Line

Tangent Line

Example 1: Find the derivative of using first principle

 

8

Let  

 

¿lim

h →0

� −� h

 

From First Principle,

 

¿ lim

h →0

0 h

 

¿

lim

h →0

0

 

¿ 0

 

∴ ��

�� = �

�� ( � ) = �

 

Example 2: Find the derivative of using first principle

 

Let  

 

¿ lim

h →0

� + h − � h

 

From First Principle,

 

¿ lim

h →0

h h

 

¿

lim

h →0

1

 

¿ 1

 

∴ ��

�� = �

�� ( � ) = �

 

Example 3: Find the derivative of using first principle

 

10

Let  

 

¿lim

h →0

(�+h)^�−�^� h

 

From First Principle, 

¿lim

h →0

{

^�

(

^1+�h

) }

^�−��

h  

 

¿lim

h →0

�^�

(

^{1+ h}�

)

^�−��

h  

∴ ��

�� = �

��

(

�^�

)

=� �^�−1

 

¿�^�lim

h →0

(

^{1+ h}�

)

^�−1

h  

��

��=�^�lim

h →0

1+�

(

�^h

)

^+�(�2⁻! ¹⁾

(

�^h

)

^{2+�(�−1}3^{) (}! ^�−2)

(

^h�

)

^{3+… −1}

h  

¿ �^� lim

h →0

�

(

^h�

)

^+�(�2⁻! ¹⁾

(

^h�

)

^{2+�(�−1}3^{) (�}! ⁻²⁾

(

�^h

)

^3+…

h  

¿ �^� lim

h →0

h

{

^�

(

^�1

)

^{+�(�2−! 1)}

(

^�h2

)

^{+�(�−13)!(�−2)}

(

^�h23

)

^+…

}

h

 

¿ �^� lim

h →0

{

^�

(

^1�

)

^{+�(�2−! 1)}

(

^�h2

)

^{+ �(�−13) (�! −2)}

(

^�h23

)

^+…

}

 

¿ �^� �

(

�¹

)

 

Example 4: Find the derivative of using first principle

 

Let  

 

¿ lim

h →0

�

^(�+h)

− �

^�

h

 

From First Principle,

 

¿ lim

h →0

�

^�

. �

^h

− �

^�

h

 

¿

lim

h →0

�^�

(

_�^h −

1 )

h

 

¿�^� lim

h →0

1+h+ h²

2 ! + h³

3 ! + h⁴

4 ! +… −1 h

 

��

�� =�^� lim

h →0

h+ h²

2! + h³

3! + h⁴

4! +… h

 

[

^�

�

²

�

³

�

⁴

]

 

∴ ��

= �

( �

^�

) =�

^�

 

¿�^�   lim

h →0

h

(

¹⁺ 2^h! + h²

3! + h³

4 ! +…

)

h

 

¿ �

^�

  lim

h →0

( 1 + 2 h ! + h

²

3 ! + h

³

4 ! + … )

 

¿ �

^�

.1

 

Example 5: Find the derivative of using first principle

 

12

Let  

 

¿lim

h →0

ln

(

^�+�^h

)

h

 

From First Principle,

 

¿lim

h →0

ln

(

^{1+ h}�

)

h

 

��

�� =lim

h →0

h

(

^{�1 − 12} �^h2 + 1 3

h²

�³ − 1 4

h³

�⁴ +…

)

h

 

[

^{∵ln (1+�)=�− �22+ �33 − �44 +⋯}

]

 

∴ ��

�� = �

�� ( ln ( � ) ) = 1

�

 

¿lim

h →0

(

^{1� − 12} �^h2 + 1 3

h²

�³ − 1 4

h³

�⁴ +…

)

 

¿ 1

�

 

¿lim

h →0

h

� − 1 2

h²

�² + 1 3

h³

�³ − 1 4

h⁴

�⁴ +…

h

 

Example 6: Find the derivative of using first principle

 

Let  

 

¿

lim

h →0

cos (

�+h

)

−

cos

� h

 

From First Principle,

 

¿lim

h →0

− 2 sin

(

^2�2^+h

)

^sin

(

^h2

)

h

 

¿− lim

h →0

sin

(

^2�2^+h

)

^{× lim}h 2 →0

sin

(

^h2

)

h 2

 

¿ −sin

(

^{2 �}2⁺⁰

)

^×1

 

�� �

 

cos  �−cos�=−2 sin

(

^�+2�

)

^sin

(

^{�−2 �}

)

limh →0

sin h

h =lim

h →0

h

sin h =1

 

  ????

 

Example 7: Find the derivative of using first principle

 

14

Let  

 

¿

lim

h →0

sin (

� �+�h

)

−

sin

� � h

 

From First Principle,

 

¿lim

h →0

2 cos

(

^2��2^+�h

)

^sin

(

^�2^h

)

h

 

¿lim

h →0

cos

(

^2��2^+�h

)

^×limh →0

sin

(

^�2^h

)

h 2

 

¿cos

(

^2��2^+�.0

)

^{× limh →0}

�sin

(

^�2^h

) (

^�2^h

)

 

��

�� =(cos ��)×�lim ⁡

�h 2 →0

sin

(

^�2^h

) (

^�2^h

)

 

sin  �−sin�=2���

(

^�+2^�

)

^���

(

^�−2 ^�

)

∴ ��

�� = �

�� ( ��� � � ) = � ��� ��

 

¿ ( cos �� ) × � ( 1 )

 

[

^{∵ h →0⟹ �2h →0}

]

 

lim

h →0

sin h

h =lim

h →0

h

sin h =1

 

Example 8: Find the derivative of using first principle

 

Let  

 

¿

lim

h →0

tan (

�+h

)

−

tan

� h

 

From First Principle,

 

¿ lim

h →0

sin ( �+h) cos (� +h) ⁻

sin � cos � h

 

¿lim

h →0

sin ( �+h) cos � − sin �cos( � +h) cos (�+h ) cos �

h

 

¿ lim

h →0

sin ( � + h ) cos � − cos ( � + h ) sin � h cos ( � + h ) cos �

 

��

�� = lim

h →0

sin h

h cos ( � +h ) cos �

 

sin  �cos�−cos�sin�=sin (�−�)

∴ ��

�� = �

�� ( ��� � ) =�� �

^�

�

 

¿ lim

h →0

sin h

h × lim

h →0

1

cos ( � +h ) cos �

 

¿ 1 × 1

cos ( � + 0 ) cos �

 

¿ 1

co s²� =se c² �

 

sin ( � + h − � )

 

Example 9: Find the derivative of using first principle

 

16

Let  

 

��

�� =

lim

h →0

tan

⁻¹

(

� +h

)

−

tan

⁻¹ �

h ⋯(�)

 

From First Principle,

 

Suppose and

 

When

Thus,

In concept of Limit

 

��

�� = lim

h →0

� + � − � h

 

��

�� = lim

�→0

�

tan ( � + � ) − tan � [ ∵ h → 0 ⟹ � → 0 ? ? ? ]

 

Example 9: Find the derivative of using first principle (cont.…)

 

sin  �cos�−cos�sin�=sin (�−�)

��

�� = lim

�→0

�

tan ( � + � ) − tan � [ ∵ h → 0 ⟹ � → 0 ]

 

¿ lim

�→0

� sin ( � +�) cos ( � +�) ⁻

sin � cos �

 

¿ lim

�→0

�

sin ( �+ �) cos � − sin � cos ( �+�) cos ( �+�) cos �

 

¿ lim

�→0

� cos ( � + � ) cos �

sin ( � + � ) cos � − cos ( � +� ) sin �

 

¿ lim

�→0

� cos ( � + � ) cos � sin ( � + � − � )

 

��

�� = lim

�→0

�

sin � × lim

�→0

cos ( � + � ) cos �

 

¿ 1 × cos ( � + 0 ) cos �

 

¿ cos

²

�

 

¿ 1

sec

²

�

 

¿ 1

1+ tan

²

� = 1 1 + �

²

 

�

cos (

� +�

) cos

�

  ∴  �� � −�

[

 

∵

�

��� � = � ]

lim�→0

sin �

� =lim

�→0

�

sin � =1

 

Example 10: Find using first principle

 

18

Let  

 

¿ lim

h →0

( √ � + h − √ � ) ( √ � + h + √ � )

h ( √ � +h + √ � )

 

From First Principle,

 

¿ lim

h →0

( √ � + h )

²

− ( √ � )

²

h ( √ � + h + √ � )

 

¿ lim

h →0

� + h − � h ( √ � + h + √ � )

 

¿ 1

√ � +0 + √ �

 

∴ ��

�� = �

�� ( √ � ) = �

� √ �

 

¿ lim

h →0

h

h ( √ � + h + √ � )

 

��

�� = lim

h →0

1

√ � + h + √ �

 

¿ 1

√ � + √ �

 

¿ lim

h →0

√ � + h − √ �

h

 

Exercise

Find the derivative of the following functions using first principle: