Differentiation Integration

7/24/2019 Differentiation Integration

1/15

MEEN 364 Parasuram

July 13, 2001

1

HANDOUT M.2 - DIFFERENTIATION AND INTEGRATION

Section 1: Differentiation

Definition of derivative

A derivative )(xf of a functionf(x) depicts how the functionf(x) is changing at the point

x. It is necessary for the function to be continuous at the point x for the derivative to

exist. A function that has a derivative is said to be differentiable. In general, the

derivative of the function y =f(x), also denoted dy/dx, can be defined as

x

xfxxf

dx

xdf

dx

dy

x

+==

)()(lim

)(

0

This means that as x gets very small, the difference between the value of the function atx and the value of the function at x + x divided by x is defined as the derivative.

Derivatives of some common functions

xxdx

d

xxdx

d

aeedx

d

xxdx

d

nxxdx

d

ttdt

d

xdx

d

axax

nn

sin)(cos

cos)(sin

)(

1)(log

)(

2)(

1)(

1

2

=

=

=

=

=

=

=

General rules of differentiation

1. The derivative of a constant is equal to zero. If y = c,

0)( == cdx

d

dx

dy

where c is any arbitrary constant.


2/15

MEEN 364 Parasuram

July 13, 2001

2

2. The derivative of the product of a constant and a function is equal to the constant

times the derivative of the function. If y = cf(x)

dx

dfcxcf

dx

d

dx

dy== ))((

Example 1

If y = 8x, then

8)1(8)(8)8( ==== xdx

dx

dx

d

dx

dy

3. The derivative of the sum or difference of two functions is equal to the sum or

difference of the derivatives of the functions. If y =f(x) g(x)

))(())(())()(( xgdx

dxfdx

dxgxfdx

d

dx

dy ==

Example 2

If y = 8x-x2, then

xxdx

dx

dx

dxx

dx

d

dx

dy28)()8()8( 22 ===

4.

Product ruleThe derivative of the product of two functions is equal to the firstfunction times the derivative of the second, plus the second function times the

derivative of the first. If y = f(x)g(x)

))(()())(()())()(( xfdx

dxgxg

dx

dxfxgxf

dx

d

dx

dy+==

Example 3

If y = xex, then

xxxxxxx exeexexdxdee

dxdxxe

dxd

dxdy +=+=+== )1()()()(

Example 4

If y = x2sin(x), then


3/15


4/15

MEEN 364 Parasuram

July 13, 2001

4

))(()(

))(()(

))(()(

))(()(

)(

)()(

3

3

2

2

xfdx

dxfy

xfdx

dxfy

xf

dx

dxfy

xfdx

dxfy

xfy

n

nnn ==

==

==

==

=

Example 7

If 5xy= then,

0)120()(

120)120()(

120)60()(

60)20()(

20)5()(

5

5

5

6

6

4

4

5

5

2

3

3

4

4

23

2

2

3

3

34

2

2

4

===

===

===

===

===

=

dx

d

dx

yd

dx

d

dx

yd

xdx

d

dx

yd

dx

d

dx

yd

xxdxd

dxyd

dxd

dxyd

xxdx

d

dx

yd

dx

d

dx

yd

xxdx

d

dx

dy

dx

d

dx

yd

xdx

dy


5/15

MEEN 364 Parasuram

July 13, 2001

5

Partial differentiation

A partial derivative is the derivative with respect to one variable of a multivariablefunction, assuming all other variables to be constants. For example if y = f(x,y), is a

function depending on two variables x and y, then the partial derivative of f with

respect to x is obtained by assuming the variable y to be a constant and taking thederivative of f with respect to x. This is represented as ),( yxf

x

.

Example 8

If y = xsin(t), then

txx

ttxxx

ysin)(sin))sin(( =

=

=

In this example, since the partial derivative with respect to the variable x is required, the

variable t is assumed to be a constant and the derivative with respect to x is obtainedby following the general rules of differentiation.

Example 9

If z = x2y

3, then

22223232 3)3()()( yxyxyy

xyxyy

z==

=

=

General rules of partial differentiation

If the function z is dependent on two variables x and y, i.e., if z =f(x,y), then

)(

)(

)(

)(

2

2

2

2

2

2

x

z

yxy

z

y

z

xyx

z

y

z

yy

z

x

z

xx

z

=

=

=

=

If z =f(x,y) and x = h(t), y = g(t), then the total derivative of z with respect to tis given by


6/15

MEEN 364 Parasuram

July 13, 2001

6

dt

dy

y

f

dt

dx

x

fyxf

dt

d

dt

dz

+

== )),((

Example 10

If z = xy and x = cos(t) , y = sin(t), then

txtytxty

tdt

dxy

yt

dt

dxy

xdt

dy

y

z

dt

dx

x

zz

dt

d

dt

dz

cossin)(cos)sin(

)(sin)()(cos)()(

+=+=

+

=

+

==


7/15

MEEN 364 Parasuram

July 13, 2001

7

Section 2: Integration

Introduction

The basic principle of integration is to reverse differentiation. An integral is sometimes

referred to as antiderivative.

Definition:Any function F is said to be an antiderivative of another function, f if and

only if it satisfies the following relation:

fF='

where

Fofderivative'=F

Note that, the definition does not say the antiderivative, it says an antiderivative. This isbecause for any given function f, if there are any antiderivatives, then there are

infinitely many antiderivatives. This is further explained with the help of the followingexample.

Example 11

For the functionf(x) = 3x2, the functionsF(x) =x

3, G(x) =x

3+ 15, andH(x) =x

3- 38 are

all antiderivatives off. In fact, in order to be an antiderivative of f, all that is required is

that the function be of the form K(x) = x3+ C, where C is any real number. This is so

because the derivative of a constant function is always zero, so the differentiation process

eliminates the C.

)(3)('

)(3)('

)(3)('

2

2

2

xfxxH

xfxxG

xfxxF

==

==

==

Although integration has been introduced as an antiderivative, the symbol for integration

is . So to integrate a functionf(x), you write

dxxf )(

It is very essential to include the dx as this tells someone the variable of integration.

Definition:The expression f(x) dx=F(x) + C, where Cis any real number, means thatF(x) is an antiderivative off(x). This expression represents the indefinite integral of f(x).


8/15

MEEN 364 Parasuram

July 13, 2001

8

The C in the above definition is called the constant of integrationand is absolutely vital

to indefinite integration. If it is omitted, then you only find one of the infinitely many

antiderivatives.

Some properties of indefinite integrals

= dxxgdxxfdxxgxf )()())()((

constantrealanyiscwhere)()( = dxxfcdxxcf

Some of the common integration formulae

cxxdx

cxxdx

cxdxx

cn

xdxx

cx

dxx

cx

xdx

cdx

nn

+=

+=

+=

++

=

+=

+=

=

+

sincos

cossin

ln1

1

3

2

0

1

32

2

The constant c can be found if some more information about the antiderivative is given.

This is explained with the aid of the following example.

Example 12

Find the function whose derivative is 13)( 2 += xxf and which passes through the point

(2,5).

In other words, the function f has to be integrated with respect to x. So

cxxxF

dxdxxdxxdxxfxF

++=

+=+== 3

22

)(

13)13()()(

It is given that, the function F(x) passes through the point (2,5), i.e., when x = 2, F(2) = 5.


9/15

MEEN 364 Parasuram

July 13, 2001

9

5

105

102)2()2()( 3

=

+=

+=++==

c

c

ccFxF

Therefore the function F(x) is given by

5)( 3 += xxxF .

Definite integration

This is very much similar to the indefinite integration, except that the limits of integration

are specified. Since the limits are specified, there is no need to put the constant ofintegration. In other words

)()()()( aFbFxFdxxf b

a

b

a

==

Example 13

Evaluate the following integral

2

1

xdx .

2

3

2

122

)1(

2

)2(

2

222

1

22

1====

xxdx

Example 14

40

}015{}025{

)}0(3)5(3{})0()5{(

332)32(

22

5

0

5

0

2

5

0

5

0

5

0

=

+=

+=

+=+=+=

I

I

I

xxdxxdxdxxI


10/15

MEEN 364 Parasuram

July 13, 2001

10

Integration by parts

Formulae

If u =f(x) and v = g(x) and iff and g are continuous, then

= vduuvudv (1)

For definite integrals

[ ] =b

a

b

a

b

a

dxxfxgxgxfdxxgxf )(')()()()(')(

The following example illustrates how the integration-by-parts formula works.

Example 15

Evaluate the integral

= dxxeI x

Let u = x and dv = exdx (2)

So

du = 1 and v = ex

(3)

Since

xx edxedvv ===

Thus substituting equations (2) and (3) in equation (1), we have

cexeI

cdxexedxxe

xx

xxx

+=

+= )1(


11/15

MEEN 364 Parasuram

July 13, 2001

11

Example 16

Find xdxex cos

Let

xvedu

xdxdveu

x

x

sin,

so

cos,

==

==

Substituting the above relation in equation (1), we get

cxdxexexdxe xxx += sinsincos . (4)

Consider again the integral xdxex sin . Let

xvedu

xdxdveu

x

x

cos,

so

sin,

==

==

Then we have

cxdxexexdxe

cdxxexexdxe

xxx

xxx

++=

+=

coscossin

)cos(cossin (5)

Substituting the second equation of equation (5) in equation (4), we get

++=

++=

++=

cxxexdxe

cxxexdxe

cxdxexexexdxe

xx

xx

xxxx

)cos(sin2

1

cos

)cos(sincos2

)coscos(sincos


12/15

MEEN 364 Parasuram

July 13, 2001

12

Leibniz rule

The fundamental theorem of integral calculus states that if fis a continuous function onthe closed interval [a, b], then

)()( tfdxxfdtd

t

a

=

for bta

The Leibniz rule can be derived from the above fundamental theorem of calculus.If

=)(

)(

),()(

tb

ta

dxtxftI (6)

The final relation, which outlines the Leibniz rule is given by

)),(())(()),(())((),(),())((

)(

)(

)(

)(

ttaftadt

dttbftb

dt

ddxtxf

tdxtxf

dt

dtI

dt

d tb

ta

tb

ta

+

== (7)

This rule is useful when one needs to find the derivative of an integral without actually

evaluating the integral. The rule is further explained with the aid of the following

example.

Example 17

Given

=

2

)cos()( 2t

t

dxtxtI

Evaluate ))(( tIdt

d.

The Leibniz rule can be applied to evaluate the derivative without actually evaluating the

integral. Comparing the above integral with equation (6), we have

)sin())(cos()),((

)(

)(

)cos(),(

222

2

2

txxtxt

txft

ttb

tta

txtxf

=

=

=

=

=


13/15

MEEN 364 Parasuram

July 13, 2001

13

Substituting the above relations in equation (7), we get

).cos()cos(2)sin())((

]))()(cos[1(]))()(cos[2()sin())((

3522

22222

2

2

tttdxtxxtIdtd

tttttdxtxxtIdt

d

t

t

t

t

++=

+=

LHospitals rule

LHospitals rule gives a way to evaluate limits of the form)(

)(lim

xg

xf

axif the numerator and

denominator both tend to zero, that is, 0)(lim)(lim ==

xgxfaxax

. The rule states that if the

numerator and the denominator both tend to zero, then the limit of the quotient )(

)(

xg

xf

is

equal to the limit of the quotient of the derivatives)(

)(

xg

xf

. In other words

)(

)(lim

)(

)(lim

xg

xf

xg

xf

axax

=

Example 18

Evaluate

x

x

x

sinlim

0.

It can be seen that both the numerator and the denominator tend to zero, as x tends to

zero. Therefore applying the LHospitals rule, we have

11

1

1

coslim

))((

))(sin(

limsin

lim000

====

x

xdx

d

xdx

d

x

x

xxx


14/15

MEEN 364 Parasuram

July 13, 2001

14

Assignment

1) Calculate the derivatives of the following functions with respect to x.

xxyc

x

xyb

xya

sin)

32

5)

42)

2

3

=

+

+=

=

2) Calculate the partial derivatives,dt

y

x

y

and for the following functions.

ttxtxyb

txya

cossintan)

sin)

63

22

+=

=

3) Calculate the following indefinite integrals analytically and verify the result UsingMATLAB.

xdxxc

xdxxb

dxxa

tan)

sin)

cos)

2

2

4)Evaluate the following definite integrals and verify the result using MATLAB.

++++

2

0

2

1

234

sin)

)9657()

xdxb

dxxxxxa

5) Evaluate the derivative with respect to time of the following integral by means of

Leibniz rule.

=2

0

2 )sin()()t

dxtxtIa


15/15

MEEN 364 Parasuram

July 13, 2001

15

6) Evaluate the following limits:

)sin(

)sin(lim)

1

1

lim)

sin

2lim)

2

3

0

21

2

0

x

xc

x

x

b

x

xxa

x

x

x

+

+

NOTE:This handout is not a comprehensive tutorial for differentiation and integration.This just deals with the very basics of differentiation and integration. It is advisable

always to go through some MATH book for various other techniques of performing

differentiation and integration.

Differentiation Integration

Documents

Transcript of Differentiation Integration